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Eng /ioo2> .10, A

HARVARD
COLLEGE
LIBRARY

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ELECTRICALENGINEERING TEXTS

A COURSE IN
ELECTRICAL ENaiNEERING

VOLUME I
DIRECT CURRENTS

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ELECTRICAL ENGINEERING
TEXTS

A series of textbooks outlined by a com-
mittee of well-known electrical engineers
of which Harry E. Clifford. Gordon Mc-
Kay Professor of Electrical Engineering.
Harvard University, is Chairman ana
Consulting Editor.
Laws —

ELECTRICAL MEASUREMENTS
Lawrence —

PRINCIPLES OF ALTERNATING-CUR-
RENT MACHINERY
Lawrence —

PRINCIPLES OF ALTERNATING CUR-
RENTS
Langadorf —
PRINCIPLES OF DIRECT-CURRENT

MACHINES
Dawes —

COURSE IN ELECTRICAL ENGINEER-
ING
Vol. I. — Direct Currents
Vol. II. — Alternating Currents

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ELECTRICAL ENGINEERING TEXTS

A COURSE IN

ELECTRICAI ENGmEEROG

VOLUME I
DIRECT CURRENTS

BY
CHESTER L. DAWES, S. B.

ASSISTANT PB0FB880B OF BLKCTBICAL EMGIMESBINO, THB HARVABD BNQIKBlSBINa

school; mxmbeb, American iNSTrrxmB of BuscTBicAii

BNOINEBBB, XTC.

First Edition
Fifth Imfbession

McGRAW-HILL BOOK COMPANY, Inc-
NEW YORK: 370 SEVENTH AVENUE

LONDON : 6 4 8 BOU VERIE ST., E. C. 4
1920

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TV.

HARVARD UNtvVRSlTY
EHUlNEERlNQ 6cH00L

•ilVl^li C»LL£«l IIMMV

TUB MAPXiK PRBSS TOKK Z*A

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PREFACE

For some time past the editors of the McGraw-Hill Electrical
Engineering Texts have experienced a demand for a comprehen-
sive text covering in a simple manner the general field of Electrical
Engineering Accordingly, these two volumes were written at
their request, after the scope and general character of the two
volumes had been carefully considered.

As the title implies, the books begin with the most elementary
conceptions of magnetism and current-flow and gradually ad-
vance to a more or less thorough discussion of the many types of
direct and alternating current machinery, transmission devices,
etc., which are met in practice. These two books are intended for
Electrical Engineering students as a stepping stone to the more
advanced Electrical Engineering Texts which are already a part
of the series.

These two volumes should be useful also to students not plan-
ning to specialize in the electrical engineering field, who are tak-
ing courses in Electrical Engineering as a part of their general
training. Such men often find difiiculty in obtaining detailed
and straightforward discussions of the subject in any one text
and the brevity of their course does not give them time to assimi-
late fragmentary information obtainable only by consulting a
number of references. Men taking foremen's and industrial
courses in Electrical Engineering, which as a rule are carried
on only in the evening, require text books sufficiently comprehen-
sive, but at the same time not involving much mathematical
analysis. Ordinarily, this type of student does not have ready
access to reference libraries and is usually out of contact with his
instructors except during the short time available for class-room
work. In preparing this work the needs of the foregoing types
of students have been carefully kept in mind and as a result, a
liberal use of figures and illustrative problems has been made.

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vi PREFACE

Also frequent discussions of the methods of making measurements
and laboratory tests are included.

In any course in Electrical Engineering, even though it be
intended for non-electrical engineers, the author feels that the
student gains little from a hurried and superficial treatment of
the subject, as such treatment tends only to develop the memoriz-
ing of certain formulae which are soon forgotten. Accordingly
the attempt has been made in this text to develop and explain
each phenomenon from a few fundamental and well-understood
laws rather than to give mere statements of facts. Such treat-
ment will develop the student's reasoning powers and give him
training that will be useful in the solution of the more involved
engineering problems that may arise later in his career.

Throughout the text, especially in the treatment of the more
abstract portions, attempt has been made to show the ultimate
bearing upon general engineering practice. The student takes
more interest in the theory when he sees that it can be appUed
to the solving of practical problems. Because this work is not
intended for advanced students in Electrical Engineering, little
or no calculus is used and the mathematics is limited to simple
equations.

The author is indebted to several of the manufacturing
companies who have cooperated in the matter of supplying pho-
tographs, cuts and material for the text; and particularly to Pro-
fessor H. E. CUfiford of The Harvard Engineering School, for his
many suggestions and for the care and pains which he has taken
in the matter of editing the manuscripts.

C. L. D.

Harvard University, Gambridqe, Mass.
January f 1920.

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CONTENTS

Paqh

Preface v

CHAPTER I

Magnetism and Magnets

1. Magnets and Magnetism

2. Magnetic Materials

3. Natural Magnets

4. Artificial Magnets

5. Magnetic Field 2

6. Effect- of Breaking a Bar Magnet. 3

7. Weber's Theory 3

8. Consequent Poles 5

9. Magnetic Force 5

10. Pole Strength 5

11. Lines of Force ... 6

12. Field Intensity, Electromagnetic 7

13. Flux Density 7

14. Compass Needle 8

15. Magnetic Figures 10

16. Magnetic Induction 11

17. Law of the Magnetic Field 12

18. Other Forms of Magnets 13

19. Laminated Magnets 14

20. Magnet Screens 14

21. Magnetizing 15

22. Earth's Magnetism 15

CHAPTER II

Electromagnetism 17

23. Magnetic Field Surrounding a Conductor 17

24. Relation of Magnetic Field to Current 18

25. Magnetic Field of Two Parallel Conductors 19

26. Magnetic Field of a Smgle Turn 20

27. The Solenoid 21

28. The Commercial Solenoid 22

29. The Horseshoe Solenoid 24

vii

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viii CONTENTS

Pagx

30. The Lifting Magnet 26

31. Magnetic Separator 27

32. The Magnetic Circuits of Dynamos 27

CHAPTER III

Resistance 31

33. Electrical Resistance 31

34. Unit of Resistance 32

35. Resistance and Direction of Current 32

36. Specific Resistance or Resistivity 34

37. Volume Resistivity 35

38. Conductance 36

39. Per Cent. Conductivity 36

40. Resistances in Series and in Parallel 37

41. The Circular Mil 38

42. The Circular-mil-foot 39

43. Table of Resistivities 40

44 1

. * > Temperature Coefficient of Resistance 41

46. Alloys 43

47. Temperature Coefficients of Resistance 43

48. Temperature Coefficients of Copper at Different Initial Tem-

peratures 43

49. The American Wire Gage (A. W. G.) 44

60. Working Table, Standard Annealed Copper Wire, Solid; Ameri-
can Wire Gage (B. & S.). English Units 45

51. Bare Concentric Lay Cables of Standard Annealed Copper.

English Units . 46

52. Conductors 46

CHAPTER IV

Ohm's Law and the Electric Circuit 48

53. Electromagnetic Units 48

54. Nature of the Flow of Electricity 49

55. Difference of Potential 51

56. Measurement of Voltage and Current 52

57. Ohm's Law 53

58. The Series Circuit . . . ' 54

59. The Parallel Circuit 55

60. Division of Current in a Parallel Circuit 56

61. The Series-parallel Circuit 58

62. Electrical Power 58

63. Electrical Energy 60

64. Heat and Energy 61

65. Thermal Units 62

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CONTENTS ix

Paqb

66. Potential Drop in Feeder Supplying One Concentrated Load . 63

67. Potential Drop in Feeder Supplying Two Concentrated Loads

at Different Points 64

68. Estimation of Feeders 65

69. Power Loss in a Feeder 67

CHAPTER V

Battery Electromotive Forces — Kirchhoff's Laws 68

70. Battery Electromotive Force and Resistance 68

71. Battery Resistance and Current 70

72. Batteries Receiving Energy 71

73. Battery Cells in Series 73

74. Equal Batteries in Parallel 73

75. Series-parallel Grouping of Cells 75

76. Grouping of Cells 76

77. Kirchhoff's Laws 77

78. Applications of Kirchhoff's Laws 79

79. Assumed Direction of Current 81

80. Further Application of Kirchhoff's Laws 82

CHAPTER VI

Primary and Secondary Batteries 84

81. Principle of Electric Batteries 84

82. Definitions 85

83. Primary Cells 86

84. Internal Resistance 87

85. Polarization 88

86A. Daniell Cell 89

86B. Gravity Cell 90

87. Edison-Lalande Cell 91

88. Le Clanch^ Cell 91

89. Weston Standard Cell 92

90. Dry Cells 94

91. Storage Batteries 96

92. The Lead Cell 97

93. Faure or Pasted Plate. . 101

94. Stationary Batteries 103

95. Tanks 103

96. Separators 104

97. Electrolyte 105

98. Specific Gravity 106

99. Installing and Removing from Service 107

100. Vehicle Batteries 108

101. Rating of Batteries 110

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X CONTENTS

Paob

102. Charging Ill

103. Battery Installations 114

104. Temperature 114

105. Capacities and Weights of Lead Cells. 114

106. The Nickel-iron-alkaline Battery 115

107. Charging and Discharging 117

108. Applications 118

109. Efficiency of Storage Batteries 118

110. Electroplating 120

CHAPTER VII

Electrical Instruments and Electrical Measurements 122

111. Principle of Direct-current Instruments 122

112. The D'Arsonval Galvanometer 123

113. Galvanometer Shunts 126

114. Ammeters 128

115. Voltmeters 134

116. Multipliers or Extension Coils 135

117. Hot-wire Instruments 136

118. Voltmeter-ammeter Method 137

119. The Voltmeter Method 139

120. The Wheatstone Bridge 141

121. The Slide Wire Bridge 144

122. The Murray Loop 147

123. The Varley Loop 148

124. Insulation Testing 150

125. The Potentiometer. 153

126. The Leeds & Northrup Low Resistance Potentiometer . . . .155

127. Voltage Measurements with the Potentiometer 157

128. The Measurement of Current with Potentiometer 158

129. Measurement of Power 160

130. The Wattmeter 161

131. The Watthour Meter 162

132. Adjustment of the Watthour Meter 165

CHAPTER VIII

The Magnetic Circuit 169

133. The Magnetic Circuit 169

134. Ampere-turns 170

135. Reluctance of the Magnetic Circuit 171

136. Permeability of Iron and Steel 173

137. Law of the Magnetic Circuit 174

138. Method of Trial and Error 175

139. Determination of Ampere-turns 176

140. Use of the Magnetization Curves 178

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CONTENTS 3d

Paob

141. Magnetic Calculations in Dynamos 170

142. Hysteresis 181

143. HysteresiB Loss 182

144. Linkages 183

145. Induced Electromotive Force 184

146. Electromotive Force of Self-induction 186

147. Energy of the Magnetic Field 191

148. Mutual Inductance 193

149. Magnetic Pull 197

CHAPTER IX

EliECTROSTATICS: CAPACITANCE 198

150. Electrostatic Charges 198

151. Electrostatic Induction 199

152. Electrostatic lines 200

153. Capacitance 202

154. Specific Inductive Capacity or Dielectric Constant 204

155. Equivalent Capacitance of Condensers in Parallel 205

156. Equivalent Capacitance of Condensers in Series 206

157. Energy Stored in Condensers 208

158. Calculation of Capacitance 209

159. Measurement of Capacitance 211

160. Cable Testing— Location of a Total Disconnection 213

CHAPTER X

The Generator 215

161. Definition 215

162. Generated Electromotive Force 215

163. Direction of Induced Electromotive Force. Fleming's Right

Hand Rule 218

164. Voltage Generated by the Revolution of a Coil. ........ 219

166. Gramme-ring Winding 222

166. Drum Winding 223

167. Lap Winding 224

168. Lap Winding — Several Coil Sides per Slot 229

169. Paths Through an Armature 230

170. Multiplex Windings 233

171. Equalizing Connections in Lap Windings 236

172. Wave Winding 238

173. Number of Brushes 243

174. Paths Through a Wave Winding 244

175. Uses of the Two Types of Windings 246

176. Frame and Cores 249

177. Field Cores and Shoes 250

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xii CONTsENTS

Page

178. The Armature 251

179. The Commutator 253

180. Field Coils 254

181. The Brushes 255

CHAPTER XI

Generator Characteristics 257

182. Electromotive Force in an Armature 257

183. The Saturation Curve 258

184. Hysteresis 260

185. Determination of the Saturation Curve 261

186. Field Resistance Line 262

187. Types of Generators 263

188. The Shunt Generator 264

189. Critical Field Resistance 265

190. Generator Fails to Build Up 266

191. Armature Reaction 267

192. Armature Reaction in Multi-polar Machines 272

193. Compensating Armature Reaction 274

194. Commutation ! 276

195. The Electromotive Force of Self-induction 280

196. Sparking at the Commutator 281

197. Commutating Poles (or Interpoles). 285

198. The Shunt Generator — ^Characteristics 288

199. Generator Regulation 292

200. Total Characteristic 293

201. The Compound Generator 295

202. Effect of Speed. . 299

203. Determination of Series Turns; Armature Characteristic . . . 300

204. The Series Generator 301

205. Effect of Variable Speed Upon Characteristics. 305

206. The Unipolar or Homopolar Generator 305

207. The Tirrill Regulator 306

CHAPTER XII

The Motor 309

208. Definition 309

209. Principle 309

210. Force Developed with Conductor Carrying Current 310

211. Fleming's Left-hand Rule . 311

212. Torque 312

213. Torque Developed by a Motor 313

214. Counter Electromotive Force 316

215. Armature Reaction and Brush Position in a Motor 319

216. The Shunt Motor 321

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CONTENTS xiii

Page

217. The Series Motor 324

218. The Compound Motor 328

219. Motor Starters 329

220. Magnetic Blow-outs 338

221. Resistance Units 338

222. Speed Control 339

223. Railway Motor Control 345

224. Dynamic Braking 347

225. Motor Testing— Prony Brake .348

226. Measurement of Speed 353

CHAPTER XIII

Losses; Eppicibncy; Operation 355

228. Dynamo Losses 355

229. Efficiency 359

230. Efficiencies of Motors and Generators 360

231. Measurement of Stray Power 361

232. Stray-power Curves 363

233. Opposition Test— Kapp's Method 365

234. Ratings and Heating 368

235. Parallel Running of Shunt Generators 372

236. Parallel Running of Compound Generators 374

237. Circuit Breakers 377

CHAPTER XIV

rBANSMISSION AND DISTRIBUTION OF PoWER 380

238. Power Distribution Systems 380

239. Voltage and Weight of Conductor 381

240. Size of Conductors 382

241. Distribution Voltage 383

242. Distributed Loads 383

243. Systems of Feeding 384

244. Series-Parallel System 385

245. Edison 3- wire System — Advantages 385

246. Voltage Unbalancing 388

247. Two-generator Method 390

248. Storage Battery 390

249. Balancer Set 391

250. Three-wire Generator 394

251. Feeders and Mains. 395

252. Electric Railway Distribution 396

253. Electrolysis .'397

254. Central Station Batteries 399

255. Resistance Control 401

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xiv CONTENTS

Page

256. Counter Electromotive Force Cells 401

267. End CeU Control 402

258. Floating Battery 403

269. Series Distribution 405

APPENDIX A
Relations op Units 407

APPENDIX B
Specific Gbayities 408

APPENDIX C
Table OP TtiBNs PER Sq. In.; Solid Layer Winding 409

APPENDIX D
Current-Carrying Capacity in Amperes of Wires and Cables . . 410

Questions on Chapter I 411

Problems on Chapter I 412

Questions on Chapter II 413

Problems on Chapter II . . 414

Questions on Chapter III 416

Problems on Chapter III 417

Questions on Chapter IV 420

Problems on Chapter IV 421

Questions on Chapter V 425

Problems on Chapter V 427

Questions on Chapter VI 430

PROBLEBfS ON CHAPTER VI 434

Questions on Chapter VII. 438

Problems on Chapter VII 442

Questions on Chapter VIII 447

Problems on Chapter VIII 449

Questions on Chapter IX 455

Problems on Chapter IX 456

Questions on Chapter X 458

Problems on Chapter X 460

Questions on Chapter XI 461

Problems on Chapter XI 465

Questions on Chapter XII 467

Problems on Chapter XII 470

Questions on Chapter XIII 474

Problems on Chapter XIII 476

Questions on Chapter XIV 477

Problems on Chapter XIV 480

Index 485

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A COURSE IN

ELECTRICAL ENaiNEERING

VOLUME I

DIRECT CURRENTS

CHAPTER I
MAGNETISM AND MAGNETS

1. Magnets and magnetism are involved in the operation
of practically all electrical apparatus. Therefore an understand-
ing of their imderlying principles is essential to a clear conception
of the operation of all such apparatus.

2. Magnetic Materials. — Iron (or steel) is far superior to all
other metals and substances as a magnetic material, and is
practically the only metal used for magnetic purposes. Cobalt
and nickel (and some of their alloys) possess magnetic properties,
which are far inferior to those of iron. Liquid oxygen is also
attracted to the poles of magnets.

3. Natural Magnets. — Magnetic phenomena were first noted
by the ancients. Certain stones, notably at Magnesia, Asia
Minor, were found to have the property of attracting bits of
iron, hence the name magnets was given to these magic stones.
The fact that such stones had the property of pointing north and
south, if suspended freely, was not discovered until the tenth or
tweKth century. The practical use of such a stone in navigation
gave it the name of Lodestone or leading stone. Natural magnets
are composed of an iron ore known in metallurgy as magnetite,
having the chemical composition FesO*.

4. Artificial Magnets. — If a piece of hardened steel be rubbed
with lodestone, it will be found to have acquired a very appreci-
able amount of magnetism, which it will retain indefinitely.

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DIRECT CURRENTS

Such a steel magnet is called an artificial magnet. Artificial
magnets commonly derive their initial excitation from- an electric
current as will be shown later. If a piece of soft steel or soft
iron be similarly treated, it retains but a very small portion of the
magnetism initially imparted to it.

These properties make it desirable to use hardened steel when a
permanent magnet is desired and to use soft iron or steel when
it is essential that the magnetism respond closely to changes of
magnetizing force. It is found that even hardened steel ages
or loses some of its magnetism with time. Where a high degree
of permanency is desired, as in electrical instruments, or even in
magnetos, the magnets are aged artificially.

Fig. 1. — Magnetic field about a bar magnet.

6. Magnetic Field. — It is found that magnetism manifests
itseK as if it existed in lines, called lines of magnetism or lines of
induction. The region in space through which these lines pass
is called the magnetic field. Further, if the lines of induction
of such a field be determined experimentally, it is found that
they seem to emanate from one region of the magnet and enter
some other region as shown in Fig. 1. These regions are called
the poles of the magnet. The two poles are distinguished by
the position which they seek if suspended freely. The one which
points north is called the north-seeking pole or north pole for
short, and the other the south-seeking pole, or south pole. In
practice it is assumed that the lines of induction leave the magnet
at the north pole and re-enter it at the south pole. Within the

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MAGNETISM AND MAGNETS

magnet the lines of induction continue from the south to the
north pole so that each line of induction forms a closed loop.
The plane half way between the poles is the neutral zone or
equator of the magnet. No magnetic force is apparent at this
point. The entire path through which the lines of induction
pass is called the magnetic circuit,

6. Effect of Breakiiig a Bar Magnet — Neither a north pole
nor a south pole can exist alone. For every north pole there
exists an equal (but opposite) south pole. If an ordinary bar
magnet be broken at the middle, or at various points, each frag-

FiQ. 2. — Effect of breaking a bar magnet.

ment will constitute a bar magnet having its north and its
south pole lying in the same respective directions as those of the
original magnet. This phenomenon is easily explained by noting
that the lines of induction still continue to pass from one frag-
ment to the next adjacent one, and in -so doing constitute north
and south poles as shown in Fig. 2. In experimental work,
this phenomenon may be easily illustrated by magnetizing a
highly-tempered steel knitting needle and breaking it at various
points.

7. Weber's Theory. — An explanation of the appearance of north
and south poles upon breaking a magnet, and other phenomena

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4 DIRECT CURRENTS

occurring in the magnetization of iron, is offered by Weber's
Theory which has been expanded by Ewing. The molecules
of a magnet are assimied to be an indefinitely great nimiber
of very small magnets as shown in Fig. 3 (a). Under ordinary
conditions these small magnets are arranged in a haphazard
way, as shown at (a), so that the various north and south poles
all neutralize one another, and no external effect is produced.
Upon the application of a magnetizing force, however, the small
magnets tend to so arrange themselves that their axes are parallel

(&)

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If s

if stf s¥

N sWaW 8

N SN a\y ss

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SiV 8

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OF!

if—S

FiQ. 3. — Weber's molecular theory of magnets.

and their north poles are all pointing in the same general direc-
tion as the magnetizing force. This is shown in Fig. 3 (b). It
is evident that if the magnet be cut along the line XX , Fig. 3 (c),
a new north and a new south pole will result, which, before the
fracture took place, neutralized each other.

This theory is further substantiated by grinding a permanent
magnet into very small particles. Each of the small particles
possesses the properties of the bar magnet, each having its own
north and its own south pole. Further, the theory offers a
rational explanation of saturation, hysteresis, etc., occurring
in iron subjected to a magnetizing force. This will be considered
later.

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MAGNETISM AND MAGNETS 5

8. Consequent Poles. — Consequent poles are occasionally
found in bar magnets where different portions have been rubbed
by a north pole, or a south pole, or when exciting coils, acting
in opposition, have been placed upon the bar. Consequent
poles are in reality due to the fact that the bar consists of two

Fig. 4, — Consequent poles.

(cm.)
(a) Repulsion

or more magnets arranged so that two north or two south poles
exist in the same portion of the magnet. This is illustrated in
Fig. 4. The magnetic field shown in Fig. 11, page 11, is in a way
illustrative of the field resulting from consequent poles. In
this case, however, two bar magnets are used and a small air-gap
exists between the adjacent north poles.

9. Magnetic Force. — When a freely suspended north pole is
brought in the vicinity of another
north pole, it is repulsed, whereas,
if a south pole is brought in the pre-
sence of a north pole, it is immediately
attracted toward the north pole.
South poles are also found to repel
one another. From this it may be
stated that like poles repel one another
and unlike poles attract one another.

10. Pole Strength.— The force of
attraction (or repulsion) between
two given poles is found to be in-
versely as the square of the distance between the poles, pro-
vided that the dimensions of the poles are small compared
with the distance between them. A unit magnetic pole is one
of such strength that if placed at a distance of one centimeter in

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^h^f

(cm.)
(5) Attraction

FiQ. 6. — Repulsion and attrac-
tion between magnetic poles.

6 DIRECT CURRENTS

free space from a similar pole of equal strength tvill repel it with a
force of one dyne,

Pole strength is measured by the number of unit poles which,
if placed side by side, would be equivalent to the pole in question.

The force /, existing between poles in air may be formu-
lated as follows:

/ = ^TT- dynes (1)

where m and m' are the respective pole strengths (in terms of a
unit pole) of two magnetic poles, placed a distance r cm. apart,
as shown in Fig. 5. This force may be attraction or repulsion
according as the poles are unlike or like.

Example. — Two north poles, one having a strength of 600 units and the

other a strength of 150 units, are placed a distance of 4 inches apart. What

is the force in grams acting between these poles, and in what direction does

it act?

4 in. = 4 X 2.54 = 10.16 cm.

. _ 500X150 _ 75,000 _ _^ ,^^

^- (10.16)« --103^-^2^^^^^'

728

^ = 0.741 gram. Poles repel each other. Ans,

11. Lines of Force. — ^Thus far the magnetic field has been
studied only with respect to the Unes of magnetism or induction.
If a single north pole be placed in such a field two effects will be
observed.

1. This pole will be urged along the lines of induction.

2. The force urging this pole will be greatest where the lines of
induction are the most dense, and, moreover, the force will be
proportional to the number of lines per unit area taken perpen-
dicular to the lines in the field in which the pole finds itself.

From these statements it can be seen that lines of force, similar
to lines of induction, can be drawn, to represent the forces at the
various points in the magnetic field. In much of the literature
on the subject lines of induction and lines of force are used indis-
criminately. The fallacy of so doing is immediately apparent
upon considering a solid bar magnet. The lines of induction
pass completely through the solid metal of the magnet, whereas
the lines of force terminate at the poles. To be sure, a magnetic
force does exist within the magnet, but this force can be deter-
mined only by making a cavity in the magnet, and the force

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MAGNETISM AND MAGNETS 7

acting under these conditions is quite distinct from that in-
dicated by the number of lines of induction passing through the
bar. In air, however, the lines of force and the lines of induction
coincide.

12. Field Intensity. — It has been stated that the force acting
upon a magnetic pole placed in a magnetic field is proportional
to the number of lines of induction at that point. Unit field
intensity is defined as the fisld strength which will act upon a unit
pole with a force of one dyne. One line of force perpendicular to
and passing through a square centimeter represents unit field
intensity. Field intensity is usually represented by the symbol
H, It is evident that if a pole of m units be placed in a field of
intensity H, the force acting on this pole is

f = mX H dynes (2)

A pole placed in such a field must be of such small magnitude
that it will have no appreciable disturbing effect upon the mag-
netic field.

13. Flux Density. — ^Flux density is the number of lines of
induction per unit area, taken
perpendicular to the induction. ^ ^^^
In free space, flux density and
field intensity are the same,

numerically, but within magnetic '^ ^<^Mm ^^ait n-po1c

material the two are entirely
different. The two should not Radius of sput
be confused. The unit of flux ^ <^™-
density (one line per sq. cm.) is
often called the gauss, but the
expression '* lines per square
centimeter" and '* lines per
square inch" are more often

used in practical work when ^°- 6-— Lines of force emanating
, . f n 1 . . from a unit N-pole.

speaking of flux density.

By definition the force exerted by a unit pole upon another
unit pole at centimeter distance in air is always one dyne. The
field intensity on a spherical surface of one centimeter radius
must then be unity and can be represented by one line per square
centimeter over the entire spherical surface as shown in Fig. 6.

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8 DIRECT CURRENTS

Since there are 47r square centimeters upon the surface of
a unit sphere, each unit pole must have radiating from it 4x =
12.6 lines of force. Fig. 6 represents a portion of a spherical
surface of one centimeter radius and shows roughly the passage
of one Une of force through each square centimeter of surface,
each line originating in the unit north pole. This also explains
the appearance of the 47r term so often encountered in magnetic
formulas. A pole having a strength of m units will radiate irm
lines of force.

Example, — A total flux of 200,000 lines passes in air between two parallel

pole faces, each 8 cm. square. The field is uniformly distributed. With

what force (grams) will a pole, having a strength of 100 units, be acted

upon if placed in this field?

200 000
Flux density = ^ ' ^ = 3,120 lines per sq. cm. or 3,120 gausses. Being

in air tl^s value of flux density also equals the field intensity, H.

/ = m X H = 100 X 3,120 = 312,000 dynes

312,000 „,„ .

— ^^:j — = 319 grams. Ans.

Example, — A pole having a strength of 400 units is placed at the center
of a sphere having a radius of 3 cm. What is the flux density at the sur-
face of the sphere and what force will be exerted on a pole of 10 units placed
at the surface of the sphere?

Total lines emanating from pole = 400 X 4ir = 5,020 lines.

Area of surface of sphere = 4irr* = 4ir9 = 113 sq. cm.

Flux density = -jj^ = 44.4 gausses.

Force upon pole of 10 units = 44.4 X 10 = 444 dynes. Ans,

As a check, the force may also be determined by the law of inverse squares
(see Par. 10).

mm' 400 X 10 ... ,
f^l^^ 3X3 =444dynes.

14. The Compass Needle. — The compass consists of a hard-
ened steel needle or small bar, permanently magnetized and
accurately balanced upon a sharp pivot. The north-seeking
end or north pole points north, and the south-seeking end points
south. The north pole of the needle is usually colored blue or
given some distinguishing mark. With the exception of a few
used for lecture purposes, the needle is enclosed in an air-tight
case for mechanical protection. Mariners' compasses are
mounted carefully upon gimbals, so that they always hang
level. Upon steel ships, heavy iron balls placed near the compass

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MAGNETISM AND MAGNETS

are necessary to compensate for the magnetic effect of the ship
itself.

By means of the compass the polarity of a magnet is
readily determined. The south pole of the compass points to

Fia. 7. — Compass needle and bar magnet.

the north pole of the magnet as shown in Fig. 7. Likewise, the
north pole of the compass points to the south pole of the magnet.
This action of the compass needle follows immediately from the
law that like poles repel and unlike poles attract each other.

Fig. 8. — Elxploring the field about a bar magnet with a compass.

This is very useful in practical work for it enables one to deter-
mine the polarity of the various poles of motors and generators
and to show if the exciting coils are correctly connected.

Fiuiiher, the compass needle always tends to set itself in the
direction of the magnetic field in which it finds itself, the north
end of the needle pointing in the direction of the lines of force or
naagnetic lines. This is illustrated in Fig. 8. By placing a

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DIRECT CURRENTS

small compass at the various points in the region of a magnet,
and drawing an arrow at each point, the arrow pointing in the
same direction as the needle, the field around the magnet may-
be mapped out as shown in Fig. 8. In mapping out a field in
this way it must be remembered that the earth's field may exert
considerable influence on the compass needle in addition to the
eflfect of the field being studied.

15. Magnetic Figures. — If a card be placed over a magnet and
iron filings be sprinkled over the card, a magnetic figure is ob-
tained. The filings at each point set themselves in the direction

Fig. 9. — Magnetic figure, unlike poles adjacent.

of the lines of force at that point, and the resultant figure shows
in very close detail the character of the magnetic field. Fig. 9
shows the magnetic field due to two bar magnets placed side by
side and having unlike poles adjacent. On the other hand, Fig.
10 shows the field due to these same bar magnets when like poles
are adjacent. It will be noted in Fig. 9 that the lines of force
seem like elastic bands stretched from one pole to the other,
acting to pull the unlike poles together. In Fig. 10 the lines of
force from the two like poles appear to repel one another, indicat-
ing a state of repulsion between the poles. Fig. 11 shows the
field obtained by placing the bar magnets end to end, having the
two north poles adjacent.

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MAGNETISM AND MAGNETS

16. Magnetic Induction. — If a magnet is brought near a
piece of soft, non-magnetized iron, the piece of iron becomes

Fig. 10. — Magnetic figure, like poles adjacent.

Fig, 11. — Magnetic figure, like N-poles adjacent.

Magnetized by induction. If the north pole of the magnet is
brought near the iron, a south pole is induced in that part of the
iron nearest the inducing magnet, and if the south pole of the

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12 DIRECT CURRENTS

magnet is brought near the u*on a north pole is similarly induced.
This is illustrated in Fig. 12 (a). From the foregoing, the ability
of magnets to attract soft iron is readily understood. An oppo-
site pole to that of the magnet is induced in the iron, and these
two poles being of unlike polarity are then attracted toward
each other.

It is sometimes noticed that if a comparatively weak north
pole be brought into the vicinity of a strong north pole, attraction
between the two results, rather than the repulsion which might
be expected. This is no violation of the laws governing the at-

Soft Iron

FiQ. 12 (a). — Poles produced by magnetic Fig. 12 (6). — Proper method of
induction. " keeping " bar magnets.

traction and repulsion of magnetic poles, but comes from the fact
that the strong north pole induces a south pole which overpowers
the existing weak north pole and results in attraction. In this
way it is easy to reverse the polarity of a compass needle by hold-
ing one end too close to a strong magnetic pole of the same
polarity.

For a similar reason, when two bar magnets are put away in a
box, the adjacent ends should be of opposite polarity, as shown
in Fig. 12 (b). They will retain their magnetism better under
these conditions. When a horseshoe magnet is not in use a
"keeper" of soft iron should be placed across the poles.

17. Law of the Magnetic Field. — The magnetic field always
tends to so conform itself that the mcmmum am^mnt of flux is
attained. This offers further explanation of the attraction of
iron to poles of magnets. The iron is drawn toward the magnet
so that the magnetic lines may utiUze it as a part of their return

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MAGNETISM AND MAGNETS

path; since iron conducts these lines much better than the air.
This is illustrated in the horseshoe magnet of Fig. 14. The
armature is drawn toward the poles of the magnet, and the return

(a) (b)

Fig. 13. — Ring magnets.

path through the air is materially shortened, so that the number
of magnetic lines is materially increased. The maximum flux
exists when the armature is against the poles.

18. Other Forms of Magnets. — The simple bar magnet fre-
quently is not suitable for practical
work. For the same amount of material,
other forms are more powerful and more
compact. Fig. 13 (a) shows a closed
ring magnet. All the magnetic flux is
contained in the ring and little external
effect is noted. This type is not very
useful. However, if the ring be cut as
shown in Fig. 13 (&), a north and a
south pole are obtained. A piece of
soft iron, if brought near this gap, will
be strongly attracted and will tend to
be drawn across the gap and thus shorten
the length of the flux path.

The horseshoe magnet, shown in Fig.
14, is very useful, for two reasons. The two poles being near
each other, a comparatively strong field exists. Fiu'ther, if
the function of the magnet is to exert a pull upon an armature,
each pole is equally effective. Fig. 118, Chap. VII, page 130
shows a horseshoe magnet such as is used in Weston direct-current
instruments. Digitized by (^OOgle

Fig. 14. — Horse-shoe
magnet attracting a soft-
iron armature.

DIRECT CURRENTS

19. Laminated Magnets. — It is found that thin steel magnets
are stronger in proportion to their weight than thick ones.
For a given amount of material a magnet made up of several
laminations, as shown in Figs.
15 and 16, is more powerful
than one made of a single piece
of metal. Fig. 16 shows the
form of horse-shoe magnet
generally used for telephone
and ignition magnetos.

20. Magnet Screens. — There
is no known insulator for mag-
netic flux. No appreciable
change in the flux or in the pull
of a magnet is noticed if glass,
paper, wood, copper, or other

Soft iron
pole pieces

Fig. 16. — Compound or laminated bar Fio. 16. — Compound horse-shoe
magnet. magnet used in magnetos.

such material be placed in the magnetic field. However, it is often
desirable to shield galvanometers and electrical measuring
instruments from the earth's field and from stray fields due to

InBtrament to be MreeneS.

Fig. 17. — Magnetic screen.

generators, conductors carrying currents, etc. This is done by
surrounding the instrument with an iron sTiell as shown in Fig.
17. This shell by-passes practically the entire flux and thus

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MAGNETISM AND MAGNETS

prevents it from affecting the sensitive portions of the instrument.
The smaller the openings in the shell, the more effective the screen-
ing becomes. Three or four shells, with air spaces between, are
found to be more effective than one shell of the same total
thickness. Such, however, are used only in connection with
the screening of the most sensitive galvanometers.

21. Magnetizing. — A magnet may be magnetized by merely
rubbing it with another magnet. The resulting polarity at
any point is opposite to that of the last pole which came in con-
tact with this point. Therefore, it is well to rub one end with the
north pole of the inducing magnet and the other end with the
south pole. This may be done sim-
ultaneously by the "divided touch"
method shown in Fig. 18. It is ad-
visable to rub both sides of the bar.
Stronger magnets may be obtained
by placing them between the poles

Fig.

18. — Divided touch method of
magnetizing.

Fig. 19. — Magnetizing a horse-
shoe magnet with an electro-
magnet.

of a very powerful electromagnet. Fig. 19 shows this method
of magnetizing a horseshoe magnet. An armature or "keeper"
should be placed across the poles of the horseshoe magnet
before removing it from the electromagnet. Magnetiza-
tion may also be produced by inserting the magnet in a
suitable exciting coil and allowing a heavy current to flow in
the coil. A few turns of low resistance wire may be wound
around the magnet and connected in series with a fuse to the
supply mains. Upon closing the switch, an enormous current
passes temporarily, but the fuse blows immediately and prevents
damage to the electric circuit. The heavy rush of current is
usually sufficient to leave the steel in a strongly magnetized
condition.

22. The Earth's Magnetism. — The earth behaves as a huge
bar magnet, the poles of which are not far from the geographical

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16 DIRECT CURRENTS

poles. The north magnetic pole (corresponding to the south
pole of a magnet) is situated in Boothia Felix, about 1000 miles
from the geographical north pole. The south magnetic pole
has never been located but experiment points to the existence of
two south poles. Due to the non-coincidence of the geographical
and magnetic poles and to the presence of magnetic materials
in the earth, the compass points to the true north in only a few
places on the earth's surface. The deviation from the true north
is called the declination, and magnetic maps are provided show-
ing the declination at various parts of the earth. At New York
it is about 9® west. The declination undergoes a gradual varia-
tion from year to year, called the variation change. A careful
record is kept of this secular variation and scientific measure-
ments, such as are used in astronomy, surveying, and naviga-
tion, must be corrected correspondingly. The needle undergoes
a very small daily variation and an annual variation, due possibly
to the influence of the sun and the moon.

A freely suspended and balanced needle does not take up a
position parallel to the earth's surface, when under the influence
of the earth's magnetism alone, but assumes a position making
some angle with the horizontal. This angle is called the dip
of the needle. At New York it is about 70* North. The dip
undergoes changes similar to those in the variation. The field
intensity (total, not horizontal) of the earth's field at New York
is about 0.61 C.G.S. units, although this value changes slightly
from time to time.

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CHAPTER II

ELECTR0MA6NETISM

23. Magnetic Field Surrounding a Conductor. — It had long
been suspected that some relation existed between electricity
and magnetism, but it remained for Oersted in 1819 to show that
this relation not only existed but that it was a definite relation.

Fio. 20. — Magnetic field about a straight conductor.

If a compass be brought into the neighborhood of a single
conductor carrying an electric current, the needle deflects,
thus indicating the presence of a magnetic field. It is further
observed that the needle always tends to set itself at right angles
to the conductor. When it is held above the conductor, the

Fig. 21. — Klines of force surrounding
a cylindrical conductor — current
inwards.

Fig. 22. — Lines of force surrounding
a cylindrical conductor — current
outwards.

needle points in a direction opposite to that which it assumes
when held beneath the conductor. Further investigation shows
that the magnetic flux exists in circles about the conductor
(if there are no other conductors in the vicinity) as shown in
2 17

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DIRECT CURRENTS

Figs. 20, 21 and 22. These circles have their centers at the
center of the conductor and their planes are perpendicular to
the conductor. If the current in the conductor be reversed,
the direction in which the compass needle is deflected will be seen
to reverse also, showing that the direction of this magnetic field
is dependent upon the direction of the current. The relation
of the two is shown in Fig. 20. The fact that the magnetic
field exists in circles perpendicidar to the conductor explains the
reversal of the compass needle when moved from a point above
the conductor to a point beneath it, for the direction of the field
above the conductor must be opposite to that beneath the con-
ductor. This is illustrated in Figs. 21 and 22. ^

The experiment shown in Fig. 23 is illustrative of this concen-
tric relation of the flux to the conductor. A conductor carrying

a current is brought vertically
down through a horizontal sheet
of cardboard. Iron filings
sprinkled on the cardboard form
concentric circles. (A current
of about 100 amperes is neces-
sary to obtain distinct figures.)
If four or more compasses are
arranged as shown in Fig. 23,
they will indicate, by the di-
rection in which their needles
point, that the magnetic lines
are circles having the axis of the
wire as a center.

24. Relation of Magnetic
Field to Current.— A definite re-
lation exists between the direc-
tion of the current in a conductor
and the direction of the magnetic
There are two simple rules by

Fig. 23. — Investigation of the magnetic
field surrounding a conductor.

field surrounding the conductor.

which this relation may be remembered.

1 A circle having a cross inside (0) indicates that the current is flowing
into the paper, and represents the feathered end of an arrow. A circle
having a dot at the center (O) indicates that the current is flowing out of
the paper, and represents the approaching tip of an arrow.

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ELECTROMAGNETISM

Hand Rule, — Grasp the conductor in the right hand with the
thumb pointing in the direction of the current. The fingers will
then point in the direction of the lines of force (Fig. 24).

Fig. 24. — Hand rule.

Corkscrew Rule. — The direction of the current and that of the
resulting magnetic field are related to each other as the forward
travel of a corkscrew and the direction in which it is rotated.

This last rule is probably the most common and the most easily
remembered. However, it must not be inferred from this ride
that the magnetic field exists in spirals about the conductor.
It exists actually in planes perpendicular to the conductor.

Fig. 25. — Magnetic field about two
parallel conductors— current in same
direction.

Fig. 26. — Magnetic field about two
parallel conductors — (furrent in oppo-
site directions.

25. Magnetic Field of Two Parallel Conductors. — When each
of two parallel conductors carries an electric current, flowing
in the same direction, there is a tendency for the two conductors
to be drawn together. The reason for this is obvious. In
Fig. 25 the linfes of force encircle each conductor in the same
direction (corkscrew rule) and the resultant field is an envelope
of lines tending to pull the conductors together. Further
reason for this attraction is given by the rule of Par. 17 stating
that the magnetic field tends to so conform itself that the

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DIRECT CURRENTS

number of magnetic lines is a maximum. The pulling together
of the conductors reduces the length of path ahcd through which
the lines must pass. The field due to each conductor separately
is still circular in form but the resultant magnetic lines are no
longer circidar, as is shown in Fig. 25.

In Fig. 26 are shown the conditions which exist when two
parallel conductors carry current in opposite directions. The
magnetic lines are circles, but these circles are not concentric
either with one another or with the conductor. The lines are
crowded between the conductors and therefore tend to push the
conductors farther apart. Again, when the -conductors sepa-
rate, the area through which the flux passes is increased, so that
the magnetic circuit in this case also tends to so conform itself
that the magnetic flux is a maximum.

From the foregoing, the following rules may be formulated.
Conductors carrying current in the same direction tend to be drawn
together; conductors carrying current in opposite directions tend to
he repelled from each other.

All electric circuits tend to take such a position as vdll make
their currents parallel and flowing in the same direction.

This effect is especially pronounced in modern large capacity

power systems. Bus-bars have
been wrenched from their
clamps; transformer coils have
been pulled out of plaice and
transformers wrecked by the
forces produced by the enormous
currents arising under short-
circuit conditions.

26. Magnetic Field of a Sin-
gle Turn. — If a wire carrying a
current be bent into a loop a
field similar to that shown in
Fig. 27 is obtained. This mag-
netic field has a north pole and a
south pole which possess all the
properties of similar poles of a short bar magnet. A compass
needle placed in this field assumes the direction shown, the
north pole pointing in the direction of the magnetic lines.

FiQ. 27.-

-Magnetic field produced by
a single turn.

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ELECTROMAGNETISM

27. The Solendid. — An electric conductor wound in the form
of a helix and carrying current is called a solenoid, A simple
solenoid and the magnetic field produced within it when current
flows through the conductor is shown in Fig. 28. The solenoid
may be considered as consisting of a large number of the turns

Fig. 28. — Magnetic field produced by a helix or solenoid.

shown in Fig. 27 placed together. The solenoid winding may
consist of several layers as shown in Fig. 30.

The relation of the direction of the flux within the solenoid
to the direction in which the ciu-rent flows in the helix may be
determined by the hand rule, or by the corkscrew rule of Par. 24.

Fig. 29. — Relation of magnetic pole to direction of exciting current.

Another simple method, is shown in Fig. 29, where the arrows
at the ends of the ' W and the "aS'' show the direction of current
in the coil. For example, when looking down upon a north pole
the current direction in the coil will be counter-clockwise as shown
by the "iV;" when looking down upon a south pole the direction
of the exciting current will be clockwise as shown by the "S."

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DIRECT CURRENTS

28. The Commercial Solenoid. — The solenoid is used in
practice for tripping circuit breakers (Par. 237), for operating
contactors in automatic motor starters (Par. 219), for operating
voltage regulating devices (Par. 207), for arc lamp feeds (Chap.
XIII, Vol. II), for operating valves, and for niunerous other
purposes. In practically all instances a soft iron (or steel) plung-

Fig. 30. — Simple solenoid and plunger.

er or armature is necessary to obtain the tractive pull required of
the solenoid. The operation of a solenoid and plunger is indicated
in Fig. 30. The flux due to the solenoid produces magnetic poles
on the plunger. The pole nearer the plunger will be of such sign
that it will be urged along the Unes of force, (see Par. 11) and in
such a direction as to be drawn within the solenoid.

Fig. 31. — ** Iron-clad "solenoid and plunger with stop.

A position of equilibrium is reached when the center of the
plunger reaches the center of the solenoid (Fig. 30). Fig. 31
shows an '* iron-clad*' solenoid commonly used for tractive
work. The iron-clad feature increases the range of uniform
pull and produces a very decided increase of pull as the plunger

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ELECTROMAGNETISM

approaches the end of the stroke. When a stop "a" is used,
the solenoid becomes a plunger electromagnet. This changes

d —

P-J

10,000 Total Amp'eM Tnrna

s20

•v/6)

J2l]

-— ^

"* — ,

:::^

■*"

===^

4 6 8

Distance Z>- Inches

Fig. 32. — Pull of solenoid on plunger.

the characteristics of the solenoid in that the maximum pull
now occurs when the end of the plunger is near the stop. Fig.

32 shows the results of solenoid tests

made by C. R. Underbill.' Curve (a)
is the pull upon the plunger of a simple
solenoid like that of Fig. 30; curve (&)
shows the pull when this solenoid is
iron-clad as in Fig. 31 but without a
stop; curve (c) shows the effect of the
"stop'' on the pull. It will be noted
that the iron-clad feature and the stop
have but little effect except near the end
of the stroke.

An important practical application of
the solenoid occurs in the braking of
elevators and cranes. When the power
is removed from the lifting motor or
when the power is interrupted due to
a broken wire or other accident, the
brake must be applied immediately.
One method of accomplishing this is shown ii^ Fig. 33.

* " Standard Handbook, Section 5."

Fig. 33. — Plunger electro-
magnet operating a crane
brake.

When

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DIRECT CURRENTS

the power, for any reason, is interrupted, the plunger P of the
solenoid A drops, due partly to gravity and partly to the action
of the springs S, The springs S immediately force the levers
L against the brake bands B, pressing these against the brake
drum Z>, thus effecting the braking action. When the power
is applied to the lifting motor, the plunger P is pulled up, thus
releasing the brake. A plunger electromagnet is most suitable
for this piu-pose because the stroke is short and the pull must
be positive.

29. The Horseshoe Solenoid. — The use of an armature m
connection with solenoids is well illustrated by the relay or the
sounder used in telegraphy, and also by electric bells, buzzers,
etc. To increase the efifectiveness of such devices two solenoids

. [lard llubbtr-
Fia. 34. — Telegraph relay.

are used, each being placed on one of the legs of a horseshoe
or U-shaped magnet. When the coils C (Fig. 34) become excited,
the iron armature A is attracted because of the tendency of the
magnetic lines to make their path of minimum length. As a rule,
the armatiu-e A is not allowed to close the magnetic circuit com-
pletely, for under these conditions the magnetic lines still exist
after the excitation is removed, preventing rapid release of the
armatiu-e. The stop S' prevents the armature making contact
with the cores FF and thus completely closing the magnetic
circuit. The contacts D close any secondary circuit that the
relay may be operating. The spring T draws the armature
back against a stop S when the excitation is removed.

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ELECTROMAONETISM

Terminal Oritj-

L«*di. mil EbelT

Ooil

Outer *nd iDiBer
Pole ShoM

On* Luff for S-Folnt 8i»pmiai<m
HaiTDAt Case

R«niDv»ble Top PUto
of Coil Spool

High Permeability
Spool for Coil

Coil of Strap

Copper

—^ Coil Shield of
Non-magnetie
aianganese Steel

Fig. 35.— Gross-section of a lifting magnet.

Fig. 36. — Cutler-Hammer 36-inch magnet, handling heavy castings.

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DIRECT CURRENTS

30. The Lifting Magnet. — ^Lifting magnets are used commer-
cially to handle iron and steel in various forms. A very ap-
preciable saving of time and labor is effected by their use, because
chains and slings for holding the load are not necessary. They
are very useful for handling steel billets in rolling mills, but
the billets cannot be picked up when red hot as they lose their
magnetic properties at this temperature. Magnets are especially
usefid in loading and unloading steel rails, for an entire layer may
be picked up and laid down again without being disarranged.
Lifting magnets effect a very great saving of labor when small
pieces of iron, such as scrap iron, are handled, for they will
pick up large quantities at every lift. Without a magnet each
individual piece would have to be moved by hand. Fig. 35
shows in cross-section a typical Cutler-Hammer lifting magnet.

Fig. 36 shows a lifting magnet in actual operation.

Formulae for the holding force of electromagnets are given in
Par. 149.

^.*-^^ Ma^nd-ic PvHtt^

Non-Ma^tffhc

Fig. 37. — Magnetic separator.

It should be understood that the magnet itself does little or
no work in the lifting, but merely serves as a holding
device. The actual work is performed by the engine or motor
which operates the steel ropes or chains attached to the magnet.

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ELECTROMAGNETISM

31. Magnetic Separator. — Another important application of
magnetic principles is found in the magnetic separator shown
in Fig. 37. It is especially designed to remove steel and iron
from coal, rock, ore, etc., but it may be used for separating steel
shot from molding sand, iron chips from machine shop turnings,
etc. The material is fed on an endless belt running at a speed of
about 100 feet per minute. The belt passes over a magnetized
pulley. The non-magnetic material immediately drops ofif into
a hopper, but the magnetic material is held by the pulley until
the belt leaves the pulley when the material drops into another
hopper. The pulley is magnetized by concentric exciting
coils, to which current is carried by means of slip-rings.

.o.

Yoke^

Pole Piece

Armiiture

PoI« Hlec«

Fig. 38. — Magnetic circuit and field windings of an Edison bl-polar generator.

32. The Magnetic Circtdts of Dynamos. — One of the most im-
portant uses of electromagnets is in the magnetic fields of motors
and generators. An early and simple type of such a magnetic
circuit is illustrated by the Edison bi-polar generator, shown
in Fig. 38. The type of magnetic circuit shown is very ineffi-
cient, because of its great length, in comparison with its sectional
area. There results a considerable magnetic leakage which re-

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DIRECT CURRENTS

duces, therefore, the amount of flux passing through the armature.
Moreover, the flux in taking the shortest path tends to crowd
through the upper half of the armature. This tends to produce
unsatisfactory commutation.

The magnetic circuit of a bi-polar generator of modern design
is shown in Fig. 39. Because of the symmetry of the magnetic
circuit the flux divides evenly through the two sides of the
armature. The long air path existing between the pole shoes

Field WiEidfAf

Toka

Fio. 39. — Magnetic circuit and field windings of a modern bi-polar generator.

reduces the magnetic leakage to a minimimi. It is to be noted
that the flux in the cores divides as it passes into the yoke.
Ordinarily the yoke need only be one-half the cross-section of the
field cores. Direct-current machines of the bi-polar type are
made usually in small units.

Fig. 40 shows the more complex magnetic circuits of a multi-
polar generator having eight poles. It is to be noted that the
poles are alternately north and south. Again the flux passing
through the field cores divides, both upon reaching the yoke
and upon reaching the armature path and the cross-section of the

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ELECTROMAGNETISM

yoke need only be one-half that of the cores. In both Fig. 39
and Fig. 40 the magnetic leakq^ge is very materially reduced

Fia. 40. — Magnetic circuits of a multi-polar generator.

Fig. 41. — Magnetic leakage produced by incorrect position of exciting coils.

by placing the exciting ampere-turns as near the armature as
possible. This result is not secured in the Edison bi-polar

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30 DIRECT CURRENTS

generator of Pig. 38. To illustrate, Fig. 41 shows the same
generator as that in Fig. 40 but with the exciting coils placed
upon the yoke. They still act upon the magnetic circuits, but
because of their remoteness from the armature, a large magnetic
leakage exists around the outside of the yoke and through the
interpolar space, resulting in a smaller percentage of the total
flux passing through the armature.

It is to be understood that of itself magnetic leakage does not
lower the efficiency of a machine, since to maintain a constant
magnetic field does not require an expenditure of energy. How-
ever, in order that the necessary number of magnetic lines may
reach the armature both the yokes and the cores must have suffi-
ciently large cross-sections to carry the leakage flux in addition
to the useful armature flux. Thi&in turn increases the amount of
field copper. Therefore a large magnetic leakage results in a
much heavier and more expensive machine than would otherwise
be necessary.

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CHAPTER III
RESISTANCE

S3. Electrical Resistance. — The current flowing through an
electric circuit depends not only upon the electromotive force
impressed across the circuit, but upon the circuit properties as
well. If, for example, a wh-e be connected across the terminals of
a battery a current will obviously flow through this wire. If a
poor contact be made at a battery terminal or at some other
point in the circuit the current wiU drop in value, even with the
same electromotive force acting. Also considerable heat will
be dissipated at the point of poor contact. This property of
tending to prevent the flow of current and at the same time
causing heat dissipation .is called resistance.

Resistance in the electric circuit may be likened in its effect
to friction in mechanics. For example, if a street-car is running
at a uniform speed on a straight, level track, friction tends to
prevent the moving of the car. The power which is used in
moving the car is converted by friction into heat. Friction tends
to impede the flow of water in a pipe or in a flume, some of the
energy of the water being expended in overcoming this friction.
The loss of energy is represented by a loss of head. This energy
loss is largely absorbed by the water and therefore careful
measurements would show a slight increase in its iemperature.

As will be shown in the next chapter, the energy loss which
occurs when an electric current passes through a resistance, is
directly proportional to the amount of resistance.

AU^ substahces have resistance, but the resistance of some sub-

1 Professor Kamerlingh-Onnes of Ley den, in a recent experiment, was able
to produce a circuit in which an electric current showed no diminution in
strength 5 hours after the electromotive force had been removed. The
current was induced magnetically in a short-circuited coil of lead wire at
—270° C. in the presence of liquid helium and the inducing source then
removed. Liquid helium has the lowest temperature known, being in the
neighborhood of absolute zero ( — 273° C . ) . This experiment indicates that the
resistance of the lead was practically zero at this extremely low temperature.

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32 DIRECT CURRENTS

stances is many times greater than that of others. This leads
to the classification of substances into either conductors or in-
sulators. Even silver, one of the best conductors, has appreciable
resistance, and glass or porcelain, among the best insulators known,
will allow a certain amount of current to pass and therefore are not
perfect insulators. The best conductors are the metals, silver
coming first and copper second. Carbon and ordinary water
also may be classed as conductors. Distilled or pure water,
however, is a good insulator. Oils, glass, silk, paper, cotton,
fiber, ebonite, paraflSn, rubber, etc., may be considered as non-
conductors or good insulators. Wood, either dry or impreg-
nated with oil, is a good insulator, but wood containing moisture
is a partial conductor.

34, Unit of Resistance. — The ohm is the practical unit of
resistance and is defined as that resistance which will allow
one ampere to flow if one volt is impressed across its terminals.

An ohm also has such a value that one ampere flowing through
it for one second dissipates as heat one joule of energy.

The resistance of insulating substances is ordinarily of the
magnitude of millions of ohms, so that it is awkward to express
this resistance in terms of a unit as small as the ohm. The
megohm, equal to 1,000,000 (10^) ohms, is the unit ordinarily used
under these conditions. (The prefix ''mega" means million.)

On the other hand, the resistance of bus-bars and short pieces
of metals may be so low that the ohm is too large a unit for con-
veniently expressing it. Under these conditions the microhm

is used as the unit, and is equal to i qqa nnn of an ohm (10""®) .

(The prefix ''micro" means one millionth.)

35. Resistance and Direction of Current. — The resistance of a
body of given material depends not only on its size and shape,
but upon the direction in which the electric current flows through
it.

This may be illustrated by the reservoirs and pipes shown in
Fig. 42. Twp equal reservoirs A and B are to be emptied through
pipes P and P' respectively. The pipe P is twice as long as
the pipe P' but of one-half the cross-section. Therefore both
pipes have the same volume. It is evident that reservoir B
will be emptied much quicker than A, because pipe P' has twice

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RESISTANCE

the cross-section of pipe P, and therefore oflFers less resistance
per unit of length. Further, the length of P' is only half that
of P, and this again makes the total friction of P' half that of -P
even if the cross-sections were equal.

Now consider the two conductors A and B (Fig. 43) each of
the same material. Conductor A has twice the length of con-
ductor B but only one-half the cross-section. Therefore each

Fia. 42. — Water discharge through different-sized pipes.

conductor contains the same amount of material. It is evident,
however, that the resistance per unit length of conductor A is
twice that per unit length of B. Then if conductors A and B
were of the same length conductor A would have twice the
resistance of conductor 5. However, conductor A is twice the
length of Bj and therefore must have 2 X 2 or 4 times the re-
sistance of 5. •

Fig. 43. — Two conductors of equal Fia. 44. — Rectangular prism as a
volume. conductor.

When specifying the resistance of a body or a substance the
direction in which the current flows must be specified. Consider
the rectangular prism of Fig. 44, composed of two cubes, each
a centimeter on edge. Assume that the current flows in the
direction of Ii from the end A through the solid to the end B.
It encounters the resistance of each of two cubes successively.
If it encounters a total resistance of 4 microhms, the resistance
of each cube then must be 4/2 or 2 microhms. If, however,

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34 DIRECT CURRENTS

the current flows in the direction of It, between the two oblong
surfaces C and D of the solid, it finds two paths in parallel, each
1 sq. cm. in cross-section, and each having a length of 1 cm.
In virtue of the two paths the resistance per cm. length of the
path CD is but one-half the resistance per cm. of AB. But, the
path AS is 2 cm. in length, therefore the path CD must have a
resistance of one-fourth that of path AB or 1 microhm. Con-
sequently in specifying the resistance of a solid, the direction of
the current flow should be specified unless this direction is obvious.

36. Specific Resistance or Resistivity. — From the deductions
of Par. 35 the following rule for resistance may be stated:

The resistance of a homogeneous body of uniform cross-section
varies directly as its length, and inversely as its cross-section.

That is,

«=a| (3)

where R is the resistance in ohms, L is the length in the direction
of the current flow, A is the area at right angles to the current
flow, and A; is a constant of the material known as its resistivity
or specific resistance.

If L is 1 cm. and A is 1 cm. square, the substance in question
must have the form of a cube, 1 cm. on an edge and

R=k. ^

1X1
or

B = Jfc

k is called the specific resistance or the resistivity of the substance;
in this case per cm. cube, k may be expressed in terms of an
in. cube or in other units as will be shown later. The resistivity
of copper is 1.724 microhms, or 1/580,000 ohm, per cm. cube.
It is evident that the cube is a perfectly definite unit of resistivity
for the resistance between any two opposite faces is the same.
The resistivities of various substances are given in Par. 43.
Knowing the specific resistance in terms of the cm. cube, the
resistance of a wire, bar, etc., may be readily computed from
formula (3),

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RESISTANCE 35

Example — Determine the resistance of 3,000 ft. of annealed 0000 copper
wire nayinff a diameter of 0.460 in., the specific resistance of copper bemg
1.724microhms ( = 0.000001724 ohm) per cm. cube (20® C). (See Par. 42.)

3,000' = 3,000 X 12 X 2.64 = 91,500 cm.

Cross-section = 7(0.460 X 2.54)« = 1.07 sq. cm.
4

R^kj^ (0.000001724) X (^y^) = 0.1472 ohm. Ana,
37. Volume Resistivity. — Since the volume of a body

where L is its length and A its uniform cross-section^, eouation (3)
may be written t^^^r^r-'

That is: J ' - .

The resistance of a conductor varies directly as the square of it$

len^h when the volume is fixed. ' z

The resistance of a conductor varies inversely o« the square of
its cross-section when the volume is fixed.

Example:

A kilometer of wire having a diameter of 11.7 mm. and a resistance of
0.031 ohm is dra^^n down so that its diameter is 5.0 mm. What does its
resistance become?

The original cross-section of the wire

Ai = 7 11.7« = 107.5 sq. mm.

4 / / ,

The final cross-section

'/■

«. 2

Aj = - 5.0 = 19.64 sq. mm.
4

Applying equation (4)

^^ - ^ (19.64)«

Since the volume of the wire does not change during the drawing process
and the resistivity constant k remains the same,

R2 ^ Ri ^ ^ (19.64)»
Ri 0.031 ^ V

(107.5)^

^' ^ ^'^^^ (Sf' ^ ^'^^^ ^^T " ^-^^ ''^"'- '^'^'

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36 DIRECT CURRENTS

38. Conductance. — Conductance is the reciprocal of resistance
and may be defined as being that property of a circuit or of a
material which tends to permit the flow of an electric current.
The unit of conductance is the reciprocal ohm or mho. Conduct-
ance is usually expressed by g.

also

9-^ (5)

? = *'^ (6)

where k' is the specific cond'uciance or the conductivity of a sub-
stance, A the uniform cross-section and L the length.

The conductivity of copper is 580,000 mhos per. cm. cube.

Example. — Determine the conductance of an aluminum bus-bar 0.5 in.
thick, 4 in. wide, and 20 ft. long.

The conductivity of aluminum is 61 per cent, that of copper and copper has
a conductivity of 580,000 mhos per cm. cube.

The conductivity of aluminum is:

k' = 0.61 X 680,000 = 354,000 mhos/cm, cube
The cross-section of the bus-bar:

A = 0.6 X 4 X 2.54 X 2.54 = 12.9 sq. cm.
The length L = 20 X 12 X 2.54 = 610 cm.
The conductance:

g = 354,000 X ^ = 7,490 mhos. Ans.

39. Per Cent. Conductivity. — Until very recently the per cent,
conductivity of copper has been based upon results obtained
in 1862 by Matthiesen, who made careful measurements of the
resistance of supposedly pure copper. He found the resistivity
to be 1.594 microhms per cm. cube at 0° C. In view of the
uncertainty of the quality of his copper, the Bureau of Standards
has recently made a large number of measurements upon com-
mercial copper. Its recommendation that the standard of re-
sistivity be 1.724 microhms per cm. cube at 20^ C. has been
adopted internationally. The per cent, conductivity for care-
fully refined copper may exceed 100. Comparison to obtain
per cent, conductivity should be made at 20° C.

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RESISTANCE

Example. — A copper rod 4 ft. long and having a diameter of 162 mils has
a resistance of 0.0016 ohm at 20° C. What is its per cent, conductivity?

Cross-section = ^ (0.162 X 2.64)« = 0.133 sq. cm.

Length = 4 X 12 X 2.54 = 122 cm.

T> • X u 0.0016X0.133 r.ru^nf^r^^^Ai^ u

Resistance per cm. cube = ^^^^ = 0.000001740 ohm

122
1.740 microhms.

Per cent, conductivity =

1.724

= or 99%.

1.740

40. Resistances in Series and in ParalleL-

Ans.

-If a number of re-
sistances ri, r2, Tzy etc., Fig. 45, are connected in series, that is,
end to end, the total resistance of the combination is:

« = ri + r2 + r, + . . . (7)

That is:

In a series circuit the total resistarhce is the sum of the individual
resistances.

K-

Fig. 45. — Resistances in series.

hWV\A/WV

-€h

FiG. 46. — Conductances in parallel.

If a number of conductances gfi, gr2, fifa, etc., are connected in
parallel, Fig. 46, the total conductance of this portion of the
circuit must be equal to the sum of the individual conductances,
that is,

G = gi + g2 + gz+ . . . (8)

Since,

G =

equation (8) may be written
R ri r2 rz

(9)

That is:

In a parcdlel circuit, the reciprocal of the total resistance is equal
to the sum of the reciprocals of the individual resistances.

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DIRECT CURRENTS

Fbr a circuit liaving two resistances in parallel, ri and rs, the
joint resistance

R = -^; (10)

ri + rj
for three resistances in parallel the joint resistance is

R = v^^V (11)

rir2 + r2r3 + Tzn

Example, — Determine the total resistance of a circuit having 4 branches,
the individual resistances of which are 3, 4, 6 and 8 ohms, respectively.

^=^+^+^ + |- 0.333 + 0.250 + 0.167 + 0.125

R =

0.875

0,875 mho.
^ 1.142 ohms.

Ans,

41. The Circular Mil. — In the English and American wire
tables the circular mil is the standard unit of wire cross-section.

The term mil means one-thousandth; for example, a milli-
volt = Ho 00 volt. A mil is one thousandth of an inch. A
square mil is a square, each side of which is one mil (0.001 in.),
as shown in Fig. 47 (o). The area of a square mil is 0.001 X 0.001
= 0.000001 sq. in.

A circular mil is the area of a circle whose diameter is one mil
(0.001 in.). Fig. 47(6), and is usually written CM. or dr. mil. As
will be seen from Fig. 47(c), a cir. mil is a smaller area than a

71*

square mil. The area in sq. in. of a cir. mil = j (0.001) ^ =
0.0000007854 sq. in.

peO-OW-*-

i«0.001->i pH).OoA

■ 0

(a) The (b) The (c) Comparison

square mU circnlar mil of the square mU
and the circular mU

Fig. 47.

Fio. 48. — Cross-section expressed
in cir. mils.

The cir. mil is the unit with which the cross-section of wires
and cables is measured, just as the square foot is the unit by
which larger areas such as floors, land, etc., are measured. The
advantage of the circular mil as a unit is that circular areas

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, RESISTANCE 39

measured in terms of this unit bear a very simple relation to the
diameters.

Ay in Fig. 48, represents the cross-section of a wire having a
diameter of 1 in. Required: to determine its area in cir. mils.

The area, A = - (1)2 gq, in^

The area, a, of a cir. mil = j (0.001) ^ sq. in.

A
The ratio of — obviously gives the number of cir. mils in A

Therefore

= 1,000,000 cir. mils.

^ 7 0.000001
4

The general relation may be written:

Cir. mils = j^^, = 1,000,000 (D^y = D^ (12)

where

Di is the diameter of the wire in inches.
D is the diameter of the wire in mils.

The matter. may be summed up in two rules:

To obtain the number of circular mils in a solid wire of given
diameter express the diameter in mils and then square it.

To obtain the diameter of a solid wire having a given number of
circular mils, take the square root of the circular mils and the resuU
will be the diameter of the wire in mils.

Example.— QQ wire (A.W.G.) has a diameter of 0.3648 in. What is its
cir. milage?

0.3648 in. = 364.8 mils

(364.8)« = 133,100 cir. rails. Ana.

Example. — ^A certain wire has a cross-section of 62,640 cir. mils. What
is its diameter?

V62,640 = 229.4 mils = 0.2294 in. Am.

42. The Circular-mil-foot. — Another convenient unit of resis-
tivity, especially in the English system, is the resistance of a cir.-

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DIRECT CURRENTS

mil-foot. This unit is the resistance of a wire having a cross-sec-
tion of 1 cir. mil and a length of 1 ft., as shown in Fig. 49. The
resistance of a cir.-mil-foot of copper at 20** C. is 10.37 ohms.

(In practical work this resistance may
9M^'i -ift-

SlfE

Fig. 49. — The circular-mil-foot.

frequently be taken as 10 ohms.)
Knowing this resistivity, the resis-
tance of any length and size of wire
may be determined by formula (3).

Example, — What is the resistance of a 760,000 CM. copper cable, 2500
ft. long?

If the cable had a cross-section of 1 cir. mil it would have a resistance of
2500 X 10.37 = 25,900 ohms. However, the cross-section is actually
750,000 cir. mils, therefore,

25,900
750,000
or formula (3) may be used directly

2,500

R =

0.0346 ohm

R = 10.37 ;

= 0.0346 ohm.

750,000
When applying formula (3), L must be expressed in feet and A in circular mds.

48. Table of Resistivities

Metals

Cm. cube (microhms) Cir.-mU-foot (ohms)

Aluminum

Bismuth

Copper (drawn)

German silver

la la

Iron:

Electrolytic

Cast

Lead

Manganin

Mercury

Nichrorae

Nickel

Phosphor-bronze

Platinum

Silver

Steel:

Soft

Glass hard

Silicon (4 per cent.)

Transformer

2.828

17.0

108.0

648

1.724

10.37

33.3 to 48.0

200 to 290

49.0

294

9.96

59.8

75 to 95

450 to 570

114

41.4 to 73.8

248 to 443

94.07

565

100 to 110

600 to 660

10.67

640

3.95

23.7

9.0tol5.5

54 to 93

1.5to 1.7

9.0tol0.2

15.9

95.4

45.7

274

51.15

308

11.09

665

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RESISTANCE 41

44. Temperature Coefficient of Resistance. — The resistance
of the non-alloyed metals increases very appreciably with the
temperature. As the temperature of the windings of electric
machinery is necessarily much higher than that of the surround-
ing airj it is important to know the relation between temperature
and resistance. The relation may be expressed as follows:

Rt « Bo (1 + aO (13)

where Rt is the resistance at the temperature t, Ro the resistance
at 0° C. and a is the temperature coefficient of resistance at 0®.
For copper a is 0.00427 and for most of the unalloyed metals
is sensibly of this value. The above is equivalent to saying
that the resistance increases 0.427 of 1 per cent, for each degree
Centigrade increase of temperatiu'e above 0°. For example,
assume that a coil has a resistance of 100 ohms at 0® C. For
every degree increase of temperature the resistance will increase

100 X 0.00427 ohm or 0.427 ohm.

At 40° C. the increase of resistance will be 40 X 0.427 = 17.08
ohms, and the resistance at 40° will be 100 + 17.08 = 117.08
ohms.

If the resistance at some definite temperature other than
0° C. is known, ordinarily the resistance at 0° C. must first be
found before the resistance at other temperatures can be deter-
mined.

For this purpose formula (13) may be put in the form

^« = iht ^''^

Example, — The resistance of an electromagnet winding of copper wire at
20° C. is 30 ohms. What is its resistance at 80** C?
The resistance at 0° C.

30 30

^° = 1 + 0.00427 X 20 ^ LOSS = ^^'^^ ^^""^
Ru = 27.66 (1 +0.00427 X 80) = 37.11 ohms. Ans.

This process of working back to 0° is a little inconvenient, although it is
fimdamental and easy to remember. Table 48 gives the temperature coef-

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DIRECT CURRENTS

ficients of copper at the various temperatures other than 0** C. With this
table available the above problem involves but one computation.

Example, — The temperature coefficient of copper at 20** initial tempera-
ture from Table 48 is 0.00393. The rise in temperature = 80** - 20** = 60**.

Then the resistance at 80** C.

fiw = 30(1 + 0.00393 X 60) « 37.07 ohms.

Ans.

45. If the resistance of copper at ordinary temperatures be
plotted against the temperature the result is a practically
straight line, Fig. 50. If this line be extended, it will intersect
the zero resistance line at — 234.5° C. (an easy number to

-234.5" o" «{ tt

Fia. 50. — Variation of resistance with temperature.

remember), as shown in Fig. 50. This is equivalent to saying
that between ordinary limits of temperature, copper behaves as if
it had zero resistance at —234.5® C. (Actually the curve bends
at these extremely low temperatures, as shown by the dotted
line, Fig. 50.) This gives a convenient method for determin-
ing temperature-resistance relations.
By the law of similar triangles. Fig. 50,

234.5« 234.5« + (i
Rn Rt2

234.5« + tx 234.6« + U

(15)
(16)

Applying this equation to the previous example,
30 Rio

234.50 +

234.5" + 80°

_ „ 234.5° + 80° on 314.5° o^ , ,.

Rm = 30 »o^ „o ■ o/^o = 30 j^FT-^ = 37.1 ohms. Ans.

234.5° + 20°

254.5°

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RESISTANCE 43

46. Alloys. — Certain alloys, notably manganin and nickel-
iron alloys, have practically zero temperature coefficients and
are very useful as resistances for measuring instruments, where
a change of resistance introduces error.

Table 47 gives the temperature coefficients at 20® C. for various
materials. Table 48 gives the temperature coefficient of copper
at various temperatures.

47. Temperature Coefficients of Resistance

Per. desree C. at 20<* C.

Aluminum 0.00388

Carbon (incandescent lamp) ( — ) 0.003

Graphite (-) 0.0006 to (-) 0.0012

German silver 0.00031 to 0.00020

la la 0.000005

Iron. 0.00635

Manganin 0.000011 to 0.000039

Mercury 0.00072

Nichrome 0.00016 to 0.00044

Nickel 0.00622

Phosphor-bronze 0.0009

Platinum 0.00367

Silver 0.00377

Steel 0.00635

48. Temperature Coefficients of Copper at Different Initial
Temperatures

(From formula 1/(234.5 + t))

Increase in resistance

Initial temperatures per 1° C.

0 0.00427

5 0.00418

10 0.00409

15 0.00401

20 0.00393

25 0.00385

30 0.00378

35 0.00371

40 0.00364

45 0.00358

50 0.00352

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44 DIRECT CURRENTS

49. The American Wire Gage (A.W.G.).— TheA.W.G.(formerly
Brown & Sharpe Gage) is based upon a constant ratio of cross-
section between wires of successive gage numbers. The follow-
ing approximate relations make it a comparatively simple matter
to determine the weight or resistance of any gage number
without reference to the table: (1) No. 10 wire has a diameter
of 0. 1 in. and a resistance of 1 ohm per 1 ,000 ft. (2) The resistance
of the wire doubles with every increase of 3 gage numbers. (3)
Therefore the resistance increases v^ = 1-26 (134) times for
each successive gage number and (1.26)^ = 1.6 times for every
two numbers. (4) The resistance is multiplied or divided by 10
for every difference of 10 gage numbers. (5) The weight of 1,000
ft. of No. 2 wire is 200 lb.

Example, — What is the resistance and weight of 1,000 ft. of 0000 wire?

The resistances will decrease as follows:

Gage No 10 7 4 1 000

Resistance 1 0.5 0.25 0. 125 0.0625 (rules 1 and 2)

Resistance of 0000 = 0.0625/1.25 = 0.050 ohm (rule 3).

Weight of 1,000 ft. No. 2 = 200 lb.

Weight of 1,000 ft. 00 = 400 lb.

Weight of 1,000 ft. 0000 = 400 X 1.6 = 640 lb. (rule 5, 2 and 3).

The example might have been worked more quickly by rule 4.

Resistance of 1,000 ft. of No. 10 = 1 ohm.

Resistance of 1,000 ft. of 0 = 0.1 ohm (rule 4)

Resistance of 1,000 ft. of 0000 = 0.050 ohm (rule 3).

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RESISTANCE

50. Working Table, Standard Annealed Copper Wire, Solid^
American Wire Gage (B. & S.). English Units

Gage

Diameter
in mils

CroBS-section

OhmB per 1000 ft.

Ohms per
mile

25* C.

(-77* F.)

Pounds

per 1,000

ft.

Circular

Square
inches

26«»C.

(-77*>F.)

66* C.

(-149T.)

0000

000

460.0
410.0
'366.0

212,000 .0
168.000.0
133,000.0

0.166
0.132
0.106

0.0600
0.0630
0 .0796

0.0677
0.0727
0.0917

0.264
0.333
0.420

641.0
508.0
403.0

0
1
2

325.0
289.0
268.0

106,000.0
83,700.0
66,400.0

0.0829
0.0667
0.0521

0.100
0.126
0.169

0.116
0.146
0.184

0.528
0.666
0.839

319.0
253,0
201.0

3
4
6

229.0
204.0
182.0

62,600.0
41,700.0
33,100.0

0.0413
0.0328
0.0260

0.201 +
0.263^
0.3ld

0.232
0.292
0.369

1.061
1.336
1.685

169.0
126.0
100.0

7
8

162.0
144.0
128.0

26,300.0
20.800.0
16,600.0

0.0206
0.0164
0.0130

0.403
0.608
0.641

0.465
0.686
0.739

2.13
2.68
3.38

79.6
63.0
60.0

9
10
11

114.0
102.0
91.0

13,100.0
10,400.0
8,230.0

0.0103

0.00816

0.00647

0.808

1.02

1.28

0.932

1.18

1.48

4.27
6.38
6.75 •

39.6
31.4
24.9

. 13

81.0

' 72.0

64.0

6,630.0
6,180.0
4,110.0

0.60513
0.00407
0.00323

1.62
2.04
2.68

1.87
2.36
2.97

8.55
10.77
13.62

19.8
15.7
12.4

16
16
17

67.0
61.0
46.0

3,260.0
2,680 .0
2,060.0

0.00266
0.00203
0.00161

3.26
4.09
6.16

3.76
4.73
6.96

17.16

21.6

27.2

9.86
7.82
6.20

18
19
20

40.0
36.0
32.0

1,620 !o
1,290.0
1.020.0

0.00128
0.00101
0.000802

6.61
8.21
10.4

7.51
9.48
11.9

34.4
43.3
64.9

4.92
3.90
3.09

21
22
23

28.6
26.3
22.6

810.0
642.0
609.0

0 .000636
0.000605
0.000400

13.1
16.6
20.8

16.1
19.0
24.0

69.1]
87.1
109.8

2.45
1.94
1.64

24
25
26

20.1
17.9
16.9

404.0
320.0
264.0

0.000317
0.000262
0.000200

26.2
33.0
41.6

30.2
38.1
48.0

138.3
174.1
220.0

1.22

0.970

0.769

27
28
29

14.2
12.6
11.3

202.0
160.0
127.0

0.000168
0 .000126
0.0000996

62.5
66.2
83.4

60.6
76.4
96.3

277.0
350.0
440.0

0.610
0.484
0.384

30
31
32

10.0
8.9
8.0

101.0
79.7
63.2

0 .0000789
0.0000626
0.Q000496

106.0
133.0
167.0

121.0
163.0
193.0

654.0
702.0
882.0

0.304
0.241
0.191

33
34
36

7.1
6.3
6.6

60.1
39.8
31.6

0.0000394
0.0000312
0.0000248

211.0
266.0
336.0

243.0
307.0
387.0

1,114.0
1,404 .0
1,769.0

0.162
0.120
0.0954

36
37
38

6.0
4.6
4.0

26.0
19.8
16.7

0.0000196
0 .0000166
0 .0000123

423.0
633.0
673.0

488.0
616.0
776.0

2,230.0
2.810.0

3,560 .0

•

0.0757
0.0600
0.0476

39
40

3.6
3.1

12.6
9.9

0.0000098
0.0000078

848.0
1,070.0

979.0
1,230.0

4,480.0
6,660.0

1.0377
0 .0299

Note 1. — The fundamental resistivity used in calculating the tables is the International
Annealed Copper Standard via., 0.16328 ohm (meter, gram) at 20* C. The temperature
coefficient for this particular resistivity is 020 - 0.00393, or ao = 0.(X)427. The density
is 8.89 grams per cubic centimeter.

NoTK 2. — The values given in the table are only for annealed copper of the standard
resistivity. The user of the table must apply the proper correction for copper of any other
resistivity. Hard-drawn copper may be taken as about 2.7 per cent, higher resistivity than
annealed copper. i^

NoTB 3. — Pounds per miXe may be obtained by multiplying the respective values abofC
br 6.28.

' From Circular of the Bureau of Standards, No. 31.

46 DIRECT CURRENTS

61. Bare Concentric Lay Cables of Standard Annealed Copper
English Units

Circular
mils

Ohms per

1,000 ft.

Pounds

i,ooo'ft.

Standard concentric stranding

A. W. G.

No.

25*^0.

(-77* F.)

65*' C.
(-149'»F.)

Number
of wires

Diameter
of wires,
in mils

Outside

diameter,

in mils

2.000.000

0.00539

0.00622

6.180

127

125.5

1.631

1.700.000

0.00634

0.00732

5.250

127

115.7

1.504

1.500.000

0.00719

0.00830

4.630

128.4

1,412

1.200,000

0.00899

0.0104

3.710

114.8

1,263

1,000,000
900.000

0.0108
0.0120

0.0124
0.0138

3.090
2,780

61
61

128.0
121.5

1.152
1,093

850,000
750,000

0.0127
0.0144

0.0146
0.0166

2,620
2,320

61
61

118.0
110.9

1.062
998

650.000

0.0166

0.0192

2,010

103.2

929

600.000
550.000

0.0180
0.0196

0.0207
0.0226

1,850
1,700

61
61

99.2
95.0

893
855

500.000
450,000
400.000

0.0216
0.0240
0.0270

0.0249
0.0277
0.0311

1.540
1,390
1.24(J

37
37
37

116.2
110.3
104.0

814
772
728

350.000
300,000
250,000

0.0308

0.0360

. 0.0431

0.0356
0.0415
0.0498

1.080
926
772

37
37
37

97.3
90.0
82.2

681
630
575

0000

000

212.000
168,000
133,000

0.0509
0.0642
0.0811

0.0587
0.0741
0.0936

653
518
411

19
19
19

105.5
94.0
83.7

528
470
418

0
1

106,000
83.700

0.102
0.129

0.117
0.149

326
258

19
19

74.5
66.4

373
332

2
3

66,400
52.600

0.162
0.205

0.187
0.237

205
163

7
7

97.4
86.7

292
260

41,700

0.259

0.299

129

77.2

232

62. Conductors. — Although silver is a better conductor than
copper, its use as a conductor is very limited because of its cost.
In a few instances it is used where a delicate but highly conduct-
ing material is necessary, such as in the brushes and occasionally
in the commutator of watt-hour meters. Copper, because of
its high conductivity aii3"»iQ^rate cost, is used more extensively

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RESISTANCE 47

as a conductor than any other material. It has many good
qualities such as ductility, high tensile strength, not easily
abraided, not corroded by the atmosphere, and it can be readily
soldered.

Aluminum has only 61 per cent, of the conductivity of copper,
but for the same length and weight, it has about twice the con-
ductance of copper. It is softer than copper, its tensile strength
is much less, and it cannot be readily soldered. It is not affected
by exposure to the atmosphere. The large diameter for a given
conductance prohibits its use where an insulating covering is
required. Aluminmn is used extensively as a conductor for
high voltage transmission Hnes, where its lightness and large
diameter are an advantage. It is used to some extent for low
voltage bus-bars as it offers much greater radiating surface than
copper of the same conductance. The price of aliuninum is
held about 10 per cent, less than that of copper of the same
conductance.

Iron and steel have about 9 times the resistance of copper for
the same cross-section and length. The large crossnsection for a
given conductance prohibits their use where an insulating cover-
ing is necessary and the increased weight prevents their use in
most cases where the conductors must be placed on poles. In
view of their low cost per pound, they are cheaper than copper
as simple conductors. They are most commonly used as re-
sistors in connection with rheostats and for third rails of electric
railways. Iron and steel ordinarily must be protected from oxida-
tioA by galvanizing or other protective covering. Copper-clad
steel consists of a steel wire coated or covered with a layer of
copper, fused or welded to the steel. The advantages claimed
for it are that it possesses the high tensile strength of steel, com-
bined with the high conductivity of copper. Further, the copper
protects the steel from corrosion. Its field is the transmission
line conductor, where long spans make high tensile strength
necessary. It is also used as an overhead ground wire on such
lines. i

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CHAPTER IV
OHM^S LAW AND THE ELECTRIC CIRCUIT

The exact nature of electricity is not known, but recent in-
vestigations indicate that it consists of infinitesimal charges
called electrons. When these electrons are forced to travel in
the same direction an electric current results. The flow of
electricity through a circuit resembles in many ways the flow of
water through pipes, for it acts as an incompressible fluid would
act, undergoes pressure drop, etc., as will be shown later.

63. Electromagnetic Units. Current. — The unit of electric
current is the ampere and represents the rate of flow of electricity.
It corresponds in hydraulics to the rate of flow of water, expressed
as cubic feet per second, gallons per minute, etc.

The ampere is defined by an act of Congress, 1894, as follows: "The unit
of current shall be what is known as the international ampere, which is
one-tenth of the unit of current of the centimeter-gram-second system of
electromagnetic imits and is the practical equivalent of the unvarying cur-
rent, which when passed through a solution of nitrate of silver in water in
accordance with standard specifications, deposits silver at the rate of one
thousand one hundred and eighteen millionths (0.001118) of a gram per
second."

Qiuintity. — The unit of quantity is the coidomb. This is equal
to the quantity of electricity conveyed by one ampere in one
second. The coulomb is analagous to the unit quantity of water
in hydraulics, such as the cubic foot, the gallon, etc.

From this definition it is evident that an electric current may
be expressed in coulombs per second rather than in amperes.

Difference of Potential and Electromotive Force (emf.). — Dif-
ference of potential and electromotive force tend to cause a
flow of electricity. The unit of potential difference or of emf.
is thi volty and is defined as that potential difference which when
impressed across the terminals of a resistance of one ohm will
cause a current of one ampere to flow. The international volt
is now more specifically defined as H. 01830 oi the voltage of a
normal Weston cell. (See Par. 89.)

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OHM'S LAW AND THE ELECTRIC CIRCUIT 49

The mechanical analogy of potential is pressure. The differ-
ence in hydraulic pressure between the ends of a pipe causes or
tends to cause the flow of water. The pressure of water behind
the dam tends to cause water to flow through the penstock or
through any leaks. The pressure in a boiler tends to cause steam
to flow through the pipes, valves, etc. Likewise difference of
potential tends to cause current to flow.

Resistance. — The ohm, the unit of resistance, has already been
defined in Chap. Ill as that resistance which will allow one am-
pere to flow if one volt is impressed across its terminals. The
international ohm is specifically defined as the resistance of a
column of mercury at the temperature of melting ice (0° C),
14.4521 grams in mass, of a constant cross-sectional area and of
a length of 106.300 cm.

64. Nature of the Flow of Electricity. — The flow of electricity
through a circuit resembles in many ways the flow of water
through a closed system of pipes. For example, in Fig. 51 water
enters the mechanically-driven centrifugal pump P at a pressure
hi (represented by the length of a column of mercury) above
the point of zero pressure shown by the line ho. In virtue of the
action of the pump blades, its pressure through the pump is
increased from hi to A2, representing a net increase of pressure
Hi. The water then flows out along pipe Fi to the hydraulic
motor W. Because of the friction loss in the pipe Fij the pres-
sure at the motor terminals hz is slightly less than A2. In other
words, a pressure of A2 — A3 is required to overcome the frictional
resistance of the pipe Fi. The line ab shows the pressure drop
along the pipe, this pressure drop being uniform.

In Fig. 52 the mechanically driven electrical generator G
raises the potential of the ciu-rent entering its negative terminal,
from vi to V2 where vi and V2 are measured from the earth whose
potential is ordinarily assumed as zero. (The various voltages
are measured with voltmeters v\jv'2) etc.) The generator, in rais-
ing the potential of this portion of the circuit from vi to V2} pro-
duces a net increase in pressure vg "" vi = Vi. The current now
flows out through the line Li to the + terminal of the motor.
Because of the line resistance, the potential drops from Vi at
the generator to vz at the motor in practically the same manner
that the water pressure dropped in pipe Fi (Fig. 51). A voltage

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DIRECT CURRENTS

V2 — vzia necessary to force the current through the line Li. The
hne a'b' shows the actual voltage at each point along the wire,
the distance of a*b' from the ground line being proportional to
the voltage at each point. This voltage drop is uniform.

Fig. 61. — Flow of water through a hydraulic motor and pipe system.

Fig. 62. — Flow of an electric current through an electric motor and the connect-
ing feeder system.

Referring to Fig. 51, the water enters the hydraulic motor W
and in overcoming the back pressure of the revolving blades its
pressure drops from A3 to A4, representing a net drop in pressure
H2. Pressure hi must necessarily be greater than hi in order that
the water may flow back through the pipe F2. The pressure
hi — hi is necessary to overcome the friction loss in the pipe F2.

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OHM'S LAW AND THE ELECTRIC CIRCUIT

It is to be noted that J?2, the net pressure at the motor terminals,
is less than the pressure Hi at the pump, by the sum of the pres-
sures necessary to overcome the friction in the two pipes Fi and F2.

In a similar manner, the pressure of the electric current in pass-
ing through the motor M,
drops from vz to v^, represent-
ing a net drop in pressiu'e 72.
A large percentage of this
voltage V2 is necessary to
overcome the back electromo-
tive force of the motor. V4 is
necessarily greater than vi, or
the current could not flow
along L2 back to the negative
terminal of the generator. It
is to be noted that, as in the
case of Fig. 51, the net poten-
tial difference V2 at the motor M is less than that at the generator,
Fi, by the amount lost in the drop in potential due to the
resistance of both the ovigoing and the return wire.

Difference of potential is therefore the equivalent of pressure
and tends to send current through a circuit; cm-rent is quantity of

Fio. 53. — Illustrating the existence of
potential difference without current.

J u

Tank

electricity per second.
Potential difference
may exist with no cur-
rent flow, in the same
manner that a boiler
may have a very high
steam pressure with
B^0:-^:;y^^:i^^^-^'^'^^'^ no steam flow, due to
^'^''^^' •'*'' all the valves being

closed. Likewise a
generator. Fig. 53,
may have a very high potential difference at its terminals, yet,
because the switch S is open, no current flows.

66. Difference of Potential. — In order that current may flow
between two points, there must be a difference of potential between
the two points as shown in Fig. 52. This is further illustrated in
Fig. 54. A large reservoir and a small tank are connected by a

Fig. 64. — Tank and reservoir at the same pressure.

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52 DIRECT CURRENTS

pipe P. The water level in the tank and in the reservoir is the
same, that is, there is pressure in each but there is no difference
in pressure between them. When the valve V is opened, no water
flows from the reservoir to the tank. However, if the valve 7'
is opened, allowing the water level in the tank to fall, a differ-
ence of pressure between the two tanks will result and water
will now flow from the reservoir to the tank.

Fig. 55 shows two batteries Ai and A 2 each having an emf.
of 2 volts. The positive terminal of ^1 has a potential of +2
volts above its negative terminal; likewise the + terminal of A 2
has a potential of +2 volts above its negative terminal. The
negative terminals of both batteries are at the same potential

0« + 2Volto^^

J^ 6-4-2 Volte

Volte

n— *

' p

C-+8 Volte >:^

N^ d»+9 Volte

Volte

n '

Fig. 55. — Two batteries having Fig. 56. — Two batteries having

equal electromotive forces. unequal electromotive forces.

because they are connected by a copper wire through which
no current flows, and consequently there can be no potential
difference between the ends of the copper wire. Therefore points
a and b must each be at the same potential of +2 volts. If
now the switch S be closed, no current will flow from a to 6,
because there is no difference of potential between a and 6.

In Fig. 56 the emf. of battery Bi is 3 volts and therefore the
potential of its positive terminal is 3 volts above that of its nega-
tive terminal. The emf. of battery B2 is 2 volts and therefore
the potential of its positive terminal is 2 volts above that of its
negative terminal. The negative terminals are at the same
potential. If this potential be assumed as zero, the point c
is at a potential of +3 volts whereas the potential of d is +2 volts.
Therefore the point c is at a potential of 3 - 2 or 1 volt higher than
d. When switch /S' is closed, a current will flow from c to d, in
virtue of c being at a higher potential than d.

66. Measurement of Voltage and Current. — ^Voltage or poten-
tial difference is ordinarily measured with a voltmeter. It is only
occasionally that absolute potential is of interest. Ordinarily
difference of potential is the quantity desired. The voltmeter

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OHM'S LAW AND THE ELECTRIC CIRCUIT

therefore should be connected across or between the wires whose
difference of potential is to be measured, as shown in Fig. 57.

Current is ordinarily measured with an ammeter. As ciu-rent
is the quantity of electricity per second passing in the wire, the
ammeter must be connected so that only the current to be meas-
ured passes through it. This is accompUshed by opening one of
the wires of the circuit and inserting the ammeter, just as a water
meter is inserted in a pipe when it is desired to measure the flow
of water in the pipe. When the ammeter is so connected, the
current passing through to the load is measured by the ammeter.
(See Fig. 57.) Never connect an ammeter across the line.

— Generator

Ivoltmlter ^^

Load

Fig. 57. — Proper method of connecting a voltmeter and an ammeter.

67. Ohm's Law. — Ohm's Law states that for a steady current
the current in a circuit is directly proportional to the electro-
motive force acting on the circuit and is inversely proportional
to the resistance of the circuit.

The law may be expressed by the following equation if the
current / is in amperes, the emf . E is in voUSy and the resistance R
is in ohms.

(17)

That is, the ciurent in amperes in a circuit is equal to the emf.
of the circuit in volts divided by the resistance of the circuit in
ohms. Potential difference may be represented by either the
letter "F" or "-B," V usually meaning terminal voltage and E
electromotive force or induced voltage.

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54 DIRECT CURRENTS

Example, — The resistance of the field winding of a shunt motor is 30 ohms.
What current will flow through the winding when it is connected across
116-volt mains?

/ - I « ^ - 3.83 amp. Am

By transformation, equation (17) becomes

"e^q JS? = /R (18)

^1-8.2x22 -70.4 V.

That is, the voltage across any

fj^Sx 48-163.6 V P*^ ^^ ^ circuit is equal to the prod-

8.2 A. uct of the current in amperes and

1 the resistance in ohms, provided the

'^~~ current is steady and there are no

FiQ. 58. — Voltage drops across a r f • j. i • j.i • ^ i?

generator field and its rheostat. SOUrceS of emf. Wlthm thlS part of

the circuit.

Example. — The resistance of the field
winding of a shunt generator is 48 ohms and the resistance of its rheostat is 22
ohms. (See Fig. 58.) If the field current is 3.2 amp., what is the voltage
across the field winding terminals, the voltage across the rheostat, and the
voltage across the generator terminals?

j^i = /i2i = 3.2 X 48 = 153.6 volts across field windmg
Ei = IR2 « 3.2 X 22 = 70.4 volts across rheostat
Total 224.0 volts at terminals.
Also

E = I{Ri +R2) - 3.2 (22 + 48) = 224.0 volts (check). An&.

Again, if equation (17) be solved for the resistance the result is

« = f . (19)

That is, the resistance of a circuit is equal to the voltage
divided by the current, provided the current is steady and there
are no sources of emf. within the circuit. This formula is very
useful in making resistance measurements. (See Par. 118.)

Example, — The voltage across the terminals of a generator field is 220
volts and the field current is 4 amperes. What is the resistance of the field
circuit?

-, E 220 _. , .

ic =s -^ = — r- = 55 ohms. Ans.
1 4

68. The Series Circuit. — As was stated in Par. 39, if several
resistances are connected in series, the total resistance is the
sum of the individual resistances. That i%igitizedbyCjOOQlc

R ^ ri + r2 + u, etc. (20)

OHM'S LAW AND THE ELECTRIC CIRCUIT

and the current

R ri + r» + rs +

(21)

Example, — ^A 50-ohm relay is connected in series with a resistance tube
of 30 ohms and with a small pilot lamp having a resistance of 5 ohms. The
operating voltage is 115 volts. What current flows in this relay circuit?

, 115 116 , „.

^=50+30+5^-85 =^-^^^"^P-

Arts.

69. The Parallel Circuit— In Par.
39, the relation of total resistance to
the component resistances in a
parallel circuit was proved by trans-
forming conductances into resistances.
This equation may be proved by
Ohm's Law as follows: Consider the

circuit of Fig. 59, consisting of resistances ri, r2, and rz in
parallel across the voltage E. Let I\ = the current in resistance
^1, /a = current in r2, and Iz = the current in r%.

Then

Fio. 59. — A parallel circuit.

h =

/3 =

E
r2
E

rz\

(equation 17)

Adding these together:

/. + 7, + /3 = ^ + ^ + ^ = ^(i + l + i)
^1 r^ rz \ri r2 rj

Let the total current be / = /i + /a + /s.
Let the equivalent resistance be R

7 = ^
R

Substituting I for 7i + 72 + 78
E

^-h<^7A)

l=i + l + i
R Ti Ti n

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(22)

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DIRECT CURRENTS

That is, the reciprocal of the equivalent resistance of a parallel
circuit is the sum of the reciprocals of the individual resistances.
If but two resistances are involved,

Ti + ra
If three resistances are involved.

R =

(23)

(24)

nrz+rirz+rzri

Example, — Determine the total current in a circuit consisting of 4 resist-
ances of 4, 6, 8, and 10 ohms respectively, connected in parallel across a
10-volt source.

R 4^6^8^10

R =

0.642

= 0.25 + 0.167 + 0.125 + 0.10
= 0.642 mho
1.56 ohms

j^= 6.42 amp.

Ans.

60. Diyision of Current in a
Parallel Circuit— In Fig. 60, two
resistances, Ri and i?2, are con-
nected in parallel across the

Fio. 60. — Division of current in a
two-branch parallel circuit.

voltage E.

Then

E/R,

i?2

E/Rt

(25)

That is, in a parallel circuit of two branches, the currents are
inversely as the resistances. (This does not apply to the division
of current through the field and armature of a shunt motor
when the motor is running.)

Example, — A current of 12 amp. divides between two paths in parallel,
part passing through a branch having a resistance of 8 ohms, the other
branch having a resistance of 12 ohms. How much current passes through
each branch?

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OHM'S LAW AND THE ELECTRIC CIRCUIT 57

Let Ji be the current in the S-ohm branch and I2 be the current in
the 12-ohm branch.

^-^ (eq.25) (1)

/,+/,= 12 (2)

Also

substituting in (2)

/, - 7, ^ = J, I from (1)

f-i.

h\^h - 12

h « 4.8 2 = 7.2 amp

If the circuit consists of three
branches (Fig. 61)

Ri, R%, and fia,

the respective currents may be f,g. ei.— Division of current in a

found as follows : three-branch parallel circuit.

Let / be the total current = /i + /2 + /a.

It can be shown that the current in each branch is given by :

J _ J / ^2^8 \ ^26)

\ RiR% -f- RiRz -\- RzRil

l^ ^ I I RzRi \ ^27)

\ R\R% -\- R^Rz "h RJtil

J^ :=: I ( ^1^2 \ /28)

\ R1R2 H" R2RZ + RzR\l

(Note the cycUc order of the subscripts.)

Example, — A current of 25 amperes passes through a parallel circuit of three
resistances of 2.6, 4 and 6 ohms, respectively. How does the current divide?
Current in 2.6 ohms.

4.0 X 6.0

- 25

(2.5 X 4.0) + (4.0 X 6.0) + (6.0 X 2.5)

= 26

10 + 24 + 15

12.25

amp.

= 25

6.0 X 2.5
49

= 7.65

amp.

= 25

2.5 X 4.0
49
Total

= 5.10
25.00

amp.
amp.

(check).

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68 DIRECT CURRENTS

61. The Series-parallel Circuit. — ^A circuit may consist of

groups of parallel resistances in series with other resistances or

groups of resistances as shown in Fig. 62.

^- In tUa case, each group of parallel re-

► »^ Ex sistances is first combined into its equivalent
■"■"f" single resistance by equation (22) and the
u«v. ioii| |wA E. whole is then treated as a series circuit.

Example, — Determine the total current in the
circuit shown in Fig. 62; determine the voltage
across each portion of the circuit ; determine the
Fio. 62. — Series-parallel current in each resistance.

circuit. Combine first the 10 and 12 ohm resistances

into a resistance R\

~ = -^ + 4 = 0.10 + 0.0833 = 0.1833 mho
lii 10 1^

Ri — 5.45 ohms

Likewise combining the group of three resistances into Rt

= 0.1567 mho

T 110 110 aKA

^ ^ 5 + 5.45 + 6.39 ' r6784 ^ 6-^»°^P-
El = 6.54 X 5.0 = 32.7 volts
E2 = 6.54 X 5.45 = 35.6 volts
Et = 6.54 X 6.39 -- 41.7 volts

Total 110.0 volts (check).

35 6

Current in 10 n = -r^ — 3.56 amp.

or £»

Current in 12 a = -j~- = 2.97 amp.

Total 6.53 amp. (check).

41 7
Current in 15 a = -7^ = 2.78 amp.
lo

41 7
Current m 20 a = -^ = 2.09 amp.

41 7
Current in 25 Q = -„V = 1-^7 amp.
Zo

Total 6.54 (check).

62. Electrical Power. — The unit of electrical power is the
watt and is defined as the power developed by one ampere in
falling through a potential difference of one volt. Watts are

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OHM'S LAW AND THE ELECTRIC CIRCUIT 69

therefore equal to the product of the volts and the amperes.
Thus the power

P = EI watts. (29)

Since E = IR in a, circuit containing resistance only, (equation
18), equation (29) may be written

P = (IR) I = im (30)

Substituting for I its value (/ = ^J in equation (29)

P = |- (31)

Equation (29) is useful when the volts and the amperes are
known; equation (30) is useful when the current and the resist-
ance are known; and equation (31) is useful when the voltage
and the resistance are known.

Example. — The resistance of a 150-scale voltmeter is 12,000 ohms. What
power is consumed by this voltmeter when it is connected across a 125- volt
circuit?

Since the voltage and the resistance are known, equation (31) is most
convenient :

^-S5 = '-'"***^- ^'"'

This may be checked by equation (29)

P = 125 X 0.0104 = 1.3 watts (check).

The watt is often too small a unit for commercial use and the
hilowaU (equal to 1,000 watts) is used when large amounts of
power are being considered. It is often necessary to transform
from mechanical horsepower to electrical power and conversely,
and a knowledge of the relation of the two is therefore useful :

746 watts = 1 horsepower (32)

0.746 kw. = 1 hofsepower (33)

and 1 hp. = 3/4 kw. very nearly (34)

1 kw. = 4/3 hp. very nearly (35)

Example. — An electric motor takes 28 amperes at 550 volts and has an
efficiency of 89 per cent. What horsepower does it deliver?
Input = 28 X 550 = 15,400 watts
Output = 15,400 X 0.89 = 13,700 watts

13,700/746 = 18.35 hp. at the pulley. Ans,

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60 DIRECT CURRENTS

63. Electrical Energy. — Power is the rate of doing worky
or is the rate of expenditure of energy. Therefore electrical
energy is equal to the product of electrical power and time.
The unit of electrical energy is the wait-second or joule.

W = E 1 1 watt-seconds

if < is in seconds, £ is in volts, and / is in amperes.

The watt-second is ordinarily too small a unit for commercial
purposes, so the larger unit, the kilowatt-hour (kw-hr.) is com-
monly used. 1 kilowatt-hour = 1,000 X 60 X 60 = 3,600,000
joules or watt-seconds.

The difference between power and energy (or work) should be
clearly borne in mind. Power is rate of doing work, just as
velocity is rate of motion. On the other hand, energy is the
total work done and is equal to the power multiplied by the
time during which it acts just as distance covered is the velocity
or rate of motion multiplied by the time. To speak of a train
traveling at a rate of 40 miles per hour gives no information
as to the total distance which the train travels. Likewise, to
speak of 50 kilowatts does not state the amount of energy that
is involved. The statement "electricity is sold for so many
cents per kilowatt" is incorrect. The correct expression is
'^electrical energy is sold for so many cents per kilowatt-/wmr."
To illustrate:

Example, — If energy is sold for 10c per kilowatt-hour (kw-hr.), how many
kilowatts may be purchased for 20c? This question as it stands cannot be
answered, since the time is not given. If, however, it is assumed that the
power is to be used for 1 hour:

20c/10c = 2 kw.-hr. available
2 kw.-hr./l hr. = 2 kw. Ans.
If used in 0.5 hr., 2kw.-hr. /0.5 hr. = 4 kw. Ans,
If used in 0.001 hr., 2 kw.-hr./O.OOl hr. = 2,000 kw. Ana.

so that the 20c could purchase any number of kilowattSy depending oil the
time during which the power is supplied.

In a similar way, horsepower is rate of doing work and is equivalent to
33,000 ft.-lb. per mirivie and not to 33,000 ft.-lb. A motor developing J^hp.
could do 33,000 ft.-lb. of work if allowed 8 minutes in which to do it.
When speaking of work in connection with horsepower, the korsepower-
hour is the unit ordinarily used.

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OHM'S LAW AND THE ELECTRIC CIRCUIT

Example. — ^How many watt-seconds are supplied by a motor developing
2 hp. for 5 hours?

2 X 5 = 10 hp.-hr.
10 hp.-hr. X 746 = 7,460 wat1>-hours
7,460 X 3,600 = 2.68 X 10^ watt-seconds. Ans,

64« Heat and Energy. — It is well known that heat may be con-
verted into mechanical and electrical energy, and, conversely, that
electrical and mechanical energy may be converted into heat.
The complete cycle of energy transformation is well illustrated
by a steam power plant. The energy is brought to the plant in
the coal, as chemical energy. The ingredients of the coal combine
with the oxygen of the air, thus converting the chemical energy
into heat energy. A certain percentage of this heat is trans-
ferred to the boiler and produces steam. The expansion of the
steam in the engine cylinders, or through the buckets and blades
of the turbine, converts the heat energy of the steam into me-
chanical energy. This mechanical energy drives the generator,
which converts a large percentage of this energy into electrical
energy, A portion of this electrical energy is transformed into
heat in the wires, bus-bars, etc. Finally, the remainder is
used to supply lamps, propel electric cars, operate motors, and
some may be used for chemical processes. Ultimately all the
energy appears again as heat or else is converted into chemical
or other forms of energy.

The following table shows approximately what becomes of each
100 heat units existing initially in the coal in the most efficient
modern power plants, using superheaters, condensers, and large
units.

EfELciency of Energy Conversion

Form of
energy

Efficiency
(per cent.)

Heat units
converted

Coal

Boiler

Turbine

Generator

Distribution system (to point of

utilization)

Small motors

Lamps

Chemical
Heat

Mechanical
Electrical

Electrical

Mechanical

Light

80
25
95

65 (av.)
2

100.0
80.0
20.0
19.0

16.2
10.5
0.32

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DIRECT CURRENTS

Fig. 63^ (by R. A. Philip) shows graphically the flow of power
from the boiler to the point of utihzation. It is apparent that
even in the most modern power plants the over-all efficiency is
very low.

Unltt 8S4 IM Its 190

UaltslllO IW 108 1«K>

Mm moo T800 B.T.U. pw MIb.

Fio. 63. — Energy flow — thermal, mechanical and electrical transmission.

66. Thennal Units. — The unit of heat in the English system
is the B.t.u. (British thermal unit) and is equal to the amount
of heat required to raise one pound of water 1® F. It is equal
to 778 ft.-lb. (called the Mechanical Equivalent of Heat).

In the C.G.S. system the heat unit is the gram-calorie and
is equal to the amount of heat required to raise one gram of
water 1 ° C. ^ A gram-calorie is equal to 4.2 watt-seconds or j oules.

By Joule's Law the heat developed in a circuit is:

TF - A PRt = 0.24 PRt calories
4.2

(36)

where t is in seconds, / in amperes and R in ohms.

Example, — 10 horsepower is delivered by a pump circulating 400 gallons
of water per minute through a certain cooling system. How many degrees
F. is the temperature of the water raised by the action of the pump?
10 hp. = 10 X 33,000 = 330,000 ft.-lb. per min.
330,000/778 = 424 B.t.u. per min.
400 gal. = 400 X 8.34 = 3,336 lb.
424/3,336 = 0.13° F. Ans,
1 A. I. E. E. Trans. Vol. XXXIV (1915), page 779.
*See Appendix A, page 407. . Digitized by GoOglc

OHM'S LAW AND THE ELECTRIC CIRCUIT

Example, — An incandescent lamp taking 0.5 amp. from 110-volt mains
is immersed in a small tank of water, containing 2,000 c.c. of water.
Neglecting radiation, by how many degrees per minute is the temperature
of the water raised?

W = 0.24 X 0.5 X 110 X 60 - 792 calories per minute.
792/2,000 = 0.396** C. Ana.

66. Potential Drop in a Feeder Suppl]ring One Concentrated
Load. — Fig. 64 shows a feeder (consisting of a positive and a
negative wire) supplying a motor load. The feeder is connected
to bus-bars having a constant potential difference of 230 volts.
The feeder is 1,000 ft. long and consists of two 250,000 CM.
conductors. The maximum load on the feeder is 250 amperes.
It is required to determine the voltage at the motor terminals
and the effi-ciency of transmission.

■^woo^

: 10J8 Volts arop In
+ Feeder

_|^_iiqj VollsdroplB
—Feeder

Fig. 64. — Voltage drop in a feeder due to a single load.

As was stated in Par. 54, the voltage at the motor must be less
than that at the bus-bars because of the voltage lost in supplying
the resistance drop in the feeder.

From Table 51, the resistance of 1,000 ft. of 250,000 CM.
cable is 0.0431 ohm. As was shown in Par. 54, the net voltage
at the receiving end of the line is less than the voltage at the
sending end by the voltage loss in both the (mtgoing and the
return wire. Therefore the drop in 2,000 ft. of cable must be
taken, the total resistance being 0.0862 ohm.

The current is 250 amperes.

By equation (18) the voltage drop in the line:

E' = 250 X 0.0862 = 21.55 volts.

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DIRECT CURRENTS

Therefore the voltage at the motor terminals is

230 - 21.6 = 208.4 volts. Am.

In Fig. 64 the voltage drop along the line is shown graphically.
The voltage at the sending end of the line is 230 volts, and there
is a uniform drop in each wire, this drop increasing uniformly
to 10.8 volts, making a total voltage loss of 21.6 volts The po-
tential difference between the two wires 500 ft. from the sending
end will be 230 - 10.8 = 219.2 volts as shown.

The power delivered to the motor = 208.4 X 250 watts.
The power deUvered to the line = 230 X 250 watts.

^^ ^ . c ^v. 1- output 208.4 X 250 208.4

The efficiency of the Ime = -— ^ = -^^n v> ocn = ~7^^?r

mput 230 X 250 230

or 90.8 per cent.

With one concentrated load the efficiency of transmission is
given by the voltage at the load divided by the voltage at the
sending end of the line.

-w>—

Bob
Bars

300,000 CM.

I 900,000 CM.
SOOA.
I 300,000 CM.

FiQ. 65. — Voltage drops in a feeder supplying two loads.

67. Potential Drop in a Feeder Supplying Two Concentrated
Loads at Diflferent Points.— In Fig. 65 a 300,000 CM. feeder
supplies 200 amperes to a load 800 ft. from the bus-bars, and
150 amperes to a load 400 ft. farther on. If the bus-bar voltage
is maintained constant at 240 volts, determine the voltage at
each load, the total line loss and the efficiency of transmission.

From Table 51, the resistance of 1,000 ft. of 300,000 CM. cable
is 0.0360 ohm. The resistance of 800 ft. = 800/1,000 X 0.0360
= 0.0288 ohm.

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OHM'S LAW AND THE ELECTRIC CIRCUIT 65

Voltage drop to the 200-amp. load

E' = 350 (2 X 0.0288) = 20.16 volts.
Voltage at 200-amp. load

i^i = 240 - 20.2 = 219.8 volts. Arts,
Resistance of one cable from the 200-amp. load to the 150-
amp. load = 400/1,000 X 0.0360 = 0.0144 ohm.
Voltage drop from 200-amp. load to 150-amp. load

E" = 150 (2 X 0.0144) = 4.32 volts.
Voltage at 150-amp. load

Ei = 219.8 - 4.3 = 215.5 volts. Arts.
The voltage distribution along this line is shown graphically
in Fig. 65.

To determine the efficiency:
Line loss to 200-amp. load

Pi = (350)2 (2 X 0.0288) ^ 7,060 watts (equation 30).
Line loss from 200-amp. load to 150-amp. load

Pi = (150)2 (2 X 0.0144) = 649 watts (equation 30).
Total line loss

P1+P2 = 7,060 + 649 = 7,709 watts or 7.709 kw.
Efficiency =
input - losses _ (240 X 350) - 7,709 _ 76,290 ^^ ^

input " 240 X 350 " 84,000 ~ ^'^ ^^"^ ''^''^'

68. Estimation of Feeders. — It was stated in Par. 42 that a
cir .-mil-foot of copper has a resistance of 10.37 ohms. In many
cases it is sufficiently exact to assume a value of 10 ohms. Assume
the current density in a feeder to be 1,000 cir. mils per ampere,
or 0.001 ampere per cir. mil. Call this the normal current den-
sity. (Bus-bars and large feeders operate at a density very
nearly equal to this.)

The voltage drop through a cir.-mil-foot carrying 0.001 ampere
is:

£ = //? = 0.001 X 10 = 0.01 volt.

Another cir.-mil-foot, carrying 0.001 ampere, will also have a
drop of 0.01 volt between its ends. If these be placed side by
side, the drop across the two will still be 0.01 volt. With any
number of wires, each having one circular mil cross-section, a

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DIRECT CURRENTS

length of one ft. and a current of 0.001 ampere, the drop across
the ends of each wire will be 0.01 volt. The wires may be
separated or they may be made into a cable.

In Fig. 66 (a) are shown four separate conductors, each of 1
cir.-mil-foot and each carrying 0.001 amp. The voltage across
ea^h must be 0.01 volt. In Fig. 66 (b) these same four conductors
are grouped together and as each carries 0.001 ampere, the total
current must be 0.004 ampere. The voltage drop across the

1 Oir. HiL

^/.\\

Pig. 66. — Voltage drop in a cir.-mil foot.

group is still 0.01 volt. If any number of cir.-mil-ft. conductors
each carrying 0.001 ampere are added in parallel to the group of
Fig. 66 (b)j the drop remains 0.01 volt.

From the foregoing the following rule may be deduced:

The voltage drop per foot of copper condicdor is always 0.01
volt provided that the current density is 0.001 ampere per circular
mil. Further J if the density is other than 0.001 ampere per circular
milj the voltage drop will be in direct proportion to the current density.
This last follows from equation (18), page 54.

Example. — A motor 800 ft. from the power house is to take 500 amperes
from 230-volt bus-bars. What size cable is necessary in order that the
voltage drop shall not exceed 20 volts?

A cable to operate at the normal density must have
500 X 1,000 = 500,000 CM.

The total voltage drop then becomes

0.01 X 800 X 2 = 16 volts.

The allowable drop is 20 volts, so a smaller cable may be used.
16

500,000 X ~ = 100,000 CM.

Ans,

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OHM'S LAW AND THE ELECTRIC CIRCUIT 67

This makes the actual current ^density

-rjr^ = 1.25 amp. per 1,000 cir. mils.
4U0

69. Power Loss in a Feeder. — The method of Par. 68 may be
used to estimate the power loss in a copper conductor. At the
normal density

P' = PR = (0.001)210 = 0.00001 (or lO^^) watt
per cir.-mil-ft.

The total power loss at the normal density

Po =^ 0.00001 X CM. X I

where CM. is the conductor cross-section in circular mils and I its
length in feet.

The actual power loss is proportional to the square of the ratio
of the actual to the normal density.

That is,

P = PoD^

where P is the actual power loss, Po the power loss at the normal
density and D is the actual current density in amperes per 1,000
cir. mils.

Example, — Determine the power loss in the example of Par. 68

Po = 0.00001 X 400,000 X 800 X 2 = 6,400 watts at the normal density.

The actual power loss

P = 6,400 X (1.25)2 = 10,000 watts = 10 kw. . Ans.

The foregoing gives an easy and rapid method of solution of
many problems. It is sufficiently exact for most practical
problems.

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CHAPTER V

Tm.

'"^

Ne«.

Battery

kvvwvvvO

BATTERY ELECTROMOTIVE FORCES— KIRCHHOFF'S

LAWS

70. Battery Electromotive Force and Resistance. — If a volt-
meter be connected across the terminals of a battery (Fig. 67),
the switch S being open, the instrument will record a certain vol tage

E, If now the switch
S be closed, allowing
the current I to flow,
the instrument will
record a voltage V which
is less than E.

The voltage E, meas-
ured when the battery-
delivers no current, is
the internai voUage or
the electromotive force
of the battery; the
voltage Vy measured
when a current I flows,
is known as the terminal,
voltage of the battery for that particular current value.

The difference between the open-circuit voltage E and the volt-
age Vy measured when current is being taken from the battery,
is the voltage drop in the battery due to the passage of current
through the battery resistance. Every cell has a certain resist-
ance, lying for the most part in the electrolyte, but partly in the
battery plates and terminals. When the external circuit is
closed so that current can flow, a certain voltage is required
to send this current through the battery resistance, just as vol-
tage is required to send current through an external resistance.
If the voltage -B, measured at the battery terminals when the
circuit is open, drops to V when the circuit is closed, the voltage

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Fig. 67.-

-Connections for measuring battery
resistance.

BATTERY ELECTROMOTIVE FORCES

e = (E — V) is the voltage drop through the cell due to the
passage of the current /. Let the cell resistance be r. Then,
by Ohm's Law,

E — V = e = Ir (by equation 18)
or

r = - = — (by equation 19)

(37)
(38)

/ I

E == V + Ir

That is, the internal resistance of the battery is equal to the
open-circuit voltage minus the closed-circuit terminal voltage
divided by the current flowing when the circuit is closed.

Example. — The open-circuit voltage of a storage cell is 2.20 volts. The
terminal voltage measured when a current of 12 amp. flows is found to be
1.98 volts. What is the internal resistance of the cell.

The voltage drop through the cell

Then

JS; — y = 2.20 — 1.98 = 0.22 volt
r = ^ = 0.0183 ohm. Ans.

In making a measurement of this character, it must be remem
bered that under open-circuit conditions even the ordinary volt
meter takes some current. If the cell capacity
is small (as in the case of a Weston cell) the
voltmeter current alone may reduce the
terminal voltage to a vajue one-half, or even
less, of the open-circuit voltage. Under these
conditions the voltmeter cannot be used to
measure the electromotive force of the cell.

Moreover, it is impossible to measure di-
rectly the internal voltage of the battery when
the battery dehvers current, for the voltage
■drop occurs vnthin the cell itself. Fig. 68
represents these conditions so far as their
effect on the external circuit is concerned. A battery cell B is
enclosed in a sealed box. Its resistance r is considered as re-
moved from the cell itself and connected external to the cell,
but within the sealed box. The cell then may be considered
as having no resistance, its resistance having been replaced by r.
The connections are brought through bushings in the box to

Fig. 68. — The internal
resistance of a cell.

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70 DIRECT CURRENTS

terminals a and 6. When no current is being delivered by
the cell, if a voltmeter be connected across the two terminals a
and 6, the instrimient will measure the emf., E. If, however, a
current / flows, the terminal voltage will drop from E to V, due
to the voltage drop in the resistance r. Under these conditions
it is impossible to measure E when the current is flowing, since
the voltmeter can only be connected outside the resistance,
through which the voltage drop occurs.

The voltage E and the resistance r are seldom constants but
are more or less dependent upon the current. They are also
affected by temperature, change in specific gravity of the electro-
lyte, polarization, etc.

71. Battery Resistance and Current. — As was shown in Par.
70, the resistance within the battery tends to reduce the flow of
current. If, in Fig. 67, the switch be closed, the cell electromotive
force E will be acting upon a circuit consisting of the internal
resistance of the cell r and the resistance of the external circuit jB.
The resistances r and R being in series, the total resistance in
the circuit is their sum. The current is

' = TTH ^^^^

The power lost in the battery is

P = 7V

If the cell is short-circuited, R becomes zero and 7 = — Under

these conditions all the electrical energy developed by the cell

is converted into heat within the cell itself.

Example — ^A battery-cell having an electromotive force of 2.2 volts and
an internal resistance of 0.03 ohm is connected to an external resistance
of 0.10 ohm. What current flows and what is the efficiency of the battery
as used?

' = o:o3¥aIo = o^j| = ^^-^*'^P- ^"''-

Power lost in the battery

P' = (16.9)2 X 0.03 = 8.67 watts.
The useful power

P = (16.9)2 X 0.10 = 28.6 watts.

P is equal to the total power developed by the battery minus the
battery loss.

2.2 X 16.9 = 37.2 watts
P = 37.2 - 8.6 = 28.6

Oft ft

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BATTERY ELECTROMOTIVE FORCES

From the above, the following rule may be deduced: The
current in a circuit is eqibdl to the total electromotive force acting
in the circuit divided by the total resistance of the circuit,

72. Batteries Receiving Energy. — If a resistance load be
connected across a battery, current will immediately flow from
the positive terminal of the battery and will return to the battery
though its negative terminal. As has already been pointed out, the
battery terminal voltage will be less than its open-circuit value,
due to the current flowing through the internal resistance of the
battery. Under these conditions, the battery is a source of
energy and is acting as a generator, that is, it delivers energy.

FiQ. 69. — Generator charging a battery.

If current is forced to erder at the positive terminal of the bat-
tery, the battery will no longer be supplying energy but will
be receiving energy. This energy must be suppUed from some
other source, as from another battery, or, as is more common,
from a generator. The cell shown in Fig. 69 has an electromotive
force of 2 volts, and a voltmeter F, connected across its terminals,
indicates 2 volts when no current flows. If another source of
electrical energy, such as a direct-current generator, supply a po-
tential difference of just 2 volts and its + terminal be connected to
the + terminal of the battery and its — terminal connected to the
— terminal of the battery, as shown in the figure, the voltmeter
V will still read 2 volts and the ammeter A will read zero That is,
the battery neither delivers nor receives energy and no effect
is noted other than those noted when the battery stood open-
circuited. Under these conditions the battery is said to be
"floating." If, however, the voltage of the generator be raised
slightly, the ammeter A will indicate a current flowing from the

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DIRECT CURRENTS

^^^^^^

400 Lb.

Fig. 70. — Force necessary to start a car.

+ terminal of the generator into the + terminal of the battery,
a direction just opposite to that which the current had when the
battery supplied energy. The voltmeter will no longer read 2
volts, but will indicate a potential difference somewhat in excess
of 2 volts.

What actually happens may be illustrated by a mechanical
analogy. Fig. 70 shows a car standing on the track. A force
of 400 lb. is necessary to overcome the standing friction of the
car on the track. At one end of the car a force F is applied.
Before the force F can move the car its value must at least equal
400 lb. When F is exactly 400 lb. the car will not move, just

as no cm-rent flowed into
the battery when the gen-
erator voltage was just
equal to that of the battery.
When the force F exceeds
400 lb., however, the car will move, the force effective in
producing this motion being the amount by which F exceeds
400 lb. Thus, if F = 450 lb., 400 lb. of this is utilized in over-
coming the 400 lb. opposing force due to friction and 50 lb. is
effective in moving the car.

In the case of the battery no current will flow until voltage in
excess of the 2 volts is produced by the generator. Thus, if the
generator voltage be raised to 2.4 volts, 2.0 volts of this is utihzed
to "buck" the 2.0 volts of the cell and 0.4 volt is effective
in sending current into the cell. Thus, if the cell resistance
be 0.1 ohm, the current will be

/ = Q— = 4.0 amp.

This assumes that the resistance of the leads is negligible.

Therefore, if JB? is the electromotive force of a battery, r its
resistance and V the terminal voltage when current flows in at
its positive terminal,

V — E

I = ^^-Z^ (40)

and E = V - Ir (41)

That is, the electromotive force of the cell is less than the termi-
nal voltage by the amount of the resistance drop in the cell itself.
These equations should be compared with equations (37) and
(38), respectively. Digitized by (^OOglC

BATTERY ELECTROMOTIVE FORCES 73

Under these conditions, the cell is receiving electric energy, as
is the case when a storage battery is being charged.

73. Battery Cells in Series. — Strictly speaking, a battery
consists of more than one unit or cell. However, the term
battery has come also to mean a single cell, when this cell is
not acting in conjunction with others.

When cells are connected in series, their electromotive forces are
added together to obtain the total electromotive force of the battery ,
and their resistances are added together to obtain the total resistance
of the baMery.

Thus, if several cells, having electromotive forces, Ei, E2, Ez,
E^j etc., and resistances ri, r2, rs, r^, etc., are connected in series,
the total electromotive force of the combination is

E = Ei + E2 + E3 + ^4, etc. (42)

and the total resistance is

r = n + r2 + rz + r^y etc. (43)

Equation (42) assumes that the cells are all connected + to
— so that their electromotive forces are additive. If any cell
be connected so that its electromotive force opposes the others,
its voltage in equation (42) must be preceded by a minus sign.

If an external resistance R is connected across these cells in
series, then by equation (39) the current is

= ^ =: El + E2 + Ez + Ea, etc. .^v

r + R ri + r2 + rz + r,, etc., + R ^^

Example, — Four dry cells having electromotive forces of 1.30, 1.30, 1.35,
and 1.40 volts and resistances of 0.3, 0.4, 0.2, and 0.1 ohm, respectively, are
connected in series to operate a relay having a resistance of 10 ohms. What
current flows in the relay?

1.30 + 1.30 + 1.35 + 1.40 ^ 5^
0.3 + 0.4 + 0.2 + 0.1 + 10 11.0

= 0.486 amp. Ana.

A battery consisting of n equal cells in series has an emf. n times
thai of one cell, but has the current capacity of one cell only.

74. Equal Batteries in Parallel. — To operate satisfactorily in
parallel all the batteries should have the same electromotive
force. The behavior of batteries having unequal electromotive
forces can be treated as special problems (see Par. 78).

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DIRECT CURRENTS

Fig. 71 shows a battery of six cells, each having an electrons o-
tive force of 2.0 volts and a resistance of 0.2 ohm. It is clear
that the emf . of the entire battery is no greater than the emf . of
any one cell. The current, however, has 6 paths through which
to flow. Therefore, for a fixed external current, the voltage drop
in each cell is one-sixth that occurring if all the current passed
through one cell. If the internal resistance of one cell is 0.2 ohm,
the resistance of the battery as a whole must be 0.2/6 = 0.033
ohm.

£;-2v.

r>o.2Q

+ Jff»2V,

"Jt0.20.

+ E'Vf,

■K0.2fi

+ JE7-2V.

4.^«2V.

"r-t).2Q

+ £7-2V.

"rs-o.2n

''ie-o.s

Fig. 71. — Parallel arrangement of equal cells.

Example. — If the external resistance connected across the terminals of
the battery in Fig. 71 is 0.3 ohm, what current flows?
Resistance of battery = 0.2/6 = 0.033 ohm.

7 2.0 2.0 ^ / om ^

^ = 0:033To:3 = 0333 ^ ^ ^°^P- ^^'^' ^^^' ^'^'

If the emfs. are equal but the resistances of the cells are not all equal, but
are ri, r2, rs, r4, etc., the battery resistance r is found by considering these
resistances as being in parallel (equation (9), Chap. Ill, page 37).

1^1,1,1,,

= 1 1 1 f- etc.

r fi r^ Ti Ti

(45)

Example. — ^A battery consists of 4 cells connected in parallel, each having
an electromotive force of 2.0 volts, but resistances of 0.30, 0.25, 0.22, and
0.20 ohm respectively. If a resistance of 0.5 ohm is connected across the
terminals of the battery, what current flows, and how much current does
each cell supply? What is the voltage across the battery terminals?

-^^^ofe+ois+o^+oi^^^-^^^^^^-

16.87

= 0.0593 ohm.

/ =

2.0

0.0593 + 0.50 0.5593
The terminal voltage

El = IR =^ 3.58 X 0.5 = 1.79 volts.

= 3.58 amp. Ans.

Ans.

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BATTERY ELECTROMOTIVE FORCES

The current in each cell may be found by means of equation (37).
2.0 - 1.79

Solving
Likewise

I2 =

h =

/* =

/i
2.0 - 1.79

'= 0.30

= 0.70 amp.
0.84 amp.
= 0.95 amp.
= 1.05 amp.

^0.21

0.30 0.3

2.0 - 1.79 0.21

0.25 0.25

2.0 - 1.79 ^0.21

0.22 0.22

2.0 - 1.79 ^0.21

0.20 0.20

Total current 3.54 (check;. Ana.
That is, the current in any cell is equal to the voltage drop in the ceU divided
by the resistance of the cell.

It will be found that the products of the current and the resistance of
each cell are all equal.

0.7 X 0.3 == 0.84 X 0.25 = 0.95 X 0.22 = 1.05 X 0.20
Cells connected in parallel must aU have the same terminal voUage since all
the positive terminals are tied together and all the negative terminals are
tied together. If the emf.'s of the cells are all equal, the total battery
emf. is equal to the emf. of but one cell. The total battery resistance
may be foimd by the equation for resistances in parallel. The current in
each cell is inversely proportional to the resistance of the cell if the electro-
motive forces are all equal. The current capacity of the battery is the sum
of the current capacities of the indi^ddual cells.

75. Series-parallel Grouping of Cells. — Rows of series-con-
nected cells may be so connected that the rows themselves are

T T J

T T -

r T T 1

E (each cell)-0.9 V.
r'<each cell)«-0.08-n.

Fig. 72. — Series-parallel grouping of cells.

grouped in parallel. Fig. 72 shows a row of 4 cells in series, and
five of these rows in parallel. If there are m equal cells in series
in each row, then the emf. of each row must be

E = mE' (by equation 42)
where E' is the emf. of one cell.

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76 DIRECT CURRENTS

The resistance of each row must be

ri = mr' (by equation 43)
where r' is the resistance of one cell.

Since there are n rows in parallel, the resistance of the whole
combination must be

r =-^ = — r'
n n

(46)

If an

external resistance R is connected to the battery-

, the

current

(47)

Example, — Let each of the cells of Fig. 72 have an emf. of 0.9 volt and

an internal resistance of 0.08 ohm. If the external resistance R is 0.5 ohm,

what current flows?

, 4 X 0.9 3.6 ^ . .

/=j «_- = 6.4amp. Ans.

I 0.08 + 0.5 "-^^
o

76. Grouping of Cells. — (a) To obtain the best economy, group
the cells so that the battery resistance is as low as possible.
This usually means a large nimxber of parallel connections.
Under these conditions the life of the battery will be prolonged
but the initial cost is excessive.

(6) To obtain the maximum current with a fixed external

m
resistance make the internal resistance (— r') of the battery

equal to the external resistance. This is not economical, since
only half of the energy developed by the battery is avfiilable
in the external circuit; the other half is lost in the cells themselves.
Under these conditions the battery delivers the maximum power.
(c) To obtain quick action for the intermittent operation of
relays, beUs, etc., group the cells in series if possible

Example, — In the example of Par. 75, how should the cells be arranged to
obtain the maximum current?

The total battery resistance — 0.08 must be equal to the external resistance.

That is,

- 0.08 = 0.5

Also

w X n = 20 71 = —

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BATTERY ELECTROMOTIVE FORCES 77

Solving -^0.08 = 0.5

m = 11 + Ana,

The best arrangement is ten cells in series, and two rows in parallel.
(Eleven cells in series would not operate satisfactorily if connected in
parallel with the remaining nine cells in series.)

77. Kirchhoff's Laws. — By means of Kirchhofif's Laws it is
possible to solve many circuit networks that would otherwise
be diflScult of solution.

(1) In any branching network of wires, the algebraic sum of the
currents in all the wires that meet at a point is zero.

(2) The sum of all the electromotive forces acting around a complete
circuit is equal to the sum of the resistances of its separate parts
multiplied each into the strength of the current that flows through it,
or the total change of potential around any closed circuit is zero.

The first law is obvious. It states that the total current
leaving a junction is equal to the total current entering the junc-
tion. If this were not so ^^^
electricity would accumulate
at the junction. Xfs ^ ^^+4a

The law is illustrated by
Fig. 73. Four currents, Ii,
I2, I3, and J4 meet at the

junction 0. The first three ^^

currents flow Umard the j,^^ ^3 _ j^^^^^^^.^^ j,.^^j^j^^^.^ g„, 1^^
junction so have plus signs as

they add to the quantity at the point 0. The last current J4
flows away from the junction, so has a minus sign as it subtracts
from the quantity at the point 0. Then

I1 + /2 + /3 - /4 = 0 (48)

Assume that Ji = 5 amp.; 72 = 8 amp. and J4 = 17 amp.

Then 5 + 8 + /a- 17 = 0

and Iz = +4 amp., the plus sign indicating that the current
flows toward the junction.

The second law is but another application of Ohm's Law (equa-
tion 18). The basis of the law is obvious; if one starts at a cer-

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DIRECT CURRENTS

tain point in a circuit, and follows continuously around the paths
of the circuit until the starting point is again reached, he must
again have the same pjotential with which he started. Therefore
the sources of electromotive force encoimtered in this passage
must necessarily be equal to the voltage drops in the resistances,
every voltage being given its proper sign.

This second law is illustrated by the following example.

SialO v. ^,a6 v,

Ui=lA B2p2A

Blie aoe to E\
10

^0.5 Volt drop in (1)

(1) (2)

RmHU

•♦-1 n 0.6 Amp.
(a)

Fig. 74. — Voltage relations in an electric circuit.

Two batteries (Fig. 74), having electromotive forces of 10
and 6 volts and internal resistances of 1 and 2 ohms respectively,
are connected in series opposing (their + terminals connected
together) and in series with an external resistance of 5 ohms.
Determine the current and the voltage at each part of the cir-
cuit.

Since the two batteries act in opposition, the net electro-
motive force of the two batteries is 10 — 6 = 4 volts*

The current is,

J 10-6 4 ^.
^=1 + 2 + 5 = 8=^'^ ^^P-

Consider the point A as being at reference potential. In
passing from A to 5 there is a 10- volt rise in potential due to the
electromotive force of battery No. 1, but around the circuit
in the direction of the current flow there occurs a simultaneous
0.5- volt drop of potential due to the current flowing through the 1-
ohm resistance of cell No. 1. Therefore the net potential at B
is but 9.5 volts greater than that at A, as is shown in Fig. 74 (6).
In passing from B to C there is a drop of 6 volts due to passing
from the -|- to the — terminal of battery No. 2, and there is

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BATTERY ELECTROMOTIVE FORCES 79

also a further drop of 1 volt due to the current of 0.5 ampere
flowing through the 2-ohm resistance of battery No. 2. This
makes the net potential at C = 9.5 — 6 — 1 = +2.5 volts.
In passing from C to A there is a drop in potential of 2.5 volts due
to the current of 0.5 ampere flowing through the 5-ohm resistance.
When point A is reached the potential has dropped to zero.

Therefore the sum of all the electromotive forces in the circuit,
taken with their proper signs, is equal to the sum of the Ir
drops. This is illustrated as follows:

Electromotive forces Ir drops

Cell No. 1 = + 10 volts Cell No. 1 = - 0.5 X 1 = - 0.5 volt "
'' No. 2 = - 6 volts '' No. 2 = - 0.5 X 2 = - 1.0 volt
Total + 4 volts 5-ohm res. = - 0.5 X 5 = -- 2.5 volt

Total - 4.0 volt

+ 4 + ( - 4) = 0

78. Applications of Kirchhoff's Laws. — In the application of
Kirchhofif's second law to specific problems the question of alge-
braic signs may be troublesome and is a frequent source of error.
If, however, the following rules are kept in mind no difficulties
should occur.

A rise in voltage should be preceded by a + sign.
A drop in voltage should be preceded by a — sign.

For example, in passing through a baitery from the — to the
+ terminal, the potential rises so that this voltage should be
preceded by a + sign. On the other hand, when passing from
the + terminal to the — terminal, the potential drops j so that
a — sign should precede this voltage. These points are illu-
strated by Fig. 74.

When going through a resistance in the same direction as the
current, the voltage drops, so that this voltage should be preceded
by a — sign. A voltage due to passage through a resistance in
the direction opposite to the current flow should be preceded by
a + sign.

This is further illustrated by the electric circuit shown in
Fig. 75. Three batteries having emf .'s ^i, E2, and Ez are con-

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DIRECT CURRENTS

nected as shown in different parts of the network of resistances,
Rij R2j Rz, Ra' The assumed directions for the various currents
are indicated by the arrows. The battery resistances are as-
sumed negUgible as compared with the other circuit resistances.
Starting at the point a, and applying Kirchhoff's second law
to the path cibcda, an equation may be written

+ El- I,R, - I2R2 + E2- LRi = 0

Startmg at / and passing along the path febcdf:

— Ez + IzRz — I2R2 "f" E2 = 0

£r,-4V.

- R.

Fig. 75. — Application of Kirchhoff's Fig. 76. — Application of Kirchhoff's
laws. laws.

This gives but two equations for the determination of three
unknown currents. Three equations are necessary. The third
may be obtained by applying Kirchhoff's first law to some
junction as b,

+ I1- I2- h = 0

since Ii is assumed to flow toward the junction and I2 and Iz
away from the junction.

With these three equations it is possible to determine the three
currents.

Example, — Fig. 76 shows a network identical with that shown in Fig. 75,
except that numerical values are used. The battery resistances are assumed
to be small compared with the circuit resistances, and are neglected.

Considering path ahcday

+ 4 - (/1O.6) - (/gS) + 2 - (/il) =0

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BATTERY ELECTROMOTIVE FORCES

1.57i +37, = 6
Similarly, path febcdfy starting at /,

-3 + (7,1) - 37, + 2 - 0

and at the junction 6,

37, - 7, - - 1

+ 7i - 7, - 7, - 0

7i = 7, + 7,
Substituting h (C) in (A)

1.5(7, + 7,) + 37, = 6
4.67, + 1.57, - 6
and combining with (B)

(fi)

(C)

97, - 37, = - 3

(B)

97, + 37, « 12

- 67, ^ 15

7, = 2.5 amp.

An8,

Substituting thin value in (B)

37, - 2.5 ^ 1

37, = 1.5

7, = 0.5 amp.

7i = 7, + 7, = 3.0 amp.

79. Assumed . Direction of

-4V.

Fig.

:i-r^

^j-2 V.

0.6-^

Current. — In the solution of this ^ _i j- ^ A

type of problem, the question of
assuming the proper direction of
current often arises. The current
may be assumed to flow in either
direction. K the assumed direc-
tion of the current is not the
actual direction, this current will
be foimd to have a minus sign
when the equations are solved.

Example, — ^This is illustrated by assum-
ing that the three currents of Par. 78
have such a direction that they all meet
at point (£ as is shown in Fig. 77. This condition is of course impossible

Considering circuit abcda, starting at a,

+ 4 + 0.57i - 37, + 2 + 7, = 0
1.57, - 37, + 6 = 0

-V\AAA — \^

3-^

■8V.

- 1-r^

77. — Amplication
chhoff's laws

Kir-

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DIRECT CURRENTS

Similarly circuit fehcdf, starting at /,

- 3 + /a - 3/2 + 2 = 0
/a - 3/2 - 1 = 0
The three currents /i, /z, /s all flow toward junction d, therefore

/i + /2 + /. = 0.
Substituting and solving

/i = — 3 amp. Ans,
I2 — 0.5 amp. Ans,
/s = 2.5 amp. Ans.

The minus sign preceding I\ signifies that this current flows in the opposite
direction to that assumed and indicated by the arrow, Fig. 77. The +
signs before 1 2 and Iz indicate that the assumed directions for these two
currents were the actual directions of flow.

80. Further Applications of Kirchhoff's Laws. — Kirchhoff's
laws might be applied to problems involving distribution systems,
electric railways, etc., where power is fed to the loads through
different feeders and from different sub-stations. In practice,
however, Kirchhoff's laws are rarely applied directly to electric
railway systems, since the widely fluctuating loads which are
constantly shifting their location make it impossible to formulate
a definite problem. Only occasionally is it necessary to apply
these laws to power and lighting systems, since the feeder layout
in such systems is usually determined by various operating
considerations.

The following problem illustrates the possible application of
these laws.

Fig. 78. — Ring-feeder system.

Exam-pie. — In Fig. 78, a 240- volt sub-station at A supplies two distributing
centers B and C, by a ring system of feeders. Between A and By a distance
of 800 ft., two 1,000,000 CM. feeders are paralleled; between A and C, a
distance of 1,200 ft., three 1,000,000 CM. feeders are paralleled; between

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BATTERY ELECTROMOTIVE FORCES 83

B and C, a distance of 600 ft., a 1,000,000 CM. feeder is connected. De-
termine the current in each feeder and the voltage at each distributing cen-
ter, when the load at B is 2,000 amp. and that at C is 3,500 amp.
Assuming 10 ohms per cir.-mil-f oot :

Resistance per wire A to B — ^^ ^^ = 0.004 ohm.

600 X 10
Resistance per wire B to C = i /x^^x ryy^ ~ 0.006 ohm.

Resistance per wire AtoC = 4 nnn aaa ~ 0.004 ohm.

o,lHX),(JUU

Going from A %o B to C^ out on the positive and back on the negative
conductor,

240 - 7i(0.004) - (/i - 2,000)0.006 - Ec (/i- 2,000)0.006 -/i (0.004) =0
240 -/i(0.02)+24 -- Ec (1)

Likewise going direct from A to C

240 - 72(0.004) - Ec - 72(0.004) = 0

240 - 72(0.008) = Ec (2)

Equating (1) and (2)

240 - 7i(0.02) + 24 = 240 - 72(0.008)

0.027i - 0.00872 = 24 (3)

At the junction at C

7i - 2,000 + 72 == 3,500

7i -f 72 = 5,500 (4)

Substituting in (3) for 7i = 5,500 - 72

0.02(5,500 - 72) - 0.00872 - 24

110 - 0.027, - 0.00872 = 24

0.02872 = 86

72 = 3,070 amp. Ans.
7i = 2,430 amp. Ana.
Voltage at C (equation 2)

Ec = 240 - 3,070(0.008) = 215.44 volts. Ans.
Eb = 240 - 2,430(0.008) = 220.56 volts. Am.

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CHAPTER VI

PRIMARY AND SECONDARY BATTERIES

81. Principle of Electric Batteries. — If two copper strips or
plates be immersed in a dilute sulphuric acid solution, Fig.
79 (a), and be connected to the terminals of a voltmeter, no appre-
ciable deflection of the voltmeter will be observed. This shows
that no appreciable difference of potential exists between the
copper strips. If, however, one of the copper strips, Fig. 79 (6),

Voltmeter

Copper- i >fER^^^S-5^

>AU>A'/^^MMMM'Ai^MVA^/y}A

Copper

._-_ _. Dilute

sz.::irz.^. ^ -Sulphuric

(a) (5)

FiQ. 79. — Simple primary cell.

be replaced by a zinc strip, the voltmeter needle will deflect and
will indicate approximately one volt, showing that a potential
difference now exists. It will be necessary to connect the copper
to the + terminal of the voltmeter and the zinc to the —
terminal in order that the voltmeter may read up scale. This
shows that so far as the external circuit is concerned, the copper
is positive to the zinc.

The above experiment may be repeated with various metals.

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PRIMARY AND SECONDARY BATTERIES 85

For example, carbon or lead may be substituted for the copper
and a potential difference will be. found to exist between each
of these and the zinc, although it will not be of the same value
as it was for the copper-zinc combination. Likewise other metals
may be substituted for the zinc, and potential differences will
be foimd to exist.

Furthermore, it is not necessary that sulphuric acid be used for
the solution. Other acids such as hydrochloric, chromic, etc.,
may be substituted for the sulphiuic; or even salt solutions
such as common salt (sodium chloride), ammonium chloride
(sal ammoniac), copper sulphate, zinc sulphate, etc., may be
used.

In order to obtain a difference of potential between the two
metal plates, but two conditions are necessary.

(1) The plates must be of different metals.

(2) They must be immersed in some electrolytic solution, such
as an acid, alkali, or salt.

Again, if current be taken from the cell shown in Fig. 79 (6)
by connecting a resistance across its terminals (Fig. SO), current
will flow from the copper through the resistance AB and into
the cell through the zinc. Inside the cell, however, the current
will flow /row the zinc through the solution to the copper as shown
in Fig. 80. For the reason that current flows from zinc to copper
vrithin the cell, zinc is said to be electrochemically positive to
copper. Therefore, when considering such an electrolytic cell,
the copper is positive to the zinc when the external circuit is
considered, but the zinc is electro-positive to the copper when
the plates and the solution alone are considered.

82. Definitions. — The metal strips or plates of a cell are called
electrodes. The electrode at which current enters the solution
(as the zinc. Fig. 80) is the anode, and the electrode at which
current leaves the solution (as the copper, Fig. 80) is the cathode.

The solution used in a cell is called the electrolyte.

If current be taken from the cell under proper conditions and
for a considerable time, the zinc plate will diminish in weight.
This is true not only in the case of this particular cell, but in prac-
tically all cells the flow of current is accompanied by a loss in
weight of at least one of the plates. Energy is stored in the cell
chemically, and the electrical energy is delivered at the expense of

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DIRECT CURRENTS

the plate which goes into solution. That is, one plate is either
oxidized or converted into another chemical compound, this
change being accompanied by a decrease of chemical energy
of the system. Therefore chemical energy is converted into
electrical energy, when the cell delivers a current.

Hence:

An electric ceU or battery is a device for transforming chemical
energy into electrical energy.

Such cells or batteries are divided into two classes: primary
cells and secondary cells.

^pP^-o

Ottthode

Anode

'//////^////^//////X/.

LAAAAr-"

Fig. 80. — Current-flow in a single cell.

In a primary cell it is necessary from time to time to renew
the electrolyte and the electrod^ which goes into solution by
fresh solution and new plates, respectively.

In a secondary cell the electrolyte and the electrodes which
undergo change during the process of supplying current are
restored electrochemically by sending a current through the
cell in the reverse direction.

83. Primary Cells. — Although it was stated in Par. 81 that
there are many combinations of metals and solutions capable
of generating an electromotive force and so forming a cell,
only a limited number of such combinations are commercially
practicable. The general requirements of a good cell are as
follows:

(a) There must be little or no wastage of the materials when
the cell is not delivering current.

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PRIMARY AND SECONDARY BATTERIES 87

(6) The electromotive force must be of such a magnitude as
to enable the cell to deliver a reasonable amount of energy with
a moderate current flowing.

(d) The internal resistance and the polarization effects must
not be excessive, otherwise the battery cannot supply even

• moderate values of current, at least for any appreciable time.

As an illustration, the cell shown in Fig. 79(6) would not be
practicable, because both the copper and the zinc would waste
away even were the battery delivering no current. Polarization
(see Par. 85) would be excessive, and therefore the battery
would be capable of delivering only a comparatively small
current.

84. Internal Resistance. — As was pointed out in Chap. V,
every cell or battery has an internal resistance, which reduces the
magnitude of the current and causes the terminal voltage to drop
when current is taken from the cell. Such resistance lies in the
electrodes, in the contact surface between the electrodes and the
electrolyte, and in the electrolyte itself. This resistance may
be reduced by changing the dimensions of the cell in the same
way as would be done for any electric conductor. The cross-
section of the path through which the current flows inside the
cell should be made as large as is practicable. This means large
area of electrodes in contact with the electrolyte. Also the cross-
section of the plates must be large enough to carry the current
to the cell terminals without excessive drop in voltage. Little
diflSculty is experienced in making this voltage drop negligible.
It will be appreciated that larger electrodes mean a larger cell,
with a greater current capacity. In addition to increasing the
area of the electrodes, the resistance of the cell may be diminished
by decreasing the distance between the plates. This reduces
the length of the path through which the current flows within
the cell and correspondingly reduces the cell resistance.

Increasing the size of the cell does not increase its electromo-
tive force. This electromotive force depends only upon the
material of the two electrodes, and the electrolyte. Thus, Fig.
81 shows two gravity ieells, made up of the same materials, but
differing materially in size. The cells are bucking each other,

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DIRECT CURRENTS

that is, their + terminals are joined and their — terminals
are joined. A galvanometer G connected in one of the leads
reads zero, indicating that no current flows from the larger to the
smaller cell.

Fig. 81. — Equality of electromotive forces in cells of unequal sizes.

85. Polarization. — If a test be made to determine the fall
of terminal voltage as current is taken from a cell, by connecting a
voltmeter, ammeter, and an external resistance as in Fig. 80,
the results will be somewhat as follows:

When the cell is on open-circuit the voltmeter will indicate

I.R. Drop
. in Cell

u___ '-^^ ______j

^laBEation

Time

Fig. 82. — Drop of voltage in a cell due to polarization.

the cell electromotive force E, represented by the distance OA,
Fig. 82. When the switch S is closed, current will flow and the
voltage will drop immediately from OA to OB. The distance
AB represents the voltage drop due to the internal resistance

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PRIMARY AND SECONDARY BATTERIES

of the cell and this has been considered earlier, in some detail.
As time elapses the terminal voltage will be observed to drop still
further, even though the current be maintained constant. This
further drop of voltage is due to polarization.

When the cell delivers current, small bubbles of hydrogen
appear upon the positive plate or cathode, practically covering
it. These bubbles have two effects:

They cause a substantial increase in the resistance at the
contact surface between the cathode and the electrolyte.

Hydrogen acting in conjunction with the cathode or positive
plate sets up an electromotive force which opposes that of the
cell.

These two effects explain the reduction in the current capacity
of many types of cells after they have delivered ciurent for
some time.

Remedies for Polarization, — These hydrogen bubbles may be
removed mechanically by brushing them off or by agitating
the electrolyte. This is impracticable under commercial
conditions. If the plate be roughened, the bubbles form at the
projections and come to the surface
more readily.

The hydrogen bubbles may be
removed chemically by bringing
oxidizing agents, such as chromic
acid or manganese peroxide, into
intimate contact with the cathode.
The hydrogen readily combines
with the oxygen of these compounds
to form water (H2O) This method
is used in the bichromate cell, in the
Le Clanch6 cell and in dry cells.

86A. Daniell Cell.— This cell.
Fig. 83, is a two-fluid cell having
copper and zinc as electrodes. It
consists of a glass jar, inside of which is a porous cup containing
zinc sulphate solution or a solution of zinc sulphate and sulphiu'ic
acid. The anode or negative electrode is immersed in this
electrolyse. The porous cup is placed in a solution of copper
sulphate with copper sulphate crystals in the bottom of the jar.

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Zinc Stilplmte
Solution

Fig. 83. — Daniell cell.

90 DIRECT CURRENTS

The copper plate, which is cathode, surrounds the porous cup.
The porous cup keeps the two solutions separated. As the
copper is in a copper sulphate solution, there is no polarization.
This cell is designed for use in a circuit which is kept continually
closed. If left idle the electrodes waste away. When the cell is
taken out of service for some time, the electrodes should be re-
moved and the porous cup should be thoroughly washed. The
electromotive force of this cell is about 1.1 volts.

86B. Gravity Cell. — ^The gravity cell is similar to the Daniell

cell, except that gravity, rather than a porous cup, is depended

upon to keep the electrolytes separated.

This cell is shown in Fig. 84. The cathode,

which is of copper, is made of strips riveted

together and placed in the bottom of the

cell together with copper sulphate crystals.

A solution of copper sulphate is then

poured to within a few inches of the top

of the jar. The connection to the copper

is usually an insulated copper wire fastened

to the copper and carried out through the

solution to the top of the jar. There should

.— ravi y ce . g^j^g^yg j^^ copper sulphate crystals at the

bottom of the cell.

The anode is zinc, is usually rather massive and is cast in

the form of a crow's foot and hung on the top of the jar. This

is surrounded by a zinc sulphate solution. The solutions are

kept separated by gravity. The copper sulphate is the heavier

of the two solutions and therefore tends to remain at the bottom.

The solutions should be poured in carefully for if the copper

sulphate solution comes in contact with the zinc, copper will be

deposited. This copper should be removed if by chance it

becomes deposited in any way. In the operation of the cell

the zinc goes into solution as zinc sulphate, and metallic copper

comes out of the copper sulphate solution and is deposited upon

the copper electrode. The cathode will therefore gain in weight

whereas the anode will lose in weight. This is the reason for

having the zinc electrode massive, and the copper electrode of

very thin sheet copper, when the cell is set up initially.

Due to capillary action the electrolyte tends to creep up

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PRIMARY AND SECONDARY BATTERIES

over the top of the jar forming a crystalline deposit. To pre-
vent this creeping, the top of the jar should be paraffined. To
prevent evaporation, the upper surface of the electrolyte may
be covered with oil. When the cell is replenished, metallic
zinc and copper sulphate are suppUed and metallic copper and
zinc sulphate are removed.

The gravity cell is a closed circuit battery, and the circuit
should therefore be kept closed for the best results. Otherwise
the copper sulphate will gradually mix with the zinc sulphate.
The cell has been found very useful in connection with railway
signals, fire alarm systems, and telephone exchanges, all closed
circuit work, although the storage battery
has replaced.it in many instances. The
electromotive force of the cell is practically
that of the Daniell cell, being about 1.09
volts, but varies sUghtly with the concen-
tration of the solutions.

87. Edison-Lalande Cell. — The Edison-
Lalande cell is still used to some extent.
The cathode is of copper oxide and is
suspended between two zinc plates which
form the anode. All the plates are fastened
to a porcelain cover by means of bolts
which serve as binding posts as well as
supports for the plates,
caustic soda (NaOH), one part by weight

of soda to three of water. To prevent the soda being acted
upon by the air, the electrolyte is covered with a layer of
mineral oil. The copper oxide of the cathode gives up its
oxygen very readily to the hydrogen which forms on it, thus
preventing any substantial polarization. These cells are capable
of delivering a very heavy current. The electromotive force
is about 0.95 volt, and when delivering current the terminal
voltage drops to 0.75 volt. There is little or no local action in
this cell and it can therefore be used to advantage on both
open-circuit and closed-circuit work. Its chief disadvantage is
its low electromotive force.

88. Le Clanche Cell. — The Le Clanch^ cell is perhaps the most
familiar type of primary battery, because of its wide applica-

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Fig. 85. — E d i s o n -
. Lalande cell or Edison
The electrolyse is primary battery.

DIRECT CURRENTS

+ Terminal — Terminml

tion. The cathode is molded Carbon and the anode is amalga-
mated zinc. The electrolyte is sal ammoniac or ammonium
chloride. This type of cell is suited only for open circuit work
because of the rapidity with which it polarizes. The electro-
motive force is 1.4 volts, but because of the drop due to its in-
ternal resistance and that due to polarization, not over 1 volt
per cell should be allowed in planning an installation. The
most common method of reducing polarization is to bring manga-
nese dioxide into intimate contact with the carbon. This

gives up oxygen readily which unites
with the hydrogen bubbles to form
water.

In one type of Le Clanch^ cell a
pencil zinc is suspended in the center
of a hollow cylinder of carbon and
manganese dioxide. An improved
type, the porous cup cell, is shown
in Fig. 86. In this form a hollow
carbon cyUnder is filled with man-
ganese dioxide, arid the zinc, bent
into cyUndrical form, surrounds the
carbon cylinder, being separated
therefrom by rubber rings.

The solution should consist of 3
ounces of sal ammoniac to 1 pint of
water. A more concentrated solu-
tion produces zinc chloride crystals on the zinc and carbon. To
prevent the solution "creeping,'' the top of the cell is dipped in
paraffin and the top of the carbon is covered with a black wax.
This cell owes its wide use to its simplicity, to the small amount
of attention that it requires, and to the fact that it contains no
injurious acids or alkalis. Its uses are for intermittent work,
such as ringing door-bells, telephone work, and open-circuit
telegraph work.

89. Weston Standard Cell. — It is essential in practical work to
be able to reproduce accurately standards of current, voltage, and
resistance^ Obviously if two of the above quantities are known,
IbRe^ third is readily obtainable by Ohm's Law. It is a matter
of no great difficulty to make and reproduce resistance standards.

Fig. 86.-

-Porous cup Le Clanch6
cell.

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PRIMARY AND SECONDARY BATTERIES

as such standards are nothing more than metals in strips and in
other forms, carefully mounted and calibrated. Such standards
are very permanent and their resistance remains constant in-
definitely.

A standard of either current or voltage is much more diflScult
to reproduce and maintain than is the standard of resistance.
Of the two, it has been found more practicable to produce and
maintain a voltage standard rather than a currenJ^..ataadanL_
This voltage standard is obtained in a standard cell. fThe electro-
motive force of a cell depends upon its materials and their im-
purities, the concentration of the electrolyte, the temperature,
the polarization effects, etc.
It is difficult; therefore, to
select such materials for a
cell as will enable it to be
reproduced at different times
and at various places with a
high degree of accuracy. The
Clark cell was the first of
the standard cells to prove
commercially successful.
This had a cathode of mer-
cury, an anode of zinc, and
an electrolyte of mercurous
sulphate and zinc sulphate.
The objections to this cell
were that the electromotive force changed very appreciably with
the temperature and that this change lagged behind the change
in temperature.

In the Weston cell, cadmium is substituted for the zinc of the
Clark cell. A cross-section of the portable form of Weston cell
is shown in Fig. 87. The cathode is mercury located at the bot-
tom of one leg of an H-tube. Above this is mercurous sulphate
paste. These materials are held in position by means of a porce-
lain tube, expanded at the bottom and packed with asbestos.
This tube extends to the top of the cell and acts as a vent for any
gases that are formed. In the bottom of the other leg of the
H-tube is the anode, of cadmium amalgam. This is held in place
by another porcelain tube packed with asbestos. The electrolyte

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MrM^i] rouii
Sollphiila

51 K^rtTir}"

Fig. 87. — Weston standard cell.

94 DIRECT CURRENTS

is cadmium sulphate. The leads from the cathode and the anode
are sealed into the tubes at the bottom. The top of the cell is
sealed with cork, paraffin, and wax. The entire cell is mounted
in a wood and metal case with binding posts at the top.

The cell is made in two forms, the normal cell and the unsatu-
rated or secondary cell. In the normal cell, cadmium sulphate
crystals are left in the bottom of the solution so that it is always
saturated. Its electromotive force is affected slightly by tem-
perature, but corrections can be accurately made. It is possible
to reproduce such cells with electromotive forces differing by only
a few parts in 100,000.

In the unsaturated cell, the solution is saturated at 4*^ C.
and as no crystals are left in the solution, its concentration
is substantially constant at other temperatures. Such cells
have practically no temperature coefficient. They are not as
accurately reproducible as is the normal cell. A certificate
should accompany each one giving its electromotive force,
which usually is about 1.0186 volts. The unsaturated type of
cell rather than the normal cell is used almost entirely ib practical
work.

The terminal voltage of any cell differs from its electromotive
force by the IR drop due to the cell resistance. As the resistance
of a Weston cell is about 200 ohms, it is evident that if any ap-
preciable current be taken from the cell its terminal voltage will
be quite different from its electromotive force. The cell must be
used, therefore, in such a manner that it delivers no appreciable
current. By means of the so-called Poggendorf method , described
in par. 125, the cell is used without delivering current. Not
more than 0.0001 amp. should be taken from the cell at any
time. If appreciable current is taken, the electromotive force
drops, but when the circuit is again opened the electromotive
force slowly recovers its initial value.

90. Dry Cells. — Dry cells are a modification of the Le Clanch^
cell and as they are very light, portable, and convenient, they
are rapidly replacing other types of cells. The word ''dry cell"
is really a misnomer, for no cell that is dry will deliver any
appreciable current. In fact the chief cause of dry cells becoming
exhausted is their actually becoming dry.

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PRIMARY AND SECONDARY BATTERIES

Sealing
Compound

Blotting Paper

A cross-section of a typical dry cell is shown in Fig. 88. The
anode is sheet zinc, made in the form of a cylinder with an open
top, and acts as the container of the cell. The binding post is
soldered to the top of the zinc. The zinc is lined with some
non-conducting material such as blotting paper or plaster of paris.
The anode consists of a carbon rod, and the mixture of coke,
carbon, etc., which surrounds
this rod. The rod itself
varies in shape among various
manufacturers. It is located
axially in the zinc container
and the binding post is se-
cured to the top of it. The
depolarizing agent, powdered
manganese dioxide, is mixed
with finely crushed coke and
pressed solidly into the con-
tainer between the carbon
and the non-conducting ma-
terial which lines the zinc.
It fills the cell to within about
an inch of the top. Sal am-
moniac, with perhaps a little
zinc sulphate, is added and
the cell then sealed with wax
or some tar compound. The

outside of the zinc is frequently lacquered, and the cells are
always set in close-fitting cardboard containers.

The electromotive force of a dry cell is about 1.5 volts when
new but this drops to about 1.4 volts with time, even though the
cell remains idle. A cell is practically useless after a year to
18 months, even if not used at all. The internal resistance of
the cell is about 0.1 ohm when new and increases to several times
this value with time. The polarization effect is large as compared
with the internal resistance so that a low value of internal re-
sistance is not important except as an indication of the condition
of the cell. A method for testing the condition of a cell is to
short-circuit it through an ammeter, when it should deliver an
instantaneous value of 1.5/0.1 or 15 amp., if in good condition.

Fig. 88. — Sectional view — dry cell.

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96 DIRECT CURRENTS

When new, the current under these conditions may reach even
25 amp. When delivering appreciable current the terminal
voltage is very nearly 1 volt.

One of the chief causes of a cell's becoming useless is the using
up of the zinc as a result of electrochemical actions in the cell.
This allows the solution to leak out and to dry up and the cell
then becomes worthless. The life of a cell may be prolonged
temporarily by introducing fresh solution, but the results are
usually far from satisfactory.

As is well known, dry cells have many applications.. Their
field is limited to supplying moderate currents intermittently,
but they are capable of supplying very small currents of the
magnitude of 0.1 amp. continuously. They are used exten-
sively for door bells, electric bells, buzzers, telephones, telegraph
instruments, gas engine ignition, flash lamps, and for many other
purposes.

STORAGE BATTERIES

91. Storage Batteries. — A storage or secondary cell (sometimes
called an accumulator) involves the same principles as a primary
cell, but the two differ from each other in the manner in which
they are renewed. The materials of a primary cell which are
used up in the process of delivering current are replaced by new
materials, whereas, in the storage cell, the cell materials are re-
stored to their initial condition by sending a current through the
cell in a reverse direction. For this reason the electrochemical
products resulting from the discharge of such a cell must remain
within the cell. Therefore if a cell in its operation gives oflf
material, usually in the form of gases, so that it cannot be brought
back to its original condition with a reverse current, it is not
suitable for a storage cell. For example, the Le Clanch^ cell
gives off free ammonia gas and therefore cannot be used as a
storage cell. The Daniell and gravity cells are both reversible
and hence are theoretically capable of being used as storage cells ;
but as the active materials go into solution and do not all return
during the reverse cycle, the Ufe of such a cell would be limited.
There are but two forms of storage cells in common use, the
lead-lead-acid type and the nickel-iron-alkali type. In both of
these cells the active materials do not leave the electrodes.

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PRIMARY AND SECONDARY BATTERIES

92. The Lead Cell. — ^The principle underlying the lead cell
may be illustrated by the following simple experiment. Two
plain lead strips (Fig. 89) are immersed in a glass of dilute sul-
phuric acid. These are connected in series with an incandescent
lamp supplied from 115-volt direct current mains, or from a
battery. When current flows through this cell bubbles of gas

Negative
Plate

Positive
Plate

Fig. 89. — Forming the plates of an elementary lead storage cell.

will be given off from each plate, but it will be found that a much
greater number come from one plate than the other. After a
short time one plate will be observed to have changed to a dark
chocolate color, and the other apparently will not have changed
its appearance. A careful examination, however, will show that
the metalUc lead at the surface of the latter plate has started
to change from solid metallic lead to spongy lead.

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98 DIRECT CURRENTS

When the current is flowing as shown in Fig. 89 the voltmeter
connected across the cell* will indicate about 2.5 volts. If the
ciurent be interrupted by pulling the switch the voltmeter
reading will fall to about 2.1 volts, and the cell w^Ul now be
found to be capable of delivering a small current. This current
is of sufiicient magnitude to operate a small buzzer for a very-
short period, but the amount of energy that such a cell can deliver
is very limited; even the small current taken by the voltmeter
is sufficient to exhaust the cell in a very short time. As the cell
discharges the voltage drops off slowly to about 1.75 volts, after
which it drops more rapidly until it becomes zero and the cell is
apparently exhausted. The color of the dark brown plate will
now have become lighter and will more nearly resemble its
initial lead color. After a short rest the cell will recover slightly
and will again deliver current for a very brief period.

The plate which is a dark chocolate color in the above experi-
ment is the positive plate or cathode and the one which is par-
tially converted to spongy lead is the negative plate or anode.
The bubbles which were noted come mostly from the negative
plate and are free hydrogen gas. When the current is passed
through such a cell the metallic lead of the positive plate becomes
converted into lead peroxide, whereas the negative plate is not
changed chemically, but is converted from solid lead into the
spongy form which is softer and more porous than ordinary
metallic lead. When the cell is discharged the lead peroxide of
the positive plate is changed to lead sulphate and the spongy
lead of the negative plate becomes a sulphate so that they both
tend to become electrochemically equivalent.

The principle of the cell is the same as that of the primary
cell. When the two lead plates are the same electrochemically,
that is, when both are lead sulphate, no current flows. When the
positive is converted to the peroxide and the negative to spongy
lead by the action of an electric current, the two plates become
dissimilar and an electromotive force exists between them. This
electromotive force is about 2.1 volts, the excess of 0.4 volt
observed in charging the cell being necessary to overcome the
internal resistance and polarization effects. This simple ex-
periment illustrates the principle underlying the operation of
lead storage cells.

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PRIMARY AND SECONDARY BATTERIES 99

The chemical reactions which take place in a storage cell are
as follows:

Battery discharged Battery charged

(+ plate) (—plate) ^_ (4- plate) (—plate)

PbS04 + PbS04 -h 2H2O "^ PbOt 4- Pb + 2H2SO4

Lead sulphate + lead + water is changed Lead + lead + sulphuric

sulphate to peroxide acid

The above equation shows the changes that occur when the
battery is charged. The reverse takes place on discharge. It
will be noted that when the battery is being charged the only
change that takes place in the electrolyte is that water is con-
verted into sulphuric acid. This accounts for the rise of specific
gravity on charge. On discharge the sulphuric acid is dissociated ,
and reacts with the lead peroxide to form water. Therefore
the specific gravity of the electrolyte decreases when the cell is
discharging. When charging, free hydrogen is given oflf at the
negative plate and oxygen at the positive plate. Because of the
explosive nature of hydrogen, no flame should be allowed to come
in proximity to a storage battery, when it is charging.

It would not be practicable to construct storage cells of plain
lead sheets such as were used in this experiment. The current
capacity of the cell would be so small that the cell could not
deliver currents of commercial value for any length of time, unless
the cell were made prohibitively large in order to secure the
necessary plate area.

If the charging of the eiementary cell, Fig. 89, were carried
further, the dark lead peroxide of the positive plate would be
observed to fall oflf in flakes and drop to the bottom of the tum-
bler. Therefore in a commercial cell provision must be made to
minimize this flaking of the active material.

It was recognized very early that in order to make the storage
cell commercial, a large plate area must be exposed to the action
of the acid and a large amount of the lead must be converted into
the peroxide and so become active material. There are two
methods of obtaining this result, the Plants process and the Faure
process. In the Plants process the active material on the plates
is formed from the metallic lead by passing a current through the
cell first in one direction and then in the reverse direction, which
procedure works the lead on the surface of the plates into active

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100

DIRECT CURRENTS

material. This process is slow but may be accelerated by-
adding certain acids to the sulphuric acid during the forming
process. The Gould plate shown in Fig. 90 is made by this

process. The plate is first
passed under revolving steel
wheels which convert its
surface into ridges and fur-
rows, increasing the surface
area of the plate. As this
process weakens the plate
mechanically, certain por-
tions of it are not acted
upon by the wheels. These
portions act as ribs which
give support and mechanical
strength to the plate and
tend to prevent buckling.
The active material is then
formed electrically by the
Plants process. The negative plate is made from the positive
by reducing the peroxide to spongy lead by an electric
current.

Another type of Plants plate, the
Manchester type^ is shown in Fig, 91.
A grid made of lead and antiniony is
perforated. The active material con-
sists of a. corrugated lead ribbon, which
is coiled into spirals and pressed into
the perforations of the grid. The per-
oxide has a greater vol n mo tlmn the lead

Fig. 90.— Gould ploughed plate, Plants
process.

Fig. 91. — Plants (Manchester) positive group and button.

from which it is derived. Therefore when the cell is charged, these
spirals expand and become more firmly embedded in the plate. The
grid itself is not acted upon to any great extent, but serves as a

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PRIMARY AND SECONDARY BATTERIES

101

mechanical support. The advantage of this type of plate is its
rigidity and mechanical strength. It has less overload capacity
than other types and possibly the life is slightly less. The
ordinary Plants positive, if properly cared for, should be good
for from 1,800 to 2,400 complete cycles of charge and dis-
charge. The negative should have about 25 per cent, greater
life than this.

93. Faure or Pasted Plate. — This type of plate consists of a
lead-antimony lattice work or skeleton into which lead oxide is

Fig. 92. — Pasted positive and negative plates.

applied in the form of a paste. The battery is then charged.
The paste on the positive grid is converted into peroxide and
that on the negative grid into spongy lead. Two types of pasted
plates are shown in Fig. 92.

The chief advantage of the pasted plate is its high overload
capacity, especially for short periods, together with its lesser
size, cost, and weight for a given discharge rate. It is therefore
very useful where lightness and compactness are necessary, such
as in electrical vehicle batteries, ignition and starting batteries
for gasoline cars, etc. The pasted type of positive has a much
shorter life than the Plants type, due to a more rapid shedding
of the active material. This life is approximately one-fourth that

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102

DIRECT CURRENTS

of the Plants plates. Cells having a pasted plate for the negative
and a Plants positive are common.

In all batteries there is one more negative than positive plate.
This allows all the positives to be worked on both sides. Were
any of the positives to be worked on one side only, the expansion
of the active material, which occurs when it is converted to the
peroxide on charge, would be unequal on the two sides of the plate
and buckling would result.

^ n»irlitf« plAli

Wood vpficcr t«p4rator

^4«g*t■vf plaTc

Fig. 93. — Cut-away of an Iron-clad Exide cell.

'^Iron-dad Exide.^^ — In order to overcome the erosion of active
material from the positive plate the iron-clad exide has been
developed. Its positive consists of a lead-antimony frame which
supports a number of perforated hard rubber tubes. An irregu-
lar lead-antimony core passes through the center of each tube and
serves as a collecting device for the current. The peroxide is
pressed into the tubes, filUng the space between the core and the
inner wall of the tube. The perforations are so small that the
peroxide does not drop out readily. An ordinary pasted plate
j§ use(l for the negative plate of this cell. Although expensive,

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PRIMARY AND SECONDARY BATTERIES 103

this type of cell has a long life and can stand considerable rough
usage. It is used principally to operate electric vehicles. A view
of an iron-clad exide, cut away to show the assembly, is given in
Fig. 93.

Storage batteries are divided into two general classes, station-
ary batteries and portable batteries.

94. Stationary Batteries. — The plates of this tjrpe of battery
may be either of the Plants type or of the pasted type, depending
on the nature of the service. For merely regulating duty, involv-
ing only moderate, though continual, charging and discharging,
the Plants plate is preferable. Where a battery is installed for
emergency service, to carry an enormous overload for a very short
period during a temporary shut-down of the generating apparatus,
the Faure or pasted plate is preferable. For a given floor area
the pasted plate can discharge at the one-hour rate, twice the
current that the Plants plate can and at less than the one-hour
rate this ratio becomes greater. This is a very important factor
in congested city districts where such batteries are usually lo-
cated and where floor area is very valuable.

96. Tanks. — The containing tanks are of three general types :
glass, earthenware, and lead-lined wooden tanks. Glass jars
are used only for cells of small capacity, as they are expensive and
have not the requisite mechanical strength in the larger sizes.
Earthenware tanks have been used more as an experiment and will
probably not come into general use. The wooden tanks must
be strong, and well made. They are lined with sheet lead. The
seams of the lead lining must be sealed by burning the lead with
a non-oxidizing flame. Solder should never be used. The wood
should be painted with an acid-resisting paint, such as asphaltum.
An occasional application of linseed oil will prevent decomposition
due to the acid.

When glass jars are used, the plates are suspended by pro-
jecting lugs which rest on the edges of the jar. (See Fig. 100.)
In the lead-lined tanks, the plates are similarly suspended upon
two glass slabs, % in. thick, which rest on the bottom of the tank.
(See Fig. 94.) The plates of like polarity are burned to a heavy
lead strip or bus-bar to which the current-carrying lead is either
burned or bolted. There should always be a liberal space be-
tween the plates and the bottom of the tank to allow the red lead

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104

DIRECT CURRENTS

peroxide to accumulate without short-circuiting the plates. All
types of stationary batteries should have a glass cover to reduce
evaporation and to intercept the fine acid spray which occurs
during the charging period.

96. Separators. — To prevent the positive and negative plates
from coming in contact with one another, several types of sepa-
rators have been tried. Very thin perforated hard rubber is still
in use for small cells, but this is unsuitable for larger cells as the
limited area of the perforations offers too much resistance to the
passage of the current to the active material. Glass rods have

.L-UL

, jcnigi^i

I'.'^ Orftifl for

■■■'■J Hupponirnf

Separ^tui^

Fig. 94. — ^Lead-lined wooden tank
storage cell.

Fig. 95. — Assembly of a wooden
separator.

been suspended between the plates, but these are unsatisfactory
because there is still opportunity for bits of peroxide dropping
from the positive plate to lodge between the plates and cause a
short-circuit. Moreover, the rods are not a complete barrier be-
tween plates so that the expansion of the active material on either
the positive or the negative plate may cause a short circuit. The
most satisfactory separators are made of wood. These are very
thin and are grooved vertically to permit the circulation of the
electrolyte. They are specially treated to remove ingredients
that would be detrimental to the electrolyte. The wood, after
being treated, is not attacked by the acid. These separators
should never be allowed to become dry, as they then decompose
very readily. After being received, they should be kept wet

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PRIMARY AND SECONDARY BATTERIES

105

until installed. In larger sizes of batteries the separators are
held in place by dowel pins. (See Fig. 95.)

97. Electrolyte. — The electrolyte should be chemically pure
sulphuric acid. When fully charged the specific gravity should
be 1.210 for Plants plates and not higher than 1.300 for pasted
plates. This solution may be made from concentrated acid
(oil of vitriol sp. gr. 1.84) by pouring the add into water in the fol-
lowing ratios:

Parts Water to One Part Acid

Specific gravity Volume

1.200 4.3

1.210 4.0

1.240 3.4

1.280 2.75

Fig. 96. — Measurement of specific gravity in a stationary battery.

A large amount of heat is evolved when acid and water are
mixed. This results in a large amount of steam being generated
if the water is added to the acid. This should be avoided as it
may scatter the acid, break the container and even cause per-
sonal injury.

The specific gravity of a solution may be determined directly
by the use of a hydrometer. This consists of a weighted bulb
and a graduated tube which floats in the liquid as shown in Fig.
96. The bulb floats in the liquid whose specific gravity is to be

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106

DIRECT CURRENTS

measured, and the specific gravity is read at the point where the
surface of the liquid intercepts the tube. Such a tube may be
left floating permanently in stationary batteries in a represen-
tative cell called a pilot cell (Fig. 96).

The small amount of Uquid and the inaccessibility of vehicle
and starting batteries make the use of such a hydrometer im-
possible. To determine the specific gravity with such batteries,

Fig. 97. — Syringe hydrometer.

the syringe hydrometer shown in Fig. 97 is used. The syringe
contains a small hydrometer and when sufficient liquid is drawn
into the syringe tube, the small hydrometer floats and may be
read directly.

Fig. 98 shows the change in specific gravity during charge
and discharge. This relation is very important, as the specific

0 12 3 4„^^^^6 6 7 8

FiQ. 98. — Variation of specific gravity in a stationary battery.

gravity of the electrolyte is an accurate indication of the con-
dition of charge of the battery.

98. Specific Gravity. — When the battery is charged, hydrogen
is given off at the negative plate and oxygen is given to the posi-
tive plate to convert it into the peroxide. The electrolyte
gives up water, which means that the solution becomes more and
more concentrated. The specific gravity will rise from the
complete discharge value of 1.160 to 1.210 when fully charged as

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PRIMARY AND SECONDARY BATTERIES 107

shown in Fig. 98. Point a is called the gassing point because
it is the point at which hydrogen gas is given off rapidly. Here
the specific gravity drops off slightly due to the presence of the
hydrogen bubbles in the electrolyte. After the charging has
ceased the specific gravity continues to rise for some time. This
is due to the very concentrated acid in the pores of the active
material workinjg out into the solution and also to the fact that
the hydrogen bubbles have escaped from the solution. The dis-
charge curve shown in Fig. 98 is very similar to the charge
curve. The specific gravity will be found to drop even after
the battery has ceased to deliver current. This is now due to
the dilute acid in the pores of the active material passing out
into the solution. The specific gravity is such a good indicator
of the state of charge of the battery that the hydrometer reading
is generally used to determine how nearly charged or discharged
the battery may be.

As the hydrogen and the oxygen gas which escape from the
battery during the charging and the discharging periods are only
dissociated water, the battery loses nothing but the equivalent of
water. Ordinarily, therefore, nothing but water need be added
to replace the electrolyte. A small amount of the acid is carried
away as a spray by the gas bubbles, but this loss is rarely of
appreciable magnitude. Acid need only be added when an
actual loss of electrolyte takes place, such as occurs with a leaky
tank. Distilled water is used, as a rule, to replace the evapora-
tion of the electrolyte. If any doubt exists as to the suitability
of local water, the battery companies upon receipt of a sample
will analyze the water and report upon the matter without
charge.

99. Installing and Removing from Service. — The plates, tanks,
electrolyte and containers of stationary batteries are packed
separately when shipped. When received the separators should
be placed immediately where they may be kept wet. The jars
should be set in sand trays which rest on insulators as shown in
Fig. 100. The plates should be handled carefully and placed
in the jars in the manner shown in Fig. 99. The separators
should be carefully slid into position as shown in Fig. 100. As
the active material on the plates is more or less converted into
lead salts during exposure to the atmosphere, these salts must

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108

DIRECT CURRENTS

be reduced electrically before the battery is ready for service.
Therefore the battery should be given an initial charge at the
normal charging rate for about 40 hours or more.

If the battery stands over a long period without being used,
the active material becomes more or less converted into inactive
lead sulphate, which is a non-conductor, and so is difficult to
reduce electrically. Therefore a battery if idle should be charged
occasionally. If the battery is to remain idle over a long period

M-

Vli

■

t:
>

1
t

i^^-Ii^ — '^''2^

I^Wi- ' 1 M N ifl

Fig.

99 . — Lowering
into position.

plates

Fig. 100. — Stationary battery in position.

and it is impracticable to charge it periodically the following pro-
cedure is necessary to prevent sulphation. Give the battery a
full charge, then siphon off the electrolyte, which may be saved
and again used. Fill the cells with water and allow them to
stand 12 to 15 hours. Siphon off the water and the cells T'vdll
stand indefinitely without injury to the plates. To put back in
service, fill the battery with the electrolyte having a specific
gravity of 1.210 and charge for 35 hours or more at the normal
rate or its equivalent.

100. Vehicle Batteries. — In the design of batteries for pro-
pelling vehicles and for automobile starting it is necessary to

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PRIMARY AND SECONDARY BATTERIES

109

obtain a very high discharge rate with minimum weight and
size. Therefore pasted plates are used for both positives and
negatives. These are made extremely thin and are insulated
from one another by very thin wooden separators. They are
then packed tightly into a hard rubber jar as shown in Fig. 101.
This jar is sealed in with an asphaltum compound to prevent the
liquid splashing out. There is a hole in the top of the jar

"Tilhr sliJi) connectors'* or "oell connectors"
When used to connect cells When used to connect cells

placed side by side is placed end to end is called

called a "side connector"" an 'end conn>-ctor'"

Soft lubbet r»f««

Sliap used to connect pLiies of a groui)

Hold-down used in keeji wood separators
fioro tloaiing

Pobiiive plate of a dark brown colur

Perforated lubher setiarator placed neJtt

to posiiive plate
(wcKivi-d wiH)d separator placed with

sniouth side next to negative plate

Negative plate of a tiray oi slate color

Hard robber jar

The positive i>late«, when burned to the
strap, as shown, are called the "posi-
tive eroup"

The negative plates, when burned lo
Ihc strap, as shown, are called (he
"negative group '

Both eroups and seitarators, assembled at>
shown, ate called the "element"

Rib or btidee for s'uppotting the element

Sediment »pjce

Fig. 101. — Assembly of an Exide veliicle cell.

which is closed with a cap. This permits the replenishing of the
electrolyte and a vent in the cap allows the gases to escape.
Because of the high discharge rates which occur where this type
of battery starts a gasoline engine, and because of the necessity
for a high ampere capacity for the weight, the specific gravity
of the electrolyte is as high as 1.280 and 1.300. Further, the
amount of electrolyte is very small and therefore it is necessary
to work it between wide limits, the lower limit being 1.185 and
the upper 1.280 and 1.300. . j

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110 DIRECT CURRENTS

The individual cells are mounted beside one another in boxes
or crates and are connected together on top by lead connectors
which may be burned or held by lead nuts. The number of cells
in such a unit depends upon the voltage which is desired.

Vehicle batteries are usually shipped assembled, charged and
complete with the electrolyte so that they are ready for use when
received. However, a preliminary charge is advisable.

Because of its ruggedness, the ''Iron-clad Exide" (see par. 93)
is used to a large extent in electric vehicles.

As the space for the electrolyte is very limited in vehicle
batteries, the level of the electrolyte falls quite rapidly, so that
frequent additions of water are necessary.

101. Rating of Batteries. — Practically all batteries have a
nominal rating based on the 8-hour rate of discharge. Thus, if
a Plants battery can deliver a current of 40 amp. continuously
for 8 hom-s, the battery will have a rating of 40 X 8 = 320
ampere-hours. The normal charging rate of such a battery would
be 40 amp. Although the above battery is just capable of
delivering 40 amp. for 8 hours, it would not be able to deliver
64 amp. for 5 hours ( = 320 ampere hours) but only 88 per cent,
of this or 56.4 amp. for 5 hours. 56.4 amp. is called the 5-hour
rate.

Below is given a table showing the percentage capacity with
various discharge rates.

Discharge rate, hours 8 5 3 1

Percentage of capacity at 8-hour rate :

Plants type 100 88 75 50

Pasted type 100 93 83 60

This falling off in capacity with higher rates of discharge is
due to the inability of the free solution to penetrate the pores
of the active material. Consequently it is not possible to reduce
all the active material during the short periods of discharge.
After such a battery has stood a short time it will be found to
have recovered to some extent and is therefore capable of deliver-
ing more current, after apparently having become exhausted.
This is due to the free solution finally penetrating the pores of
the active material.

Batteries are able to discharge at enormous rates for very short
intervals. For instance, a starting battery having an 8-hour

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PRIMARY AND SECONDARY BATTERIES

111

rating of 10 amp. is often called upon to supply 450 amp. when
doing starting duty.

102. Charging. — ^There are two general methods of charging
a battery, the constant current method and the constant poten-
tial method. In the constant current method the current is
kept at its nominal 5-hour or 8-hour value until the gassing period
begins. (See Fig. 98.) If the plates are of the pasted type
the current should be reduced about one-half when gassing
beginS; for gassing represents a waste of energy because a con-
siderable portion of the charging energy is used in merely break-

Pio. 102. — Charging a starting battery from 110-volt mains.

ing up the water into hydrogen and oxygen. In addition, gassing
causes the battery to become heated, the acid is carried out in
a fine spray by the bubbles and active material may be carried
from the plates by the mechanical agitation of the bubbles.

The charging rate with Plants plates is much in excess of the
above. The charge may be started at the 3-hour rate and ended
at not less than the 8-hour rate.

A common example of the constant current rate is the charg-
ing of low voltage batteries from 110-volt mains. This is illus-
trated by Fig. 102, which shows the charging of a 6-volt starting
battery. It should be definitely determined that the mains sup-
ply direct current and it is also necessary to know which main is
positive. If doubt exists as to the polarity and a voltmeter is
not available, dip the two ends of the wires which connect the
mains to the battery into a glass of slightly acidulated water or in
salt water. Bubbles form about the negative wire. When using

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112

DIRECT CURRENTS

the constant current method of charging one must reduce the
charging rate as the battery approaches the fully charged
condition.

The constant potential method of charging is to be preferred as
the charging current automatically tapers off due to the rise in
the cell electromotive force as the cell approaches the charged
condition. The source of potential should be about 2.3 volts
per cell when there is no series resistance in the circuit.

When a battery floats on constant potential bus-bars, ready to
take load as occasion demands, it is necessary to have a series
booster for raising the charging potential to a value sufficiently

Fig. 103. — Booster method of chaiging a storage battery.

high to force current into the battery. The booster ordinarily
consists of a low voltage, separately excited shunt generator,
driven by a shunt motor. Fig. 103 shows the connections of
the set when the battery is being charged. The booster raises
the voltage just enough to send the necessary current into the
battery.

As an example, consider a 110-volt installation with a floating
battery. As the average cell voltage is about 2 volts, 55 cells are
necessary. Assume that the battery has a 320-ampere-hour
rating. The charging current will be 320/8 or 40 amp. (the
nominal 8-hour rating). The voltage of each cell should be
boosted to 2.3 volts on charge. Therefore the total voltage
necessary will be 2.3 X 55 = 126.5 volts. Of this 126.5 volts,
the bus-bars can supply 110 volts. The booster supplies the
remaining 16.5 volts and its rating will be
16^X40
1,000

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= 0.66 kw.

PRIMARY AND SECONDARY BATTERIES

113

The total power utilized in charging the battery is, however,
126.5 X 40

1,000

= 5.06 kw.

The terminal voltage of a cell rises on being charged, as
is shown in Fig. 104. The terminal voltage is about 2 volts
at the beginning of charge and rises slowly to about 2.4 volts,
after which it rises very rapidly to 2.6 volts. This last rise occurs
in the gassing period. This final rise of voltage also indicates

2.7
26

QaBsii

«-.

2.4
23

\^*^

i>^

5
"e 2-1

""■^

2.0

1 Q

Ki,

scha

■ge

1.8
1.7
16

aalV<

Itage

1.6

012S466789 10
Hours

Fig. 104. — Voltage curves on charge and discharge for lead cell.

that the cell is nearing the completion of charge. It is this rise
of voltage which automatically cuts down the charging rate when
the constant potential method is used. The voltage does not rise
so rapidly when the charging rate is reduced toward the end of
charge, because of the lesser IR drop in the cell itself. The
drop of voltage at various rates of discharge is shown in Fig. 104.
It will be noted that the battery voltage curve at the &-hour
discharge rate is fairly flat, which is a very distinct advantage
if the battery is used to supply incandescent lamps.

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114 DIRECT CURRENTS

103. Battery Installations. — Batteries should be installed in
dry, well-ventilated rooms. Small glass jars may be mounted on
wooden racks painted with asphaltum paint. (See Fig. 100.)
The jars are set in glass trays containing sand, which are in turn
set on glass insulators. The. larger battery jars should be set
on porcelain pedestals 6 in. or so above the floor. The floor
should be of acid-resisting tile or vitrified brick. All wooden
surfaces should be covered with asphaltum paint. The room
should be well ventilated, as the spray which is carried out
of the jars on charge settles on horizontal surfaces and attracts
other moisture. Therefore it is desirable to have a stream of
air sweeping along the floor. As hydrogen gas is given off, no
flame should be allowed in the room and no switches should be
installed in the room. In addition to the danger due to the
arcing at the switch contacts, the acid in the air will corrode
the copper.

104. Temperature. — Below is given the relation between the
freezing point of the electrolyte and its specific gravity. It will
be noted that the freezing point is very considerably reduced
with increasing values of the specific gravity, so that if a battery
is well charged there is no danger of freezing in the temperate
zone.

Specific gravity Freezing temp. F.

1.180 - 6**

1.200 - 16°

1.240 - 5r

1.280 - 90**

At the higher temperatures the rate of diffusion of the acid
throughout the pores of the active material is increased so that
the rating of a battery increases very appreciably with increasing
temperature. Above 70° this increase is of the order of from
0.5 to 1.0 per cent, per degree Fahrenheit.

106. Capacities and Weights of Lead Cells. — Below are given
the relations of weights to kilowatt capacity for the various types
of cells which have just been described.

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PRIMARY AND SECONDARY BATTERIES

115

Kilowatt Capacity (as Related to Weight) op Representativb Cells

Manufactured by

The Electric Storage Battery Co.

Size of plates H G MV

No. of plates per cell 81 41 13

Type of plates "Exide'' "Chloride "Iron-clad"

Accumulator"

Weight of plates in cell, lb 1,988 896 26 . 8

*Weight of cell, lb 3,790 1,841 40.0

Kilowatts — ^per cell:

Fori hour 10.98 2.82 0.237

For 4 hours 4.16 1.22 0.0904

For 8 hours 2.42 0.764 0.0533

ICilo watts — per lb. of plates:

Fori hour 0.00552 0.00314 0.00883

For 4 hours 0.00209 0.00136 0.00337

For 8 hours 0.00122 0.000853 0.00199

Kilowatts — ^per lb. of cell:

Fori hour 0.0029 0.00152 0.00592

For 4 hours 0.001096 0.000663 0.00225

For 8 hours 0.000638 0.000379 0.00138

*The necessary insulating supports for the "H" and "G" cells and
tray for the "MV" cell are not included — ^these add approximately 2 per
cent, and 10 per cent, respectively to the cell weights.

Type H "Exide" batteries used in central station stand-by service.
Largest battery 150 cells, 169 plates per cell, capacity 3,460 kw. for 1 hour;
two other batteries 3,420 kw. each.

T3rpe G "Chloride Accumulator" used in power plants for peak, regulat-
ing and exciter bus service, also in telephone exchanges, large isolated
plants, etc. Largest battery 288 cells, 85 plates per cell, capacity 1,700
kw. for 1 hour; two other batteries same size and capacity.

Type MV "Iron-clad Exide" used in electric vehicles, locomotives,
industrial trucks and tractors, and for yacht lighting, etc.

THE EDISON BATTERY

106. The Nickel-iron-alkaline Battery. — Instead of using
acid as an electrolyte, the Edison cell uses an alkali, consisting
of a 21 per cent, potassium hydrate solution. The positive
plate consists of nickel pencils about 3^ in. diameter and 43^^
in. long, filled with green nickel oxide. As the nickel oxide is. a
very poor electrical conductor, very fine metallic nickel flakes
are mixed with it to produce sufficient conductivity. The
negative plate consists of flat perforated nickel-plated steel

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116

DIRECT CURRENTS

stampings, containing iron in a very finely divided form. These
flat pockets are mounted on a nickel-plated steel frame for sup-
port. Both the positive and the negative plates are shown in
Fig. 105.

The chemical reaction in the cell is complex, but its nature
is indicated by the following chemical equation:

Positive Plate Negative Plate

8K0H + GNiOa + 3Fe = Fe304 + 2Ni304 + 8K0H

The above read from left to right indicates discharge, and read
from right to left indicates charge. It is to be noted in the above

Fig. 105. — Positive and negative
plates of an Edison storage cell.

Fig. 106. — Assembly, Edison battery
plates removed from container.

reaction that the same quantity of potassium hydrate solution
(KOH) appears on both sides of the equation. This indicates
that ultimately all the reaction occurs between the electrodes
themselves, and also that no water is formed. Therefore the
specific gravity of the solution does not change during charge
or discharge.

The plates all have a perforated lug by which they are fastened
together with a steel bolt and to a binding post. The bolt is
threaded and steel nuts clamp the plates together. Steel washers
between the plates act as spacers. The positive and negative
plates are insulated from one another by hard rubber grids.
An Edison cell assembly is shown in Fig. 106. The positive
and negative assembly is placed in a corrugated, nickel-plated.

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PRIMARY AND SECONDARY BATTERIES

117

welded steel tank. The top is then welded to the rest of the
container. The binding posts are insulated from the cover by
hard rubber bushings. In the top is a valve which allows the
gases to escape during charging and through which water may be

Fig. 107. — Five Edison storage cells mounted in a tray.

added to the electrolyte. This valve should never be allowed
to become so encrusted with a potash deposit that it sticks,
because the internal pressure may become suflScient to cause th^
sides of the container to bulge.

The individual cells are usually mounted in wooden racks,
as shown in Fig. 107, the cells being connected together by
nickel-plated steel connectors.

2.00

—

1.80

(

2Ij

Irg

— •

■^

^ 1 fiO

■^

—

s ^-w

K 1*40
o 1.20

■""

■^

—

1 00

0.80

12 3 4 5 6

Hours Charge or Discharge at Normal Rate

Fig. 108. — Voltage changes during the charge and discharge of an Edison cell.

107. Charging and Discharging. — The Edison cell is rated on
the basis of a 5-hour charging rate. Fig. 108 shows typical
charge and discharge curves for the Edison battery. It will be

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118 DIRECT CURRENTS

noted that the average voltage on discharge is about 1.2 volts
per cell. The specific gravity of the electrolyte changes but
slightly so that it cannot be used to indicate the condition of
charge^ as with the lead cell. Moreover, there is no sharp voltage
rise near the completion of charge. If doubt exists as to the
condition of charge, it is advisable to give an overcharge in order
to be on the safe side. The overcharge does not injure the cell
although it may slightly reduce the efficiency-

The electrolyte in an Edison cell evaporates rapidly and
frequent additions of water are necessary. As the electrolyte
is changed to potassium carbonate very readily, only freshly
distilled water should be used in replacing the electrolyte, as
tap water usually contains carbonates in solution. In spite of
the usual precautions, the electrolyte is slowly converted into
potassium carbonate by contact with the air, and it should be
replaced by fresh electrolyte every 250 complete cycles of charge
and discharge.

The Edison cell has many advantages. It is light, rugged,
and can stand for a long time in a discharged condition without
chemical deterioration. The plates do not buckle and the active
material does not ''flake" or drop from the plates.

108^ Applications. — Edison cells are used for vehicle lighting
and ignition, and are also much used in motor boats.
They are also used in various types of electric trucks and for
battery street cars. In automobiles they are not generally
used for starting, as their comparatively large internal resistance
does not permit a sufficiently high discharge rate.

Below is given the relation between the battery weight and
capacity.

The figures are based upon the capacity obtainable on normal
charge :

Discharge rate,
hours

Watt-hours per
pound of cell

Watt-hours per
pound of plates

9.75

18.35

14.95

28.15

16.33

30.80

109. Efficiency of Storage Batteries. — The efficiency of a
storage battery is the ratio of the watt-hour output to the watt-
hour input.

For example, a normally discharged cell is charged at a uni-

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PRIMARY AND SECONDARY BATTERIES 119

form rate of 40 amp. for 6 horn's at an average voltage of 2.3 volts.
The cell Is then completely discharged at a uniform rate of 38
amp. for 6 hours, the average voltage being 1.95 volts. What is
the efficiency of this cell?

Watt-hours output = 38 X 1.95 X 6 =445
Watt-hours input = 40 X 2.3 X 6 = 552

445
Efficiency = ^^ or 80.7 per cent.

One often hears of the ampere-nour efficiency of a storage
Dattery. As amperes do not represent energy, the ampere-hour
efficiency is not a measure of a battery's ability to store energy.
In the above example the ampere-hour efficiency may be found
as follows:

Ampere-hours output «= 38 X 6 = 228
Ampere-hours input »= 40 X 6 = 240

Ampere-hour efficiency = ^^ or 95 per. cent.

The much lower watt-hour efficiency is due to the great
difference between the voltage of charge and that of discharge,
as shown in Figs. 104 and 108.

The efficiency of a storage battery varies with the rate, both
of charge and discharge, and somewhat with the temperature.
As high charge and discharge rates produce relatively high PR
and polarization losses, the efficiency is lowered under these
conditions. Further, a cell may be charged at the 8-hour rate
and discharged at the 3-hour rate and have an apparent efficiency
of 60 per cent. This does not represent the true efficiency as
the cell actually will not be completely discharged, even though
it appears to be. Owing to the inability of the free acid to
permeate the active material, much of the active material has not
been reduced, and after a short time the cell will be found to have
recuperated to a considerable extent and to be able to deUver
more energy.

The ampere-hour efficiency of a storage battery is of the order
of magnitude of 95 per cent. For a complete cycle the watt-
hour efficiency of a stationary battery of moderate size is about
80 per cent, at the 8-hour charge and discharge rates. The watt-

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120

DIRECT CURRENTS

hour efficiency of a large stationary battery is about 86 per cent,
under the same conditions. Where a battery merely ''floats''
and the cycle of charge and discharge is a matter of minutes
or perhaps of seconds even, the watt-hour efficiency may be as
high as 95 or 96 per cent.

The ampere-hour and the watt-hour efficiency for the Edison
cell are less than for the lead cell. This is due partly to the fact
that the Edison cell has a lower electromotive force and the IR
drop is proportionately greater. For the Edison cell the ampere-
hour efficiency is about 82 per cent, and the watt-hour efficiency
about 60 per cent.

In selecting a battery, the efficiency is but one of the factors to
be considered. The first costs and the maintenance of batteries
are high so that these factors, as well as the efficiency, should be
given due consideration.

Note, — ^The uses of storage batteries in the generation and
distribution of power are considered in Chap. XIV.

-aaM^

.Copper
Plated

Fig. 109. — Copper plating bath.

110. Electroplating. 1 — Electroplating is a very important
electrical industry and is closely related to the subject of batteries.
The principle is very simple. Assume that it is desired to copper
plate a carbon dynamo brush. The portions of the brush to
be plated are immersed in a solution of copper sulphate as shown
in Fig. 109. A copper strip is also immersed in the solution and
is connected to the + terminal of a dynamo or some other
source of direct current supply. The article to be plated is

iSee "Standard Handbook," Fourth Edition, Section 19, Pars. 186 to
206, for a more complete discussion.

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PRIMARY AND SECONDARY BATTERIES 121

connected to the negative terminal of this supply. Under
these conditions the current will carry copper from the solution
and deposit it on the carbon brush. This copper which leaves
the solution is replaced by copper which is carried from the
copper strip (the anode) into the solution so that there is no
change in the solution itself. The current should be such that
the density is about 0.02 amp. per sq. in. of the surface to be
plated.

It is not necessary that the anode be of the metal which it is
desired to deposit. Other metals may be used. Under these
conditions, however, the solution in time becomes contaminated
by the going into solution of the anode. If an inert substance
such as carbon is used, as anode, acid is formed in the solution.

The only opposing electromotive force in the bath just de-
scribed is the IR drop in the solution. This may be reduced by
bringing the electrodes close together, but if the electrodes are too
close together the deposit will not be uniform. The amount of
metal deposited per second is proportional to the current. Because
of the nature of electroplating baths, they are naturally low
voltage devices. When practicable, several are connected in
series. A low voltage and high current generator is generally
used for plating purposes. In practice there are many refine-
ments to be observed.

Acid is added to the solution to prevent impurities from de-
positing. A cyanide solution of copper is found to give better
results than the sulphate. Nickel, tin, zinc, silver, gold, etc.,
may be deposited by the use of suitable baths and electrodes.

A gravity cell is an example of electroplating in which the source
of current is derived from the bath itself. The current flows
from the zinc to the copper within the solution, zinc is carried
into the solution as sulphate and copper is deposited or plated
from its sulphate on the positive electrode.

Electrotyping is another common example of electroplating.
An impression is made in wax with the type or object to be re-
produced. The surface of the wax is made conducting by ap-
plying a thin coat of graphite. Copper is then plated on this
surface. It is later backed by type metal to give it the neces-
sary mechanical strength.

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CHAPTER VII

ELECTRICAL INSTRUMENTS AND ELECTRICAL
MEASUREMENTS

111. Principle of Direct-current Instruments. — If a coil like
that shown in Fig. 110 carries a current, a magnetic field results
(Chap. II) with a north and a south pole at opposite ends of the
coil. If the coil carrying current be placed in a magentic field,
the coil will tend to turn in such a direction that:

The resulting magnetic field due to both the main field and that
of the coil will be a maximum (see Par. 17, Chap. I), and the north

Pig. 110. — Magnetic field produced by an instrument coil.

pole of the coil will be attracted toward the south pole of the mag-
netic field and the south pole of the coil will be attracted to the
north pole of the magnetic field.

This tendency of the coil to turn is shown in Fig. Ill (a) where
the coil attempts to turn in the direction indicated by the arrows.
If the coil is pivoted and free to turn it will reach the position
shown in Fig. Ill (6). Under these conditions the coil has
placed itself in such a position that its flux is acting in the same

122

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 123

direction as that of the main field. Also the unlike poles are as
near each other as possible and the like poles are as far away
from each other as possible.

This behavior of a coil canying a current and placed in a mag-
netic field should be thoroughly understood, for it is the under-
lying principle of most current measuring instruments and is
in addition the principle upon which all electric motors operate.

Current Up ®

Current Down ®

(cO Coil tending to turn in a magnetic field

Current Up ®

Current Down -^

(b) Ultimate position of coil
Fig. 111. — Turning moment of an instrument coil.

112. The D'Arsonval Galvanometer. — ^A galvanometer is a
sensitive instrument used for measuring and detecting small
electric currents. The D'Arsonval galvanometer, which is
based on the principle of a coil turning in a magnetic field, is the
most common type of galvanometer. Due to its simplicity it
has superseded practically all other types. In addition it is
comparatively rugged and is not appreciably affected by stray
magnetic fields. Fig. 112 shows the principle of its construc-
tion. A coil of very fine wire is suspended between the poles of
a permanent magnet by means of a filament, usually a flat strip

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124

DIRECT CURRENTS

of phosphor-bronze. The coil may be wound with or without
a bobbin. The bobbin is usually of fiber, or of aluminum. The
advantage of an aluminum bobbin will be considered later.

Between the poles of a magnet a soft iron core is usually placed
(Fig. 112 and Fig. 113). The addition of this core results in two

distinct advantages. The
length of the air path is re-
duced so that the amount of
flux linking the coil is in-
creased, thus making the
galvanometer more sensitive;
the flux tends to enter the
core radially. This last effect
makes the deflections of the
galvanometer almost directly
proportional to the current
flowing in the galvanometer
coil.

The coil is usually sus-
pended by a phosphor-bronze
filament. Any turning of the
coil produces torsion in the
filament which opposes the
turning of the coil and is
called the restoring force.
When the moment of the re-
storing force and the turning
moment due to the current
are equal, the galvanometer
assumes a steady deflection.
For all practical purposes
the galvanometer deflection
is proportional to the current,
usually serves as one of the
The other leading

FiQ. 112.-

-Principle of the D*Arsonval
galvanometer.

This phosphor-bronze filament

leading in wires carrjdng current to the coil

in wire consists of a very flexible spiral filament fastened to the

bottom of the coil, as shown in Fig. 112.

There are two common methods of reading the deflection
of a galvanometer. A plane mirror is mounted on the coil system

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 125

and a scale and telescope are mounted about 3^ m. from
the galvanometer. The reflection of the scale in the mirror
can be seen with the telescope (Fig. 114). When the mirror
turns, the reflection of the scale in the mirror deflects. The

Coil

Core

Fig. 113. — Effect of core upon the magnetic field of a galvanometer.

value of this deflection is determined by means of a cross hair
in the telescope.

Another method is to use a concave mirror on the galvanom-
eter moving system. A lamp filament is placed some distance
from the mirror and its image focused on a ground glass to

FiQ. 114. — Telescope and scale method of reading a galvanometer.

the

which a scale graduated in centimeters is fastened. As
mirror deflects, the beam of light travels across the scale.

Damping, — If a galvanometer coil, which is hung freely,
starts to swing, it will continue swinging for some time unless it
is in some way retarded or damped. One method of damping
is to attach an air vane to the coil. This air vane is enclosed so

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126

DIRECT CURRENTS

that it swings in a restricted space and damps any swinging
movement of the coil. The most satisfactory method is elec-
trical damping. If the coil be wound on an aluminum bobbin,
the motion of the bobbin through the magnetic field will induce
currents within itself, and these will be in such a direction as to
put an electric load on the moving coil as in an electric generator.
This opposes the motion of the coil. The same result may be
obtained by binding short-circuited copper coils on the main coil,
by shunting the galvanometer externally with a resistance (see
Ayrton Shunt), or even by short-circuiting it.

113. Galvanometer Shunts. — When galvanometers are used
to detect small currents as in null methods (see Wheatstone

<2>

^^ [JJvvvWWVWAAA/VNA^

(a) GAlvanometer shunt (&) Ayrton shunt

FiQ. 115. — Types of galvanometer shunt.

Bridge), the apparatus may be so far out of adjustment that a
comparatively large current flows through the galvanometer.
This causes a violent deflection of the coil, and may result in
injury to the galvanometer. In certain other measurements
the current that it is desired to measure with the galvanometer
may be so large that the deflection is considerably beyond the
scale.

In either case the galvanometer may be protected by the
use of a shunt, or a resistance which by-passes a certain known
proportion of the current from the galvanometer. There are
two common types of shunt. One type is shown in Fig. 115 (a).

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 127

It consists of three or four separate resistances which are plugged
across the galvanometer one at a time. These are so adjusted
in value that with a given current to be measured the successive
-galvanometer currents are in the ratio of 10 to 1. For in-
stance, if the galvanometer is to measure Jfo the external cur-
rent, the top resistance, Fig. 115 (a), is of such a value that it
shunts ^f 0 of the current away from the galvanometer when it
is plugged across the galvanometer, etc.

To determine the values of these resistances proceed as follows:

Let Rg = galvanometer resistance.

Ig = galvanometer current for full scale deflection.
/ = circuit current.
/, == shunt current.
Rt = shunt resistance.

To reduce the galvanometer deflection to one-tenth the value
which it would have if all of the current / passed through the
galvanometer, Ig must be one-tenth of /. That is,

T = ro (1)

The shunt current

7. = 7-7. (2)

But the shunt current and the galvanometer current are in-
versely as their respective resistances. Hence:

^g la J- J-0 /o\

R,~

' /. ~ la

= ^ from(l)

R,_
R.~

" 7/10 ~ ^

R.=

■^«.

For a reduction of 100 to 1

R, _
R.

7/100 ^

R. =

99^'

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128 DIRECT CURRENTS

Example, — ^A galvanometer has a resistance of 600 ohms. What re-
sistances should shunt it in order to reduce its deflections in the ratio of
10 to 1 and 100 to 1 ?

Ri = -^ = 66.7 ohms. Ans.
R2 = -QQ- = 6.06 ohms. Ans.

Ayrton Shunt — The Ayrton shunt is shown in Fig. 115 (6).
A permanent resistance A Bis connected across the galvanometer
terminals. One line terminal is permanently connected to one
end of this resistance, and the other line terminal, C, is movable
and* can be connected to various points along AB. With a fixed
line current the maximum deflection is obtained when C is at B.
If point C be moved to a, where resistance Aa is Hooo *^® total
resistance AB, the galvanometer deflection will be Hooo of i^s
maximum value. If C be moved to 6, where A6 is Koo of the
resistance AB, the galvanometer deflection will be Jfoo of its
maximum value, etc.

The advantages of the Ayrton shunt are :

(1) A shunt is applicable to any galvanometer, regardless of
the galvanometer resistance.

(2) A fixed resistance is shunted across the galvanometer,
which gives a constant value of damping in open circuit ballistic
measurements. (See par. 159.) When the shimt is adjusted to
give the maximum galvanometer deflection, this deflection for
the same value of external current is less than it would be were
the shunt not used. That is, the maximum sensitivity of the
galvanometer is reduced by the addition of the shunt. If the
shimt has a resistance of only 5 times that of the galvanometer
the sensitivity will be reduced only in the ratio of 6 to 5, which
is not usually objectionable.

114. Ammeters. — ^An ammeter is an electricalinstrument which
measures the current flowing in an electric circuit.

There were many early types of ammeters, most of which
depended for their operation upon the pull exerted by a solenoid
on some type of iron plunger. The amount of pull is dependent
upon the current strength in the solenoid, so by restraining the
motion of the plunger by gravity or by means of a spring,
the deflection of a pointer attached to the plunger might be
made to read amperes. Fig. 116 shows a typical instrument of

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 129

Fio.

116. — Early type of
ammeter.

plunger

this class. Such an instrument is inaccurate, due: (1) To mag-
netic hysteresis or lag, which for a given current results in a higher
reading for decreasing values of current than for increasing values.
(2) The weight of the plunger makes it impossible to mount the
moving system so that the fric-
tion error is negligible. (3) The
instrument is not damped, and
fluctuates violently on sUghtly
fluctuating loads.

For direct current measure-
ments, the Weston instrument,
developed by Edward Weston,
has come into almost universal
use. The instrument is based
on the principle of theD' Arson val
galvanometer, but it is so con-
structed that it is easily portable

and it is provided with a pointer and scale for indicating the
deflections of the moving coil.

The essential parts of the instrument are shown in Fig. 117.
As in the D'Arsonval galvanometer, a permanent magnet is nec-
essary, being made in horse-
shoe form. Two soft-iron
pole pieces are fitted to
the magnet poles and a
cylindrical core is held be-
tween these pole pieces by
a strip of brass. This gives
a uniform air gap and a
^i radial field. The length of
• ' \ the air gap is very much
i shorter than is usual with
/ D'Arsonval galvanometers.

Fig. 117.-Mov^ent of a Westoninstru^nt. The moving Coil is made of

very fine silk-covered
copper wire wound on an aluminum bobbin. The aluminum
bobbin, besides supporting the coil mechanically, also makes
the instrument highly damped. This damping is due to the
currents set up in the aluminum because of its cutting the
magnetic field.

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130

DIRECT CURRENTS

To Lower Sprius
Fig,

To Upper Spring

118. — A typical Weston direct-
current milli-voltmeter.

Instead of suspending the coil by a filament, it is supported
at the top and bottom by hardened steel pivots turning in cup-
shaped jewels, usually sapphire. This method of supporting

the moving coil is almost fric-
tionless and makes the instru-
ment portable, whereas the
D'Arsonval galvanometer is not
so. The current is led in and
out of the coil by two flat spiral
springs, one at the top of the
coil and the other at the bottom.
These springs also serve as the
controlling device for the coil.
That is, any tendency of the coil
to turn is opposed by these two
springs. The top and the
bottom springs are coiled in opposite directions so that the
effect of change of temperature, which causes a spiral spring
to coil or uncoil, will not cause the needle to leave its zero
position. A very light and delicate
aluminum pointer is attached to the
moving element to indicate the deflection
of the coil. This is carefully balanced
by very small counter-weights so that
the whole moving element holds its zero
position very closely, even if the instru-
ment is not level. The pointer moves
over a graduated scale, which may be
marked in volts or in amperes as the
case may be. Because of the radial
field, the deflection of the moving coil
in this type of instrument is practically
proportional to the current in the mov-
ing coil, so that the scale of the instru-
ment has substantially uniform gradu-
ations, which is desirable. The internal
connections of a Weston instrument are shown in Fig. 118.
Instruments of this construction having very weak springs are
often used for portable galvanometers. Although lacking the

Fig. 119. — Weston portable
galvanometer.

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 131

extreme sensitivity of the suspended tjrpe, they can be made
sufficiently sensitive for certain classes of work and their rugged-
ness and portability make them very useful. Such a galvanom-
eter is shown in Fig. 119.

The moving coil of Weston portable instruments deflects to the
full scale value with about 0.01 amp. in the coil. Therefore, to
measure currents greater than this, the larger portion of the
current must be diverted from the moving coil by a shunt.
The shunt is merely a low resistance, usually made of manganin
strip (Af) brazed to comparatively heavy copper blocks (c) as

Ammeter or
MiUivoltmetor

Fig. 120. — Ammeter with an external shunt.

shown in Fig. 120. Two sets of binding nuts are fastened to the
copper blocks. The heavy wing nuts (BB) are for carrying the
main current through the shunt. The small posts (bb) are used
to connect the ammeter leads. The copper blocks serve two
purposes. They are an excellent conductor of heat, so carry
the heat away from the manganin strip, and their low resistance
keeps all parts of each copper block at very nearly the same
potential. The ammeter is in reality a voltmeter reading the
voltage drop across a resistance. A complete set of shunts,
with their current ratings, is shown in Fig. 121. The heavy cop-
per terminals for connection to bus-bars should be noted.
The voltage drop across the shunt is

V»h = Ish R>

8h itthf

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132

DIRECT CURRENTS

where Ith and R^h are the shunt current and the shunt resist-
ance respectively. If Rsh is constant, the voltage drop across
the shunt is proportional to the current in the shunt, so that the
instrument readings are proportional to the current in the shunt.
For this reason the ammeter itself (Pig. 118) is often marked
"Milli voltmeter." For full scale deflection the drop across a
shunt is about 50 millivolts. The current taken, by the instru-
ment itself is usually about 0.01 amp. so that it is almost always

Fig. 121, — Ammeter shunts.
No. 1 from 25 to 200 amperes.
No. 9 from 4500 to 6000 amperes.

negligible as compared with the main current. Therefore, in
most cases the line current equals the shunt current, practically.
An ammeter and its shunt may also be considered as a divided
circuit. In Fig. 122 let Rsh and 7,a be the shunt resistance and
the shunt current respectively, and let.i^^ and Im be the instru-
ment resistance and the instrument current respectively. By
the law of divided circuits:

Ish Rm

Im R<

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 133

That is, the current divides between the instrument and the
shunt inversely as their resistances.

Example. — Assume that an instrument has a resistance of 4 ohms, the
shunt a resistance of 0.0005 ohm, and that the line current is 90 amp.
What is the value of the instrument current?

As the current in the line differs from the shunt current by a very small
amount, the two may be assumed equal. Then,

90
/«

In.

' 0.0005

•' 0.0113 amp.

l!2=»i

AAAAAAA-

Rih

Fig. 122. — Division of current between

an ammeter and its shunt.

For accuracy, the current must always divide between the
instrument and the shunt in a fixed ratio. This means either
that the resistance of the
shunt and the resistance of the
instrument must not change
at all or that both must
change in the same ratio.
As the shunt operates at a
higher temperature than the
instrument, it should be made
of a metal whose resistance

does not change appreciably with the temperature, such as
manganin. The resistance of the instrument circuit should also re-
main constant. The resistance of the leads connecting the shunt
to the instrument should remain constant and the leads with which
the instrument is calibrated should always be used to connect the
shunt to the instrument. Ihe lugs and binding post contacts
should be kept clean from oxide and dirt. A low adjustable re-
sistance (the spiral, Fig. 118) is connected inside the instrument.
By varying this resistance the instrument is adjusted to its shunt.

An ammeter with an external shunt may be made to have a
Ikrge number of scales or ranges. Assume in the example just
given that the instrument gives full scale deflection when the
instrument current is 0.0125 amp. The volts across the instru-
ment terminals are 0.0125 X 4 = 0.050 volt or 50 millivolts.
Dividing this voltage by the shunt resistance, the shunt current
is

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134 DIRECT CURRENTS

The instrument then deflects full scale with 100 amp. in the
line.

If a shunt having a resistance of 0.005 ohm be substituted,
the 50 millivolts drop across the shunt may be obtained with 10
amp. (10 X 0.005 = 0.050). Therefore a 10 scale ammeter
results. By the choice of suitable shunts the same instrument
may be made to give full scale deflection with 1 amp., and with
5,000 amp. For instance, all the shunts shown in Fig. 121
could be used with the same instrument and as many different
scales thereby obtained.

In the smaller sizes of instruments up to 50 amp. and where
only one scale is desired, the shunt is usually placed within the
instrument. For ranges between 50 and 100 amp. the use of
an internal or an external shunt is optional. Above 100 amp. it
is usual to have the shunt external to the instrument on account
of its size and its heating loss.

An ammeter can usually be distinguished from a voltmeter
by the fact that its binding posts are heavy and are of bare metal,
except in the case of an instrument having an external shunt.
The posts of millivoltmeters and voltmeters are of much lighter
construction and the metal posts are covered with hard rubber,
mostly for insulation purposes.

Any instrument when connected in a circuit should disturb the
circuit conditons as Uttle as possible. An ammeter shunt, as it
goes in series with the line, should have as low a resistance as is
practicable, so that when it is connected, very Uttle additional
resistance is introduced into the circuit. To protect anuneters
from heavy currents, etc., provision may be made for short-
circuiting them when readings are not being taken.

116. Voltmeters. — ^The construction of a voltmeter does not
differ materially from that of an anameter in so far as the move-
ment and magnet are concerned. (See Fig. 117.) The moving
coil of the voltmeter is usually wound with more turns and of
finer wire than that of the ammeter and so has a higher resist-
ance. The principal difference, however, lies in the manner of
connecting the instrument to the circuit. As a voltmeter is
connected directly across the line to measure the voltage, it is
desirable that the voltmeter take as little current as is practicable.
Because of its comparatively low resistance, the moving coil of

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 135

the voltmeter cannot.be connected directly across the line, as it
would ordinarily take an excessive current and might be burnt
out. Therefore it is necessary to connect a high resistance in
series with the moving coil. This is shown in Fig. 123. By
Ohm's Law the current through the instrument is proportional to
the voltage, so that the instrument scale can be graduated in
volts. The resistance required is easily determined. Assume
that an instrument gives full scale deflection with 0.01 amp. in

-ufison-

L^y/VVW^^VNA/WWVVWVVVVNA^ ,
B A

-14,9900-

|148a)j

160 i i

0 16

(a) (6) .

Fia. 123. — Methods of connectiDg resistance in a voltmeter.

160

B = -V =

= 15,000 ohms.

the moving coil, and that the coil resistance is 20 ohms. If it is
desired that the instrument indicate 150 volts, full scale, then the
total resistance of the instrument circuit must be

V ^ 150
/ 0.01

As the instrument has a resistance of 20 ohms, this means that
14,980 ohms additional are necessary.

If it be desired that this same instrument also have a full scale
deflection with 15 volts, the resistance of 14,980 ohms may be

15
tapped so that the resistance OB (Fig. 123 (a)) = q-^t = 1,500

ohms, and this tap can be brought to a binding post. Another
method of securing the same result is showninFig. 123 (6). Wind
another resistance equal to 1,500 — 20 = 1,480 ohms and cojinect
it from a binding post to the junction of the resistance and the
moving coil. This last method is advantageous as it permits
independent adjustment of each resistance; also injury or repair
in one resistance does not affect the other.

116. Multipliers or Extension Coils.— The range of a voltmeter
having its resistance incorporated within the instrument, may be

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136

DIRECT CURRENTS

increased by the use of external resistance connected in series
with the instrument.

Example, — A 150 scale voltmeter has a resktance of 17,000 ohms. What
external resistance should be connected in series with it so that its range is
(a) 300 volts? (6) 600 volts?

(a) In order to maintain the same current through the instrument at 300
volts as flows at 150 volts, the resistance of the circuit must be doubled.

Therefore, 17,000 X 2 = 34,000 ohms are necessary.

As the instrument already has 17,000 ohms, the added resistance wiU be.
34,000 - 17,000 = 17,000 ohms, Ans.

(6) The total resistance must now be,

Ygg X 17,000 = 68,000 ohms.

As 17,000 ohms is already within the instrument, 68,000 — 17,000 = 51-,
000 ohms must be added external to the instrument. Ans.

External resistances used in this manner are called mnUipliers,
or sometimes extension coils. They are usually placed within a
perforated .box and the terminals brought out to binding posts.
The multiplying power of the multiplier is marked near a terminal.

The equation giving the relation between the resistance of
the multiplier jB*, the resistance of the instrument jRm, and the
multiplying power M is as follows:

M = ^t-^!^ (49)

Example, — In the above problem
(6) the multiplying power of the
multiplier is as follows:

61,000 + 17,000 .

^ = — i7;ooo ^

117. Hot-wire Instruments.

In the instruments hereto-
fore considered, the action of
the instrument depends on
the electromagnetic action of
a current. There is another
type of instrument which de-
pends for its indications upon
the heating action of the ciui'ent. A diagram of this instrument
is shown in Fig. 124. AB is a fine wire of platinum-silver through
which the current passes. At C, a wire CF is attached to AB,

Fig.

124. — Principle of Hartmann and
Braun hot-wire instruments.

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 137

At Ej on CFj a silk fiber EH is attached. This passes around
the pulley W, and is held in tension by the spring H. When a
current flows through AB^ the heat expands the wire ABj re-
ducing the tension in the wire CF, and allowing the spring H
to pull the silk fiber to the left. This fiber, acting on the pulley
Wy moves the pointer P over the scale.

When used as an anmieter, a shunt is necessary unless the
current is very small. When used as a voltmeter, a high re-
sistance is connected in series with the wire AB.

This type of instrument is *'dead beat," that is, it is very slug-
gish in its behavior and only reaches its ultimate deflecton after
the lapse of considerable time. This is an advantage in the
measurement of fluctuating currents as the needle follows the
fluctuations very slowly so can be accurately read. Another
advantage of the hot wire type of instrument is that it can be
used for alternating as well as for direct currents. It is often
used as a transfer instrument to measure alternating currents in
terms of direct current. This type of instrument is particularly
useful for the measurement of high frequency alternating currents,
as its indications are independent of the frequency if a shunt is
not used. For this reason this type is very useful in radio teleg-
raphy. Such instruments are affected by temperature and do
not hold their calibration for very long periods. Therefore, for
accurate work they should be calibrated at the time of using.

ELECTRICAL MEASUREMENTS

Measurement of Resistance

118. Voltmeter-ammeter Method. — The resistance of any
portion of an electric circuit is, by Ohm's Law,

where V is the voltage across that portion of the circuit and / is
the steady current flowing in that portion of the circuit. Ob-
viously, the voltage V may be measured with a voltmeter, the
current / measured with an ammeter, and the resistance R
computed.

Let it be required to determine the resistance R in the circuit
shown in Fig. 125. The source of power is the 110-volt supply.

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138

DIRECT CURRENTS

The resistance R is comparatively small and if connected directly
across 110 volts would take an excessive current. Therefore,
it is necessary to insert a resistance R' in series with R to limit
the current. The voltmeter, however, must be connected directly
across R as it is desired to know the resistance of this portion
of the circuit only.

Fig. 125. — Voltmeter-ammeter method of measuring resistance.

Example, — The voltmeter (Fig. 125) reads 19 volts when the ammeter
reads 24 amp. What is the value of the resistance R*t
The resistance:

i2 = ^i = 0.792 ohm.
24

As a matter of interest let it be required to determine the resistance of

R', The voltmeter terminals are transferred from across R to across R\

Under these conditions the voltmeter reads 91 volts and the ammeter still

reads 24 amp. Therefore:

Ql
/2'=|^ = 3.79 ohms.

It is sometimes desired to measure resistances of such low value
that, if a voltmeter were connected directly across their termi-
nals, the contact resistance, which may be comparatively large,
would introduce considerable error and might even exceed in
magnitude the resistance which it is desired to measure. To
eUminate this error due to contact resistance, the voltmeter
terminals are connected well inside the terminals BB (Fig. 126)
through which the current is led to the specimen. As the volt-
meter takes but a very small current, small sharp pointed con-

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 139

tacts CC may be used. As the resistance of the voltmeter is
comparatively high, it is only necessary that the contact resis-
tances at CC be negligible compared to the resistance of the
instrument. This condition is easily met. As these contacts
are small and sharp the points of contact on the specimen can
be determined very accurately.

Snpply

1 I ■*■ K6,i.lort V^

Ammetov Millivoltmetcar

FiQ. 126. — Measuring the resistance of a metal rod.

Example. — When the ammeter (Fig. 126) reads 60 amp., the milli-
voltmeter indicates 40 millivolts. The contacts CC are 23 in. apart.
What is the resistance per inch length of the rod?
The resistance for 23 in. is:

0.040
50

R =

= 0.00080 ohm.

The resistance per inch:
R =

0.00080
23

= 0.0000348 ohm. Arts.

119. The Voltmeter Method. — It is possible to measure a
resistance by means of a voltmeter alone provided the resistance
to be measured is comparable with that of the voltmeter. In
Fig. 127 (a) let it be required to measure the resistance R. The
voltmeter is first connected across the source of supply and a
reading Vi taken. It is then transferred so that the resistance
R is in series with it across the source of supply and the voltmeter
reading is again taken. Let this reading be F2.

As Vi is the total circuit voltage and V2 is the voltage across
the instrument, the voltage across the unknown resistance R
is obviously Vi — ¥2- When the voltmeter is in series with /?,

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140

DIRECT CURRENTS

the same current i must flow through each so that the voltages
are as follows:

F2 - iRv (1)

Fi - F2 = iR (2)

where R, is the resistance of the voltmeter.
Dividing (2) by (1) and solving for iJ,

R = R^y^I^ (60)

This method of measuring resistance is particularly useful
in determining insulation resistance of dynamo windings^ cables.

{a) (h)

Fig. 127. — Measurement of resistance by the voltmeter method.

etc. As such resistances are very high they are usually ex-
pressed in megohms (1 megohm = 1,000,000 ohms). It will
be seen from equation (50) that the greater the value of Rvy
the greater the resistance that can be measured by this method.
For this reason special 150 scale voltmeters, having resistances
of 100,000 ohms (one-tenth of a megohm) are available. These
give a sensitivity about six times as great as can be obtained with
the ordinary 150 scale voltmeter.

Fig. 127 (b) shows the application of this method to the
measurement of the insulation resistance of a cable.

Example, — When a 100,000-ohm. voltmeter is connected across a direct
current line it reads 120 volts. One terminal of the voltmeter is then con-
nected to the core of a lead-covered cable and the sheath of the cable is
connected to the other side of the line as in Fig. 127 (6). The voltmeter
now reads 10 volts. What is the insulation resistance of the cable?

^ _, 120 - 10 , , ,

A = 0.1 :r^ = 1.1 megohms.

iim.

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 141

120. The Wheatstone Bridge. — In distinction to the fore-
going methods of measuring resistance, the Wheatstone Bridge
method is one in which the unknown resistance is balanced
against other known resistances. The bridge in its simplest
form, is shown in Fig. 128. Three known resistances ilf , iV, P,
and the unknown resistance X are connected to form a diamond.
Current from a battery B feeds the two opposite corners o and c
of the diamond. Across the
other two corners a and 6, is
connected a galvanometer.

To make a measurement,
the two arms M and N are
each set at some fixed value
of resistance, usually 1, 10,
100, 1,000 ohms, etc. The
arm P is then adjusted until
the galvanometer does not

deflect. If the galvanometer Fiq. 128.— Elementary Wheatstone bridge.

does not deflect, no current

flows through it and therefore the two points a and b must be
at the same potential. Also the currents /i = h and h = /4,
as no current passes through the galvanometer.

If the points a and b are at the same potential, the voltage
drop oa = ob and:

IiM = hX (1)

Also the voltage drop ac = be and

hN = I4P

And since
Dividing (1) by (2)

/i = Is and I2 = I A
IiN = I2P

IiM

hX M
^hP'^'N

X
P

(2)

(51)

which is the equation of the Wheatstone Bridge. M and N are
called thQ ratio arms and P the balance or rheostat arm. Ob-
viously the battery and the galvanometer may be interchanged
without affecting the relation given in equation (51).

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142

DIRECT CURRENTS

The many types of Wlieatstone Bridge found in practice do
not differ in principle from that shown in Fig. 128. The dif-
ferences lie in the positions of the arms Af, iV, and P on the
bridge as well as in the manner in which the coils in these arms
are cut in and out of circuit.

A common plug type of bridge is shown in Fig. 129. M
consists of three resistances of 1,000, 100, and 1 ohms respectively,

FiQ. 129. — Massachusetts Institute of Technology pattern of Wheatstone bridge.

and N consists of three of 10, 100, and 1,000 ohms respectively.
P consists of a number of resistances ranging from 5,000 ohms
to 1 ohm and of such values that with the proper combinations
P may be made equal to any whole number between 1 and 11,110
ohms. Between the outer ends of N and P are two infinity plugs
( 00 ) and a 10,000-ohm coil. The infinite plugs mean that the bridge
can be open circuited at these points and by their position
the 10,000-ohm coil may be made a part of iV or a part of P.

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 143

The unknown resistance X may be connected across any one of
the infinite resistances, if it is found advisable to do so. Between
M and P is another infinite resistance, across which the unknown
resistance may also be connected, the infinite plug being removed.
In this type of bridge the resistance coils are connected across
gaps cut in heavy brass or composition bars. When it is desired
to insert a resistance the plug is removed, and when it is desired
to remove a resistance it is short-circuited by the plug. These
plugs have hard rubber tops and are tapered. As the principle
source of error in this type of bridge lies in the contact resistance
of these plugs, they should be made to fit tightly when used.
This is accomplished by exerting a slight pressure and simul-
taneously twisting them, thus giving a wiping contact. As dirt
and oxide are a frequent source of error the plugs should be
kept clean.

In using the bridge, much time may be saved if a systematic procedure
is followed in obtaining a balance. Assume that it is desired to measure a
certain unknown resistance. Connect the bridge as shown in Fig. 129,
placing keys in the battery and in the galvanometer circuits and a shunt
around the galvanometer to protect it from deflecting violently when
the bridge is considerably out of balance. Make the ratio arms M and N
each 1,000 ohms, a 1 to 1 ratio. With the galvanometer well shunted and
all the plugs in P (Res. = 0), depress first the battery and then the gal-
vanometer key. The galvanometer is observed to deflect to the left. Now
remove the 6,000-ohm plug and the galvanometer deflects to the right.
From, these observations, two facts are determined. The unknown re-
sistance is less than 5,000 ohms and when the galvanometer deflects to the
left the value of resistance in P is too small, and when it deflects to the right
the value of P is too large. By inserting the 5,000-ohm plug and removing
the 1,000-ohm plug the galvanometer still deflects to the right, indicating
that 1,000 ohms in P is too large. This is repeated with 500 ohms, 200 ohms,
etc. By proceeding in this manner, it is found that the galvanometer does
not reverse until a 2-ohm plug is removed. This means that the unknown
resistance lies between 2 and 5 ohms. By removing the two 2-ohm
plugs and then a 1 and a 2 the unknown resistance is narrowed down
to between 2 and 3 ohms. To get a more precise value the ratio arms
must be changed. M is now made 1 ohm and 2,000 ohms unplugged
in P. By successive trials, all the time reducing the shunt S, a balance
is obtained at 2,761 ohms in P. Then:

X = ^P = -^2,761 =2.761 ohms.

In obtaining a balance the battery key should always be depressed before
the galvanometer key, so that the current in the bridge has time to reach

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144

DIRECT CURRENTS

a constant value. Otherwise the electromotive force of self-induction may
introduce an error.

A more convenient arrangement of the resistance units of the rheostat
arm P is shown in Fig. 130. The resistances are arranged in groups of
equal resistances, one group consisting of ten 1-ohm coils, the next of ten
10-ohm coils, the next of ten 100-ohm coils, etc. Each group is called a
decade. Only one plug per decade is necessary. This arrangement has the
advantage that the plugs are always in service, so are not so likely to be
mislaid or to become dirty; there is less probability of error in reading; it
is a simple matter to see that the few plugs used are fitting tightly, and

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rWi rivi rWl rTn rWf rWl rWI n»i »»V1 rwi
100 100 100 100 100 100 100 100 100 100

iHnHunjinrt^

fwfl rWI rVVi rWl rfW WYl rVtl IWl rVi flT%
1111111111

FiQ. 130. — Arrangement of rheostat arm resistances in a decade bridge.

a balance can be quickly obtained. It is obvious that nine coils per decade
are sufficient for obtaining any desired resistance, although ten coils per
decade are often used.

The decade principle has been extended to an even more convenient type
of bridge, the dial bridge. Instead of using plugs, a dial arm similar to the
type used in rheostats is employed to select the required resistances. Be-
cause of its ease of manipulation this type has come into extensive use.
Care should be taken to keep the dials and contacts free from dirt and oxides.
Fig. 131 shows a dial bridge of the Leeds & Northrup type.

121. The Slide Wire Bridge. — The slide wire is a simplified
Wheatstone Bridge, in which the balance is obtained by means
of a slider which moves over a German silver or manganin re-
sistance wire. A typical slide wire bridge is shown in Fig. 132.
The resistance wire A J5, 100 cm. long, is stretched tightly between
two heavy copper blocks CD, 100 cm. apart. A meter scale is

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 145

Fig. 131. — Leeds <& Northrup dial bridge.

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146 DIRECT CURRENTS

placed along this wire. A contact key K' is movable along
the scale and when the key K' is pressed a knife edge makes
contact with the wire. The rest of the bridge consists of a heavy
copper bar E, a known resistance fl, and the unknown resistance
X. Rib connected between D and E and X between C and E, al-
though the positions of R and X are interchangeable.

The galvanometer is connected between the key K' and E and
the battery terminals are connected to C and D, A balance is
obtained by moving K^ along the wire until the galvanometer
shows no deflection.

TiQ. 132. — Slide-wire bridge.

Let I be the distance in centimeters from one end of the scale
to K' when a balance is obtained. Then 100 — Z is the distance
from K' to the other end of the scale. Let r be the resistance
per unit length of the wire. Then the resistance of I is Ir and
that of the remainder of the wire is (100 *- l)r.

By the law of the Wheatstone Bridge:

X R

Ir (100- l)r
r cancels out and (52) becomes :

(52)

^ = '(100 - n ^^^

(52) may also be written

R " 100^/ ^^*^

This is equivalent to stating that when a balance is obtained
the slide wire is divided into two parts which are to each other as
X is to R.

The slide wire is not as accurate as the coil bridge, because the

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 147

slide wire may not be uniform; the solder at the points of contact
at C and D makes the length of the wire uncertain; the slide
wire cannot be read as accurately as the resistance units of a
bridge can be adjusted.

Example. — Assume that R (Fig. 132) equals 10 ohms and that a balance
is obtained at 74.6 cm. from the left-hand end of the scale. Find the un-
known resistance X.

From equation (63)

10 ^. ^ 10

X = 746

100 - 74.6

74.6 j^. = 29.37 ohms.

An8.

CABLE TESTING

122. The Murray Loop. — The slide wire bridge ofiFers a very
convenient method of locating grounds in cables and wires.
Fig. 133 shows a cable AB which has become grounded at the

Fig. 133. — Murray-loop test.

point 0, owing to a defect in the insulation. CB is the return
conductor and is similar to AB except that it has no ground or
*' fault." The two conductors are looped together at 5, the far
end of the two conductors, which may be at some power station,
telephone exchange, etc.

The slide wire is then connected to the home ends of the cable
as shown. It will be noted that the battery and the galvanometer
are not in the positions shown in Fig. 132, but have been inter-
changed. This is done in order that earth currents shall not dis-
turb the galvanometer readings. Also if the ground has a high
resistance, the electromotive force of the battery B may be in-
creased until sufficient current to operate the bridge is sent through

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148 DIRECT CURRENTS

this resistance. The resistance of the fault to ground does not
produce any error in the measurement so long as the conductor-
is not broken. If the conductor is broken with both ends lying on
the ground, the resistance of the conductor is increased and a false
location of the fault may result.

In Fig. 133, the distance X to the fault may be found as follows :

X L + (L- X)

V I

(55)

where L is the length of one cable.

The slide wire is split into two sections which are to each other
as the two lengths of cable on each side of the fault.

Solving (55) for X,

X = ^, (56)

This assumes that the resistance per foot of both conductors
is the same, and is uniform. The jumper tieing the cable ends
together at B should make good connection, as contact resistance
at this point may introduce an appreciable error. A ratio and
rheostat arm of a bridge box may obviously, be used instead of
the slide wire.

Example, — A cable 2,000 ft. long consists of two conductors. One con-
ductor is grounded at some point between stations. A Murray loop test,
with a 100-cm. slide wire bridge, is connected as in Fig. 133 to locate the fault.

A balance is obtained at 85 cm. How far from the station is the ground?
From equation (56) :

L = 2,000 Z' = 15 Z = 85

X = — — rrrr = 600 ft. from the station at which the measurement is

made.

123. The Varley Loop. — The Varley loop is also used to locate
cable faults. It is similar in principle to the Murray loop, a
bridge box being necessary, however. The connections are shown
in Fig. 134. M and N are the two ratio arms of a bridge and P is
the rheostat arm. It is necessary that the battery and the gal-
vanometer occupy the positions shown, in order to avoid dis-
turbances in the galvanometer due to earth currents. A balance
is first obtained by means of P, with the switch S at a. Let r be
the resistance per foot length per conductor, assumed uniform.

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 149

(It will be noted that P and X together form one arm of the
bridge.)

M ^r(L +L - X)

(57)

P -{-rX

Before X can be found it is necessary to know r. To obtain this,
the switch S is thrown to position b. This throws both lengths

Fig. 134. — Varley-loop test.

of cable in series and makes them the fourth arm of a bridge.
A simple bridge measurement is then made of the total loop
resistance. Call this resistance R, Then the resistance per foot
of cable

r =

(This measurement is not necessary if the resistance per foot
or the total resistance of the cable is already known.)
Substituting this value of R in (57)

M ^ R/2L(2L - X)
N P + RX/2L

Solving for X,

' '_ 2L/NR - MP\

R\ M + N I

This equation gives the distance in feet to the fault,
equation is frequently given as follows:
^ NR - MP

(58)
The

(59)

M + N

In this case Rx is not the distance to the fault, but rather the
resistance along the grounded conductor to the fault.
If Af = iV in equation (58)

X = ^ (i? - p)

which is simpler in form than (58) .

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150

DIRECT CURRENTS

Example. — In locating a fault by the Varley loop test, the connections
shown in Hg. 134 were used. Each conductor is 2,800 ft. long. With the
switch at (o) and Af = 10, iV = 1,000, P was found to be 137 when a bal-
ance was obtained. Switch 8 is then thrown over to (6). Under these
conditions, a balance was obtained when Af = 10, iV = 1,000, P = 221,
making R » 2.21.

By equation (58) the distance in feet to the fault

^ = ' iirC'"^ "" Toio '° "" '"') ° ^'^»»^^- ^-

124. Instilation Testing. — In practice it is necessary to meas-
ure the resistance of the insulation of cables, both at the factory
and after the cable is installed. A low value of insulation re-
sistance may indicate that the insulation is of an inferior grade.

Ayrton Shnnt

Battei%SI

FiQ. 135. — Measurement of the insulation resistance of a cable.

A low insulation resistance after installation may indicate im-
proper handling or faulty installation. The voltmeter method
described in Par. 119 is applicable in many cases, but where the
insulation resistance is high even a high resistance voltmeter is
not sufficiently sensitive.

To make the measurement, a sensitive galvanometer is utilized.
A considerable source of potential, from 100 to 500 volts, is also
usually necessary. Such potential may be secured from direct
current mains, although dry cells, silver chloride cells, and test-
tube batteries connected in series are more satisfactory. A
simple diagram of connections is shown in Fig. 135.

The method is one of substitution. A known resistance, usu-

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 151

ally 0.1 megohm (100,000 ohms), is first connected in the circuit
and the galvanometer deflection noted. The unknown resistance
X is then substituted and the galvanometer reading again noted.
As the currents in the two cases are inversely proportional to the
circuit resistances, the unknown resistance can be determined,
the galvanometer deflections being used rather than actual
values of current. Let Di be the deflection with the 0.1 megohm
and D? be the deflection with the unknown resistance.

0.1 D2
X = 0.1 g (61)

Under ordinary circumstances it would not be possible to obtain
accurate results under these conditions alone, because the unknown
resistance may be in the hundreds of megohms and the known
resistance is but 0.1 megohm. This would make the deflection D2
so many times smaller than Z>i that it would not be readable.

This difficulty is overcome by the use of the Ayrton shunt
described in Par. 113. When the 0.1 megohm only is in circuit,
the galvanometer sensitivity ordinarily is such that it would
deflect ofif the scale unless the galvanometer were shunted.

Therefore the shunt is adjusted to some low value as 0.0001.
Call this reading of the shunt Si and the galvanometer deflection
Di. The multiplying power of the shunt equals Mi = 1/Si.
The cable is now introduced into the circuit and the shunt ad-
justed until a reasonable deflection is obtained. Call this read-
ing D2 and the value of the shunt S2. Its multipljdng power is
now M2 = I//S2.

The current in the circuit in the two cases

h^MiDi
I2 M2D2
Therefore the unknown resistance, from (61), is

Z = 0.1 ^ = 0.1 ^ (62)

I2 M2D2

In practice, instead of substituting the cable for the 0.1 meg-
ohm, the cable is first short-circuited by the wire shown dotted,
Pig. 135, and the constant determined. This wire is then removed,
placing the cable in circuit. Ths 0.1 megohm is left permanently

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152

DIRECT CURRENTS

in circuit to protect the galvanometer in case of accidental short-
circuit of the cable. Its resistance is usually not appreciable
compared to that of the cable, so that no correction is ordinarily
necessary for it.

A switch or key S is ordinarily provided. When in position
(a) the circuit is closed through the cable. When thrown over to
(6) , the cable, which is charged electrostatically, discharges through
the galvanometer.

When the switch (a) is first closed there is a rush of current
which charges the cable electrostatically. (See par. 153, Chap.
IX.) It takes time to charge the cable, so for some time this

Fig. 136. — Charge and discharge curves of a cable.

charging current flows, decreasing continuously. This is shown
in Fig. 136, giving the relation of the galvanometer deflection to
the time. As it is often inconvenient to wait for the galvanometer
to reach a steady deflection, it has been arbitrarily agreed to
take the deflection at the end of one minute as the value to be
used in determining insulation resistance.

When the switch S is thrown to (6) the electrostatic charge
in the cable rushes out through the galvanometer in the reverse
direction. Due to absorption it requires considerable time for
the cable to become totally discharged. This is also shown in
Fig. 136.

In making insulation resistance measurements, precautions
must be taken to insulate thoroughly the apparatus itself. Hard
rubber posts should be used for supports and, wherever possible.

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 153

the leads should be carried through the air rather than be allowed
to rest on the ground. The insulation resistance varies enor-
mously with temperature, so the temperature at which the
measurements are made should be carefully determined and
stated.

Example, — The cable whose insulation curves are shown in Fig. 136 was
tested for insulation. The deflection with 0.1 megohm only in circuit was
20 cm. and the shunt read 0.0001. When the curve shown in Fig. 136 was
obtained the shunt read 0.1. The cable was 2,200 ft. long.

(a) What is its insulation resistance?

(6) What is its insulation resistance per mile?

Ml = 1/0.0001 = 10,000

M2 = 1/0.1 = 10

Dt (from curve) =11 cm.

. . _, ^,10,000 X 20 ,„^ , .

(a) X =0.1 — ' — = 182 megohms. Ans.

(b) The resistance per mile will be less than that of the 2,200-ft. length
because the amount of leakage current is directly proportional to the length
of the cable. Therefore the resistance of this leakage path is inversely pro-
portional to the length of cable. The cross-sectional area of the leakage
path for the mile length is greater than it is for the 2,200-ft. length. There-
fore the resistance per mile

R = 1^ 182 = 75.0 megohms. Ans,

POTENTIOMETERS

125. The Potentiometer. — The potentiometer is an instrument
for making accurate measurements of voltage. Its standardiza-
tion depends primarily upon the Weston standard cell. (See
Par. 89, Chap. VI.) The principle is as follows:

Assume in Fig. 137(a) that a standard cell S has an electro-
motive force of exactly 1 volt. Let a storage cell Ba supply
current to a wire AB through a rheostat R, Let the wire ABhe
divided into 15 divisions each of 1 ohm resistance, making the total
resistance oi AB equal to 15 ohms. The standard cell is con-
nected with its negative terminal to the negative terminal of the
storage cell and its positive terminal is connected to the tenth
1-ohm coil C through a key and galvanometer. If 0.1 amp.
flows through the wire AB, the voltage drop through each^resist-
ance will be 0. 1 volt and the voltage drop across A C will be 1 .0 volt.

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DIRECT CURRENTS

If the key be depressed no current will flow through the galvano-
meter, as the standard cell emf. is in exact opposition to this
1-volt drop. If, however, the current in AB is not exactly 0.1
amp., current will flow through the standard cell circuit due
to the voltage AC being either greater or less than 1 volt. If
the current is less than 0.1 amp. the galvanometer deflects in
one direction, and if it is greater than 0.1 amp. the galvanometer

i\^

0 .1 .9 .S .4 .S .« .7 .8 .0 1.0 1.11.21.S1.4 1Ji

(o) Standardizing the wire A-B

B».

- il +

(b) Measuring an unknown emf.

S (c)

Fig. 137. — Simple potentiometer.

deflects in the reverse direction. Obviously it is possible to so
adjust the current in AB that the galvanometer deflection is
zero. Under these conditions the current in AB is exactly 0.1
amp. and the potential drop across each resistance in AB is 0.1
volt. Therefore A B may be marked in volts as shown.

Let it be required to measure some unknown electromotive
force E whose value is known to be less than 1.5 volts. Its
negative terminal is connected to the end A of the wire AB,
Fig. 137(6). The positive terminal of the electromotive force
is connected through the galvanometer and key to a movable

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 155

contact b. It is assumed that the current in AS has been ad-
justed to exactly 0.1 amp. as just described. Contact b is
moved along AB until the galvanometer deflection is zero.
This means that the electromotive force E is just balanced against
an equal drop in the wire AB. As A B is calibrated in volts,
the value of E may be read directly on AB. This method of
measuring voltage is the Poggendorf Method and is the funda-
mental principle of the potentiometer.

The two diagrams (a) and (6) (Fig. 137) may be combined
into one by the use of the single-pole, double-throw (S.-P.D.-T.)
switch Sw, Fig. 137(c). When the switch is in its left-hand posi-
tion the standard cell is in circuit for calibration as in (a). When
it is in its right-hand position the unknown emf . is in contact
with the wire AB ^o that its value may be determined, j

Std.

>iiO-

|E.M.F. ^^mtm.

^- ^

Fig. 138. — Connections of Leeds & Northrup low-resbtance potentiometer.

126. The Leeds & Northrup Low Resistance Potentiometer. —

Fig. 138 shows the Leeds & Northrup low resistance potenti-
ometer. In most respects it is similar to the simple potenti-
ometer shown in Fig. 137. The battery W supplies current
through the rheostat R and then through the low resistance
OA, to the potentiometer wire AB. AB consists of two parts,
fifteen 6-ohm coils and a slide wire BB of 5.5 ohms. As each
contact represents 0.1 volt, the working current is 0.1/5 = 1/50

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156 DIRECT CURRENTS

amp. As the slide wire DB is 5.5 ohms, the voltage drop
across it when in adjustment is 1.1 volts. This slide wire consists
of 11 turns of resistance wire mounted on a marble cylinder.
Each turn represents 0.01 volt and the entire wire is divided
into 1,100 divisions.

The standard cell has a voltage slightly in excess of 1.0 volt
so that instead of connecting the standard cell exactly as in
Fig. 137 (a), an added resistance -40 is necessary to allow for
this small excess voltage. A contact T is movable on ^4.0 so that
the setting can be made to correspond with the electromotive
force of the standard cell used. M and JIf' are the movable
contacts, which are adjusted to balance the unknown emf.

Fig. 139. — Leeds & Northrup potentiometer without accessories.

M moves over the 15 contacts, each corresponding to 0.1 volt,
and M' moves over the slide wire. A double-pole, double-throw
(D.-P.D.-T.) switch (corresponding to Swj Fig. 137(c)) changes
the connection of the galvanometer from the standard cell to
the unknown emf. There are three galvanometer keys, Ri,
R2, and Ro, Ri should first be depressed as it inserts a high resist-
ance in series with the galvanometer and prevents a violent
deflection if there is considerable unbalancing. R2 inserts less
resistance and there is no resistance in series with Ro which is
depressed when the final balance is obtained.

A resistance S shunts 0.9 of the current from OB^ when the
plug at K is changed. The resistance K is automatically put
in circuit, keeping the total potentiometer resistance, and there-
fore the load on the battery, constant. By this arrangement,

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 157

the readings on the potentiometer are all made one-tenth their
previous values.

An external view of this potentiometer is shown in Fig. 139.

127. Voltage Measurements with the Potentiometer. — ^Poten-
tiometers are designed to measure potentials up to 1.6 volts only.
For the measurement of potentials in excess of this value a
voU box is necessary. A volt box is merely a very high resistance
from which suitable taps are brought. This is illustrated by
the resistance AD, Fig. 140. Assume AD to have a resistance

Volt Box

,((00 JL ^ To Potentiometer

J5$ J 5 4" E.M.F. Termimdfl

G A '•

Fig. 140. — Volt-box and drop- wire connections.

of 10,000 ohms and AB a. resistance of 100 ohms. If no current
leaves the wire at B, the voltage drop across AB will be ^^%^o,ooo
= Koo that across AD. If leads be carried from AB to the po-
tentiometer, the potentiometer will measure J^oo the voltage across
AD, since the potentiometer principle is an opposition method
so that no current is taken from B. Therefore, if a voltmeter
V is being calibrated it should be connected in parallel with AD.
If the voltmeter reads 119.0 volts and the potentiometer reads
1.184 volts, the true line voltage across the voltmeter will be
1.184 X 100 = 118.4. Therefore the correction to the voltmeter
is - 0.6 volt.

In a similar manner, voltages from 1.5 to 15 volts are connected
across AC, the multiplying factor in this case being 10.

The Drop Wire. — GH is a resistance connected directly across
the line. One voltmeter terminal and one terminal of the volt
box are connected to the end G of this wire. The other terminal
of the voltmeter and the remaining terminal of the volt box are
connected to a movable contact K. By sliding K along GH any

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158 DIRECT CURRENTS

desired voltage may be obtained. When used in this manner,
GH is called a drop wire. It is not necessary to the operation
of the volt box, but is merely a convenient means for adjusting
the voltage.

128. The Measurement of Current with Potentiometer. —
As has just been pointed out, a potentiometer is designed to meas-
ure voUage: It may also be used to measure current by merely
applying Ohm's Law. Let an unknown current / flow through a
known resistance R, If E, the voltage drop across ft, be meas-
ured, the current / is immediately determined, since for this
part of the circuit both the voltage and the resistance are known.
Therefore:

The method of making the measurement is shown in Fig. 141.,
It is desired to know the exact current passing through the am-

To E.M.F. Terminals
Potentiometer

FiQ. 141. — Calibration of an ammeter with a potentiometer.

meter, in order to determine its errors, if any exist. The am-
meter is connected in series with the standard resistance, and
also with a rheostat to control the current. Standard resistances
are provided with four terminals as a rule, two heavy ones for
current and two smaller binding posts for potential. The two
potential binding posts are connected to the potentiometer, the
proper polarity being observed. The voltage across the standard
resistance is then measured by means of the potentiometer.

Standard resistances are usually adjusted to even decimal
val^es such as 10, 1, 0.1, 0.01, etc., ohms. They are ordinarily
rated to carry a current that will give 1.0 volt drop. Thus

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 159

the 1 ohm can carry 1 amp., the 0.001 ohm, 1,000 amp., etc. To
keep the resistances cool they are often immersed in oil. The
type shown in Fig. 142 (a) is set in a water-jacketed oil bath
provided with a motor-driven stirrer. The type shown in (6) is

(a) 0.01 ohm. (6) Self-contained 0.001 ohm.

Fig. 142. — Standard resistances.

rated for larger currents, 1,000 amp. and more. The water
jacket, the stirrer, etc., are included within the unit itself.

Knowing tlikt the potentiometer is Umited to 1.5 volts, it
is easy to select the proper standard resistance. An instrument

Fig. 143. — Ammeter calibration curve.

haying a range of 100 amp., would require 1.5/100 = 0.015
ohm. 0.01 ohm would be used. Likewise a 15-scale instrument
would require 1.5/15 = 0.1 ohm.

When instruments are calibrated, they should be checked at ten
or fifteen points on the scale and the corresponding corrections
at each point are plotted as ordinates. (The instrument read-

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160

DIRECT CURRENTS

ings are plotted as abscissas.) As an instrument scale is subject
to scale errors, etc., it is customary to connect successive points
by straight lines as shown in Fig. 143. For instance, (Fig. 143),
the correct current when the instrument reads 50 amp. is 50 4- 0.8
= 50.8 amp.

129. Measurement of Power. — Direct current power is usu-
ally measured by means of a voltmeter and an anmieter. Since
the power is the product of the volts and the amperes {P = BJI),
it is merely necessary to multiply the volts by the amperes to
obtain the power in watts. Certain precautions may be nec-
essary in measuring the power, however.

Correct

C D

r-VWW^

: Ainineier

Am meter

Incorrect

Volt meter
(a) High Resistance (b) Low Resistance

Fio. 144. — Correct and incorrect methods of connecting voltmeters and am-
meters in power measurements.

Assume that it is desired to measure the power taken by an in-
candescent lamp. If the voltmeter is connected as shown by the
dotted line in Fig. 144 (a), the current taken by the voltmeter is
being registered by the ammeter. In other words, the voltmeter
is a load connected in parallel with the lamp. As the current
taken by the lamp is small, this voltmeter current, although of
itself small, may introduce a very appreciable error into the
measurement. That is, the power taken by the voltmeter will be
included in the measurement. There are three methods of
eUminating this error. The voltmeter power may be calculated,
knowing the voltmeter resistance, and proper correction made.
The voltmeter may be open-circuited when the ammeter is being
read if it is certain that this will not alter the voltage across the
lamp. The voltmeter lead may be connected as shown by the

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 161

solid line so that the voltmeter current does not pass through
the ammeter. In this last case the voltmeter is not reading the
true voltage across the lamps, but its reading is too high by the
drop through the ammeter. As the resistance of the lamp is
high and that of the ammeter low, this last error is usually
negligible.

However, if a low resistance CD is being measured, Fig. 144(6) ,
the drop across the resistance is necessarily low, and if the volt-
meter in this case is connected outside the ammeter, a very ap-
preciable error may be introduced, as the voltmeter reading
includes the voltage drop in the ammeter. The voltmeter should
now be connected inside the ammeter. This will not introduce
an appreciable error, for presumably a large current is required
for the measurement of the low resistance, and the addition of
the very small voltmeter current to the ammeter reading is
negligible.

The above precautions should be observed also in making re-
sistance measurements .

Example. — It is desired to measure the power taken by a 40-watt tungsten
lamp. A 0.5 scale ammeter having a resistance x)f 0.15 ohm and a 150
scale voltmeter having a resistance of 16,000 ohms are used for the measure-
ment. When the voltmeter is connected inside the ammeter it reads 120
volts and the ammeter reads 0.35 amp. What is the true power taken
by the lamp and what is the apparent power if the voltmeter loss is neglected?

Apparent power = 120 X 0.35 = 42 watts.

Power taken by voltmeter = ^ _ _ : _ = 0.9 watt.

lo,OOU

True power to lamp =41.1 watts.

The voltmeter introduces a 2 per cent, error in this case.

If connected outside the ammeter, the ammeter will now read:

The voltmeter will now read :

120 + (0.15 X 0.3425) = 120.05
and the apparent power = 120.05 X 34.25 = 41.12, an error of 0.05 per
cent., which is negligible.

130. The Wattmeter. — 1 he wattmeter measures power directly .
It cojisists of fixed coils FF and a pivoted coil M, free to turn
within the magnetic field produced by coilsFi^as shown in Fig. 145.
The coils FF are wound with comparatively few turns of wire
which are capable of carrying the entire current of the circuit.
11

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162

DIRECT CURRENTS

The moving coil M is wound with very fine wire and the current
is led into it through two control springs in the same manner that
current is led into the coil of a Weston instrument. The fixed coil
is connected in series with the load in the same manner as an
ammeter is connected. The moving coil is connected across the
line in series with a high resistance R in the same manner as a
voltmeter coil is ordinarily connected.

The field of the coils FF is proportional to the current and the
current in the coil M is proportional to the voltage. Therefore
the turning moment is proportional to the power of the circuit
and it also depends on the angular position of M with respect
to FF, which is taken into consideration when the scale is
marked.

QQQ^

Fig. 145. — ^The iodicating wattmeter.

Owing to the high degree of accuracy obtainable by the use
of the voltmeter and ammeter, the wattmeter is seldom used for
direct current measurements. As it is subject to stray fields, re-
versed readings should be taken, that is, both the current and
voltage should be reversed and the average of the two readings
used. The wattmeter is used more extensively for alternating
current than for direct current. A more complete description
together with its uses is found in C'lap. Ill, Vol. II.

131. The Watthour Meter. — The watthour meter is a device
for measuring energy. (See Par. 63, page 60.) As energy is the
product of power and time, the watthour meter must take into
consideration both of these factors. As power is usually sold on
an energy basis, many dollars may depend upon the accuracy of

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 163

such a meter. Therefore a proper understanding of its mech-
anism and the method of adjustment is very often essential.

In principle the watthour meter is a small motor whose in-
stantaneous speed is proportional to the power passing through
it, and whose total revolutions in a given time are proportional
to the total energy or watt-hours delivered during that time.

FiQ. 146(a). — Connections of the watthour meter.

■1

Fig. 146(6). — Interior of Thomson watthour meter.

Referring to Fig. 146 (a), the line is connected to two terminals
on the left-hand side of the meter. The upper terminal is con-
nected to two coils FF in series, wound with wire sufficiently heavy
to carry the maximum current taken by the load, which should

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164 DIRECT CURRENTS

not greatly exceed the rated current of the meter. This line
terminates at the upper binding post on the right-hand side of
the meter. These coils FF are wound so that they aid each
other and they supply the field in which the armature rotates.
The other line wire runs straight through the meter to the load.
A shunt circuit is tapped to the upper line on the left-hand side.
It runs first to the armature, through the silver brushes By which
rest on the small commutator C. From the brushes the line
passes through coil F\ and through a resistance R to the lower
line wire. This resistance R is omitted in certain types of meters.

As the load current passes through FF, and there is no iron.
in circuit, the magnetic field produced by these coils is propor-
tional to the load current. As the armature, in series with re-
sistance, is connected directly across the line, the current in the
meter armature is proportional to the line voltage. Neglecting
the small voltage drop in FF, the torque acting on the armature
must then be proportional to the product of the load current and
the load voltage or, in other words, it is proportional to the
power passing through the meter to the load.

It can be proved that if the meter is to register correctly, there
must be a retarding torque acting on the moving element which is
proportional to its speed of rotation. To meet this condition
an aluminum disc D is pressed on the motor shaft. This disc
rotates between the poles of two permanent magnets MM. In
cutting the field produced by these magnets, eddy currents are
set up in the disc, retarding its motion. As the strength of these
currents is proportional to the velocity of the disc and they are
acting in conjunction with a magnetic field of constant strength,
their retarding effect is proportional to the speed of rotation, so
that the condition for correct registration is fulfilled.

Friction cannot be entirely eliminated in the rotating element,
even with the most careful construction. Near the rated load
of the meter the effect of friction is practically negligible, but
at light loads the effect of friction, which is constant, is a much
greater proportion of the load. As the ordinary meter may-
operate at light loads during a considerable portion of the time,
it is desirable that the error due to friction be eliminated. This
is accomplished by means of coil F' connected in series with the
armature. F' is so connected that its field acts in the same direc-

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 165

tion as that due to coils FF, Therefore it assists the armature
A to rotate. Being connected in the shunt circuit, it is acting
continuously. The coil is movable so that its position can be
so adjusted that the friction error is just compensated.

To reduce friction and wear, the rotating element of the meter
is made as hght as possible. The element rests on a jewel bearing
/, which is a sapphire in the smaller sizes and a diamond in the
heavier types. The jewel is supported on a spring. A hardened
steel pivot rests in the jewel. In time the pivot becomes dulled
and the jewel roughened, which increases friction and causes
the meter to run more slowly unless F' is readjusted. The mov-
ing element turns the clock work of the meter dials through a
worm and the gears G,

Fig. 146 (5) shows the interior view of a Thomson watthour
meter.

132. Adjustment of the Watthour Meter. — Even if the initial
adjustment be accurate the registration of a watthour meter
may, in time, become incorrect. This is due to many causes, such
as pitting of the conmiutator, roughening of the jewel, wear on
the pivot, change in the strength of the retarding magnets, etc.
As the cost of energy to consumers is largely based on the reg-
istration of such meters, it is important that they be kept in
adjustment, as a small error in the larger sizes may ultimately
mean a difference of many dollars one way or the other.

To adjust the meter it may be loaded as shown in Fig. 146(a).
The power taken by the load is measured by a caUbrated volt-
meter and ammeter. The revolutions of the disc D are counted
over a period of time which is measured with a stop watch. The
relation between watt-hours and the revolutions of the disc, in
most meters, is as follows:

WXH =^ KXN (63)

where TT is in watts
/f is in hours

K is the meter *' constant^' usually found on the disc
N is the revolutions of the disc.

This equation means that the meter constant multiplied by the
revolutions of the disc gives the watt-hours registered by the meter.
The gear ratios and clockwork take care of the dial registration.

When checking a meter, the time is usually measured in seconds.

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166 DIRECT CURRENTS

Equation (63) then becomes

where t is the time in seconds.

When the meter is tested, the voltmeter and ammeter are read
intermittently while the revolutions of the disc are being counted.
A run of about a minute gives good results.

Let the average watts determined from the corrected voltmeter
and ammeter readings be TFi.

The average watts as indicated by the meter during the same
period are, from (64),

^ ^ Ji: X JNT X 3,600 ^^^

The per cent, accuracy of the meter is

100 W/Wi

Example, — In the test of a 10-amp. watthour meter having a constant

of 0.4, the disc makes 40 revolutions in 53.6 seconds. The average volts

and amperes during this period are 116 volts and 9.4 amp. What is'

the per cent, accuracy of the meter at this load?

Average standard watts TTi = 116 X 9.4 = 1,090.

Average meter watts from (65)

^ 0.4 X 40 X 3,600 _

^ " 53.6 ~ '^

T, . 100 X 1,074 oe- .
Per cent, accuracy = ^ ' = 98.5 An».

This means that the meter is 1.5 per cent, slow and should be speeded
up slightly. With calibrated indicating instruments and careful adjust-
ment, a meter may easily be brought within 0.5 per cent, of accurate regis-
tration.

There are two adjustments to be made. Near full load the
magnets are moved. If the meter is running slow the magnets
are moved nearer the center of the disc where the effect of the
retarding currents is reduced, and if the meter is running fast
the magnets are moved farther from the center. If the meter
has been correctly adjusted near full load, and is found to be in
error near light load, the error is obviously due to friction. The
light load adjustment (made at from 5 to 10 per cent, rated load)
is effected by moving the friction compensating coil F\ If
the meter is slow the coil F' is moved in nearer the armature, and

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ELECTRICAL INSTRUMENTS AND MEASUREMENTS 167

if the meter is fast it is pulled out further from the armature.
This adjustment of F' may affect the full load adjustment slightly

Load

Neutral

Fia. 147. — Diagram of a 3- wire watthour meter.

so that the meter should be re-checked at full load and then again
at light load.

+ \ BU8

Fig. 148. — Astatic heavy current watthour meter.

Other Types of Watthour Meters, — The three-wire meter is
designed to register energy upon a three-wire system. It does

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168 DIRECT CURRENTS

not differ materially from the meter shown in Fig. 146 except
that the two coils FF are connected in opposite sides of the line
as shown in Fig. 147. The armature circuit may be connected
to the neutral as shown or it may be connected across the outer
wires. If this latter connection is used the neutral connection
to the meter is omitted. In the former case the meter does not
register accurately unless the voltages between the two outer
lines and neutral are equal This error is usually small.

The meters already described should not be installed near
bus-bars carrying heavy currents because the strength of the
meter field and of the retarding magnets may be affected by
the stray fields. To eliminate the effect of stray fields an astatic
type of meter is used, Fig. 148. There are two armatures on
the spindle which rotate in the magnetic field created by a single
heavy conductor. One armature is above and the other is
below the conductor. Any stray field will presumably strengthen
the field in which one armature rotates as much as it will weaken
the field in which the other armature rotates so that the result-
ing effect will be nil. There are two sets of retarding magnets.
These magnets are so placed that if the strength of one is in-
creased the strength of the other is decreased. For further pro-
tection these magnets are surrounded by an iron box.

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CHAPTER VIII
THE MAGNETIC CIRCUIT

133. The Magnetic Circuit. — Although the general nature
and characteristics of magnetism were discussed in both Chapters
I and II, no quantitative relations were considered. If the mag-
netic properties of a circuit and the ampere-turns linked with this
circuit be known, the magnetic flux can be calculated in the same
manner that the current in the electric circuit may be calculated
if the resistance and voltage be known. In this respect the two
circuits are similar. The magnetic circuit differs from the elec-
tric circuit in three respects, which makes it difficult to attain the
same degree of precision in magnetic calculations as are obtained
in electrical calculations.

The electric current has been considered as confined to a
known path, for example, a wire. The surrounding air and the
insulating supports for the wire have an extremely high resistance,
so that any leakage current which escapes from the wire is negli-
gible compared with the current flowing in the wire or conductor.
In the magnetic circuit there is no known insulator for magnetic
flux. In fact, the air itself is a fairly good magnetic conductor.
Therefore it is impossible to restrict magnetic lines to definite
paths in the same way that electric currents are restricted. This
is niustrated by the fact that even in the best designed dynamos
from 15 to 20 per cent, of the total flux produced leaks across air
paths where it cannot be utilized. The presence of this leakage
flux may be detected with a compass, and its intensity is often
suflScient to magnetize watches even when they are several feet
distant from the machine.

Magnetic paths are usually short and have large cross-sections
in proportion to their length. They are often so complicated in
their geometry that only approximations to their magnetic re-
sistance can be obtained. This often causes errors of consider-
able magnitude in magnetic calculations.

169

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170 DIRECT CURRENTS

Under ordinaxy conditions of use the resistance of most elec-
tric conductors is substantially constant, although temperature
changes may cause variations of several per cent. Correction for
the effect of temperature changes can be accurately made. The
magnetic resistance of materials, however, is not constant but
varies over wide ranges. This resistance depends to a large
extent on the magnetic history of the material. The magnetic
resistance of iron may easily increase fifty times when the flux
alters from a low to a high magnetic density.

MAGNETIC UmTS

134. Ampere-turns (IN). — The ampere-tiu'ns acting on a
circuit are given by the product of the turns linked with the cir-
cuit and the amperes flowing through these turns. For example,
10 amperes flowing through 150 turns give 1,500 ampere-turns.
The same result is produced by 15 amperes flowing through 100
turns. If any ampere-turns act in opposition, they must be
subtracted.

Magnetomotive Force (mmf. also F). — Magnetomotive force
tends to drive the flux through the circuit and corresponds to
emf. in the electric circuit. It is directly proportional to the
ampere-turns of the circuit and only differs from the value of the
ampere-turns by the constant factor 0.4t = 1.267. That is,
F = 0.47r7iV = 1.257 IN.

The magnetomotive force of a circuit is measured by the work
done in carrying a unit north pole through the entire circuit.

The unit of magnetomotive force is the gilbert, but the name
gilbert is seldom used in commercial work. The gilberts acting
on a circuit are obtained by multiplying the ampere-turns by
0.4t or 1.267.

Reluctance ((R). — Reluctance is resistance to the passage
of magnetic flux and corresponds to resistance in the electric cir-
cuit. The unit of reluctance is that of a centimeter-cube of air.
This unit is called the oersted. The name oersted is seldom used
in commercial work.

Permeance ((P). — The permeance of a circuit is the recipro-
cal of the reluctance ((P = -^) and may be defined as that prop-
erty of the circuit which permits the passage of the magnetic flux

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THE MAGNETIC CIRCUIT 171

or of the lines of induction. It corresponds to conductance in the
electric circuit.

Permeability (m) . — The permeability of a material is the ratio
of the flux or of the number of Unes of induction existing in that
material to the flux or the number of lines of induction which would
exist if that material were replaced by air, the mmf . acting on
this part of the circuit remaining unchanged. The permeability
of air is taken as unity and with the exception of iron, steel,
nickel, liquid oxygen, and certain iron oxides, all materials
may be considered as having a pwmeability of unity. The
permeability of commercial iron and steel ranges from 50 and
even lower to about 2,000. In special investigations, vacuum-
treated iron has attained a permeability of 5, 000 and even greater.

Example, — In a ring solenoid wound on a core similar to that of Fig. 13a,
page 13, the magnetic flux is found to be 4,000 lines or maxwells. When
the iron core is removed the flux in air is but 20 lines. What is the per-
meability of the iron?

Removing the iron core does not change the ampere-turns and the flux
path does not change appreciably. Therefore

M = -^ = 200. Ana.

Flux (0). — The magnetic flux is equal to the total number of
lilies of induction existing in the circuit and corresponds to
current in the electric circuit. The um't of flux is the maxwell,
but "line of induction" or simply ''line" is more often used.

Flvx Density (B), — The flux density is the number of max-
wells or of hues of induction per unit area, the area being taken
at right angles to the direction of the flux. The unit of flux
density in the C. G. S. system is one line per sq. cm. and is called
the gauss. Flux density is usually expressed in ''lines per square
centimeter" or "lines per square inch."

?= <I>/A

where A is the area and <t> the flux through and normal to this
area.

135. Reluctance of the Magnetic Circuit. — The unit reluctance
is defined as that of a centimeter-cube of air. If a portion of a
magnetic circuit between pole faces a and 6, Fig. 149 (a),^con-

* The actual flux path between pole faces would not exist as shown in Fig.
149 (a), but the flux would "fringe" as shown in Figs. 136 and 14, page 13.

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172

DIRECT CURRENTS

sists of a path in air having a length of 3 cm. and a crossHsection
of 1 sq. cm. as shown in the jSgure, this path is equivalent to
three centimeter-cubes placed in series. As the total flux must
pass successively through each cube, it is evident that the total
reluctance is 3 units (oersteds). The reluctance is proportional
to the length of the flux path.

(a) Path whose reluctance is 3 oersteds. (&) Path whose reluctance is ^z

oersted.
Fio. 149. — Reluctance of simple magnetic paths.

On the other hand, if the path has a length of 1 cm. and a
cross-section of 3 sq. cm., as shown in Fig. 149 (6), the reluctance
of the path through which the flux passes is one-third that of one
cube alone, or 3^^ oersted. The reluctance is inversely pro-
portional to the cross-section of the path.

Moreover, if these paths were in iron, having a permeability

ju, the flux would be /a times its
value in air, provided the same
mmf . were maintained between
the two poles faces. This
means a lower reluctance. The
reluctance of any portion of a
magnetic circuit is proportional
to its length, inversely propor-
tional to its cross-section and
inversely proportional to the
permeability of the material.
The constant of proportionahty is unity, since the reluctance of a
path in air 1 cm. long and 1 sq. cm. cross-section is one oersted.
Hence,

Fig. 160. — Reluctances in series.

(Ri =

Aim

where ii = length in cm. of that part of the circuit under
consideration; Ai = the uniform cross-section in sq. cm. of

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THE MAGNETIC CIRCUIT

173

that portion of the circuit; and /xi = the permeability of that
portion of the circuit.

If a magnetic circuit consists of several parts in series as shown
in Fig. 160, the total reluctance is:

(R = (Ri + (R? + (Ra + (R4
= Zi/AiMi + /2M2M2 + /3/ASM3 + U/Aah^, (66)

Permeances in parallel are added together to find the total
permeance just as conductances in parallel are added together to
find the total conductance.
The total permeance

(P = (Pi + (P2 + (P3 + (P4

and reluctances in parallel combine just as resistances in parallel.

1/(R = l/(Ri + 1/(R2 + 1/(R3 + 1/(R4

136, Permeability of Iron and Steel. — The permeability of

iron or steel depends on the quahty of the material, the flux

density and the previous magnetic history.

30 40

H = Gilberts per Cm.

Fig. 151. — Magnetization curve for cast steel.

The relation of the flux in iron or steel to the magnetomotive
force cannot be expressed in simple form. It is necessary to show
this relation by a curve called the "magnetization curve." Such
a curve for one grade of cast steel is shown in Fig. 151. Abscissas
are magnetomotive force in gilberts per centimeter (fl"), and
ordinates are the corresponding flux densities (^).

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174

DIRECT CURRENTS

From A to B the curve is practically a straight line. Beyond
B the flux density increases much less rapidly for a given increase
in magnetomotive force and the iron approaches saturation. The
point C, where the bend in the curve is very decided, is the "knee
of the ciu^e." Beyond C the flux can be increased but slightly
even with a very great increase in the magnetomotive force.
The iron is then said to be saturated. The type of curve shown
in Fig. 151 is called the normal saturation or induction curve.
Fig. 164 shows normal induction curves for other commercial
grades of iron.

leoo

1400

1200

f-iooo

I 800

eoo

400
200

PERMEABILITV CURVE
CAST STEEL

0 2000 4000 6000 8000 12,000 16,000 20.000

BclElux Density - Linei per Sq. Cm«

Fia. 162. — Permeability curve for cast steel.

Fig. 152 shows the permeability curve for this same steel.
Each ordinate is obtained by dividing 5 by H^ for each point of
the curve in Fig. 151. It will be noted that the permeability
varies over a wide range. It begins at a comparatively low
value, increases to a maximum at the point p, and then decreases
to about one-fifth its maximum value.

137. Law of the Magnetic Circuit. — ^The relation between
flux, magnetomotive force, and reluctance, for the magnetic cir-
cuit, is identical with the relation between current, emf., and
resistance for the electric circuit.

0 = ^

- (R

(67)

^ H, the gilberts per cm., is also equal to the lines per sq. cm. in air, since
in air 0 = H/Gi, (R is unity, being a centimeter-cube, so <^ = Hand <f> ^ B,
since the cross-section of the cube is 1 sq. cm.

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THE MAGNETIC CIRCUIT

176

The jlvx is proportional to the magnetomotive force and inversely
proportional to the reluctance of the circuit.

If the magnetic circuit consists of several distinct parts having
reluctances (Ri, 6I2, etc., in series and magnetomotive forces Fi,
Fi, from equation (66)

. _ F1+F2+FZ + ... OAirlN (68)

FiQ. 153. — Ring-type electromagnet.

Example. — ^The ring magnet, Fig. 153, is wound with 250 turns of wire,
through which a current of 1.5 amp. flows. Assume the permeabiUty of
the iron to be 800. Neglecting fringing, determine the flux in the rin&
and also the flux density.

F = 0.4jr X 1.5 X 260 = 471
li = 18 in. - 18 X 2.64 = 46.7 cm.
U « Ke in. = He X 2.64 = 0.476 cm.
Ai = A, = 0.2 sq. in. = 0.2 X 2.54 X 2.64 = 1.29 sq. cm.

From equation (68)

471 471

0 =

46.7 0.476 0.0443 + 0.369

•" 1

1.29 X 800 ' 1.29 X 1.0
The flux density:

1,140 lines (maxwells). Arts.

1 140
B « ~t~nQ — 884 lines per sq. cm. (gausses)

= 6,700 lines per sq. in.

138. Method of Trial and Error. — Magnetic problems cannot
be solved readily by the method used in Par. 137. This is due

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176 DIRECT CURRENTS

to the fact that the permeability (which is a variable but is
given in the problem as a constant value of 800) is not ordinarily
known until the flux density is known and curves similar to those
of Figs. 151 and 152 have been consulted. Therefore the per-
meability is not known until the answer has been determined.
As the answer in turn depends upon the permeability, it is
usually necessary to resort to trial and error.

Example, — The iron ring of Fig. 163 and Par. 137 is made of cast steel
whose permeability curve is given in Fig. 152. The air gap is reduced to
He in. Determine the flux and the flux density.

Assume that the permeability is 800.

__ 18.13 X 2.54 _
^' - 1.29 X 800 - ^-^^

(Ra=^2|:^= 0.123-

^^^ = 2,810 maxwells

^ 0.0446 + 0.123

B = ^^ = 2,180 gausses.

From Fig. 152 the permeability at this density is 980. Therefore (Ri
must be recalculated using the new value of permeability.
^ 18.13 X 2.54 _
^' - 1.29 X 980 " ^-^^^
471
^ = 0.0365 + 0.123 ^ 2,950 maxwells.

The new value of B = 2,290 gausses.

As the value of m corresponding to this flux density is 990 or sufliciently
close to the value 980 just used, the last two values of flux and flux density
are substantially correct.

139. Determination of Ampere-turns. — It was shown in Par.
68, Chap. IV, that the voltage drop per unit length of a conductor
is independent of the total current but depends only upon the
current density and the resistivity of the conductor. In a similar
manner the magnetomotive force per unit length depends only
upon the flux dermty and the reluctivity of the material. This is
proved as follows :

Writing equation (68) for one portion of the circuit,

* = T-

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THE MAGNETIC CIRCUIT

177

Since 0 =

BA,

where A is the cross-«ection of the path,

F
BA = -j-

A^ J) ^ .
F B

F=^

(69)

The magnetomotive force is equal to the product of the flux
density and the length of the magnetic path, divided by the
permeability of the material. To determine the magnetomotive

-X^'^ "^^ '^

i^ ^ " ^j: * ,^ j"^"^' 4;"TnirL

iif,HT t ^l^m^rrlS^ *^"^H¥fHinn J

110- "- :=-' ^.-""'-^ :^, ^u^ ^ ±---^

^^ ^^ ^-'t--"' ^*^^--^

, 100 ;i^^' T: "("i-" If

-5 Art / f^ ^^ ^

fl 90 ^2 ~ ^-"^ _|_

^ e„ , ci^' ^ _ „

afsof^ ^^ __ -

tt ^aI j'^ ^ • "-

i 70 - - - ^ -" -h r _

Z t^ t ^ - ^

1 HI ■ — tii " .-ib<!i-^ '^ ^

•7 W ^ ^ f- — 1 ' ^ ^^^^, ^ - — ]-

5 _ - " ^

3 ^ "' --^-- - __ ^ _„

^ - ^ ~L MAGNETJZATION CURVES

30 ;i^ ' FOR

^^ ' STANDARD iFtON AMD STtCl,

20 C"^'' ^ ^ , IT „„

1-1 Z

^S i: : i.

' fc^ -- ^ - -^

10 20 30 40 GO 60 TO so 90 100 HO 120 130 140 160 100 ITO ISO 190200
Amper* -tarns per Inch

Fig. 154. — Typical magnetization curves.

force for a unit length of a circuit it is only necessary to know the
flux density and the permeability. Instead of plotting the per-
meability against flux density the magnetization curve is usually
plotted with ampere-turns per unit length as abscissas and the
corresponding flux density as ordinates. This is more convem'ent
and avoids using 0.47r and also the permeability. Such curves
are shown in Fig. 154 for various commercial steels used in the
manufacture of electrical machinery.

In problems where flux and the cross-section of the magnetic
paths are known, and it is desired to find the requisite ampere-
turns to produce this flux, the curves just referred to, enable the
solution to be readily obtained.

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178

DIRECT CURRENTS

140. Use of the Magnetization Curves. — ^To illustrate the
use of the magnetization curves the following problem is given.

Example. — Detennine the ampere-turns necessary to produce an air-gap
flux of 750,000 lines in the electromagnet of Fig. 155. The cores are cast
iron and the yoke and pole pieces are cast steel. Neglect fringing and
leakage.

^^^

CMt steel

Fig. 155. — Typical electromagnet.

The flux density in the lower yoke:

«,.^.,,«„

The ampere-turns per inch for a density of 62,500, from Fig. 154 (cast
steel), is 23.

The mean length of flux path is (approximately) 16 in.
IiNi = 16 X 23 - 368

or 368 ampere-tums is required to produce a flux of 750,000 lines in the
lower yoke.

The density in the cores is

750,000

= 46,900

4X4

From the curve (cast iron) the ampere-tums per inch = 118.
As there are two cores, the total length will be 16 in.
I2N2 = 16 X 118 = 1,890

The pole pieces are in every way identical with the yoke, except that the
path is 0.25 in. shorter. This small difference will not make any appreciable
error, so the amperes-turns for the two poles pieces are:

IiNi = 368

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THE MAGNETIC CIRCUIT

179

For the air-gap, the mmf. = 0.4ir/iNr

where B is in lines per sq. cm. and I is in cm.

IN =

0.4t

0.796 Bi

as /x » 1 for the air-gap.
In inch imits

IN = 0.313B'r

i (2.54)

(70)

(71)

Ans,

where B' =» lines per sq. in. and V the length of the path in inches.
The ampere-turns for the air gap then become

/4iV4 » 0.313 X 62,600 X 0.25 = 4,900 (from equation 71).
As all the various parts are in series the total ampere-turns —

368 + 1,890 + 368 + 4,900 = 7,526.

141. Magnetic Calcttlations in Dynamos. — The magnetic
circuits of dynamos have akeady been discussed in tfHP II-

niefalEliix
Leakage Elnz

Axial length of armature stampings and pole-faces = 16 in.
Fig. 156. — ^8-pole 100 R.P.M. 250-volt D.C. generator.

The calculation of the exciting ampere-turns is somewhat com-
plicated by the irregular nature of the air-gap, due to the arma-

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180 DIRECT CURRENTS

ture teeth, air ducts, fringing, etc. The amount of leakage
flux between poles introduces another factor which must be
considered.

As a simple example of such calculations, consider the dynamo shown in
Fig. 156. It is desired to send a flux of 7,500,000 lines from each pole into
the annature. The air gap has an effective length of 0.235 in., after cor-
rection has been made for armature teeth, fringing, etc. The leakage co- *
efficient (ratio of core flux to armature flux) is equal to 1.15.

The paths of the fluxes from the various poles, including the leakage flux,
are shown in the figure. The lengths of path are easily determined. Con-
sider the flux path ahcdef.

The length ob = 0.235 = 10.8 in. (approximately)

5c (approximately one-eighth the mean circumference of the yoke, less
5 in.) = .^^^ - 5" = 24.7" - 5" = 19.7 in.

^A /e =^= 12.6 in.

The flux densities are as follows:

Flux in cores = 7,500,000 X 1.15 = 8,630,000 as the flux in the core is

equal to the armature flux plus the leakage flux.

T:n J x • 8,630,000 - . ^^

Flux density m cores = ^^ J if^ = 54,000.

Flux density in yoke = o'/ifi v '\) " 90,000 as the pole flux divides,
one-half going each way in the yoke.

Flux density in armature = ^' J ia\ ~ 39,000.

This must be increased about 25 per cent, to allow for the air duct space
and the spaces between laminations.

This makes the density in the armature:

39,000 X 1.25 = 48,800
The air-gap density = ^'^^'^ = 39,000

Knowing the above factors, and utilizing the magnetization curves of
Fig. 154, it is a comparatively simple matter to determine the total ampere-
turns per pole.

For 54,000 lines per sq. in., 19 ampere-turns per inch are necessary for
cast steel (Fig. 154). Therefore for ah:
^Core ah IiNi = 19 X 10.8 = 205 (cast steel).

Yoke 6c I2N2 = 64 X 19.7 = 1,260 cast steel).

' Core cd hNz = hNi = 205 (cast steel).

^Gap de UNi = 0.313 X 39,000 X -'

0.235 = 2,870 (air). (See equation 71)
Arm. efl^i = 3 X 12.6 = 38 (O. H. sheet steel).

- Gap fa ItNi = IaNa = 2,870 (air).

Total = 7,448 ampere-turns.

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THE MAGNETIC CIRCUIT

181

As two poles in series supply the excitation for this flux the ampere-turns
per pole are

IN = 7,448/2 = 3,724. Am,

As this machine is symmetrical, each complete magnetic cir-
ciiit requires this same number of ampiere-turns per pole. The
design of the exciting coils themselves is not a diflScult matter.

142. Hysteresis. — If the magnetomotive force acting on an iron
sample begins at zero and increases, the relation between mmf,
and the flux (or flux density)
will be similar to that shown
by curve Oa (Fig. 157). This
curve is called the normal
saturation or magnetization
curve and has already been
discussed.

If the magnetomotive force
now decreases, the flux will
not decrease along the line
aOy but will decrease less
rapidly along ab. When
point b is reached, the mmf.
is zero but the magnetic in-
duction has not reached zero.
The flux density Ob is called
the remanence. Before the
flux density can be reduced
to zero, the magnetizing force
must be reversed in direction.
That is, it requires a negative

magnetizing force Oc to reduce the flux density to zero.
The magnetizing force Oc is called the coercive force.

If now the magnetizing force be increased in the negative direc-
tion to df where Odf = 0<x', the flux density will be carried to a
negative maximum d'd The negative maximum flux density d'd is
equal to a'a. If the magnetizing force is now increased toward
zero, the curve will pass through point e when the magnetizing
force is again zero and the negative remanence Oe = Ob. A posi-
tive coercive force Of = Oc is necessary to bring the flux density
again to zero. When the magnetizing force again becomes Oa'

Fig. 157. — Hysteresis loop.

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182

DIRECT CURRENTS

the flux density will return to its original value at a, closing the
loop.

This is one complete cycle of magnetization, and the curve is
called a hysteresis loop. Such a loop shows that the magnetiza-
tion in iron lags behind the magnetomotive force per centimeter
or the magnetizing force, and that an expenditure of energy is
required to carry the iron through a cycle of magnetization.

If several loops are taken, each having different maximum flux
densities, they will have the appearance of the three loops shown
in Fig. 168. The maximum points a, ai, a^ all lie along the nor-
mal saturation curve Oa2.

^^^

^Ut

■^ i4 .-- ;

-^^^

** J^ '"^ /

U^11-/i

-Ulmf-

lltiti-

itUtt-

{. rtl'^_

H-ail iwrt

ipJ

60 4

[) 80 20 1^

uuW-

20 ii

I'-i

U7^ -

-^W -

- it-

-W -

1P-

1 A

3 — ="

1<?

FiQ. 158. — Hysteresis loops for three maximum flux densities.

143. Hysteresis Loss. — The hysteresis loss is proportional to
the area of the hysteresis loop, Figs. 167 and 158. In fact the
hysteresis loss may be obtained by finding the area of the loop
to scale, and dividing by 4jr. This gives the loss in ergs per cycle.
For example, let the area of the smallest loop, Fig. 168, be A
sq. in. The scale is such that 1 in. on the abscissa scale re-
presents 10 gilberts per cm., and 1 in. on the ordinate scale
represents 4 kilogausses. The ergs loss per cycle is:

T^, A X 10 X 4,000

^'^ ^

ergs.

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THE MAGNETIC CIRCUIT 183

To convert this energy loss into joules or watt-seconds divide
by 107.

The hysteresis loss per cycle depends upon two factors, the
magnetic material and the maximum flux density. The loss
within certain limits may be expressed by the Steinmetz Law as
follows:

Wk = nB^'^ (72)

Wh is the hysteresis loss per cu. cm. in ergs per cycle, 17 is a
constant depending on the material, and B is the maximum flux
density in gausses.

Below are given a few typical values of »;:

Hard cast steel 0.025 Sheet iron 0.004

Forged steel 0.020 Silicon sheet steel 0.0010

Cast iron 0.013 SiUcon steel 0.0009

Example, — What will be the ergs loss per cycle in a core of sheet iron
having a volume of 40 cu. cm., in which the maximum flux density is 8,000

Wh = 0.004 X 8,000i.«

log 8,000 = 3.9031
1.6 X 3.9031 = 6.2449
log 1,767,000 = 6.2449

Wh = 0.004 X 1,757,000 = 7,028 ergs per cu. cm. per cycle.
Total loss W = 7,028 X 40 = 281,QP0 ergs per cycle. Ava.

INDUCTANCE

144. Linkages. — If a current flows in a conductor, a magnetic
flux is set up about the conductor. This magnetic flux completely
encircles the conductor and the current in the conductor com-
pletely encircles the flux. Some familiar examples of this are
given in Fig. 159, where the currents and related fluxes are shown.
As a current and the resulting flux always completely encircle
each other they are said to link with each other. This is shown
particularly^ well in Fig. 159 (c), where a conductor carrying a
current is linked with an anchor ring.

• The product of the turns of conductor and the number of lines
of flux linking these turns is called the linkages of the circuit.

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184

DIRECT CURRENTS

Example, — A certain solenoid has 800 turns. A current of 5 amp.
flowing in the winding produces a flux of 2,500,000 lines. What are the
linkages?

800 X 2,500,000 = 20 X 10» linkages.

The number of these linkages per unit current in a circuit is
called the inductance of the circuit and is represented by the sjon-
bol ''L, " implying linkages. The unit of inductance is the henry.

Oarrent

(5) Gurren

Fig. 159. — Illustrations of flux-current linkages.

Inductance from definition :

L =

N<t>

I X 10»

(73)

where L is the inductance in henrys, <t> is the flux in maxwells,
and / is the current in amperes.*

Note, — It is necessary to divide by 10* because 10* magnetic
lines are equal to one line in the practical system of volts, am-
peres, etc.

Example. — What is the inductance of the above circuit?

, 20 X 10« .^,
^=-5OO0i- =4.0 henrys.

145. Induced Electromotive Force. — If the terminals of an in-
sulated coil, Fig. 160 (a), be connected to a galvanometer, and a
magnetic field be set up through this coil, either by .thrusting a
bar magnet into the coil or by some other means, the galvanom-
eter will be observed to deflect momentarily and then to return

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THE MAGNETIC CIRCUIT

185

to rest. This shows that an emf . has been temporarily induced in
the coil. When the flux through the coil has ceased to change,
this electromotive force also ceases. If investigation be made, it
will be found that the direction of this induced electromotive
force is that shown in the figure and that this direction is such
that if the emf. be allowed to produce a ciurent, this current will
tend to push the bar magnet ovi of the coil, or what is the same
thing, will oppose its entering the coil.

If the magnet be withdrawn from the coil. Fig. 160(&), the
galvanometer will be observed to deflect again, momentarily as
before, but the deflection is opposite to its direction in the first

(a) North pole inserted in coil

(6) North pole withdrawn
Fig. 160. — Induced electromotive force.

case. The direction of the induced electromotive force is now
such that if the emf. produces a current, this current will tend
to prevent the magnet from being withdrawn from the coil. The
electromotive force in each case is transient and ceases when the
chmge of flux through the coil ceases.

If careful measurements be made, the value of this electro-
motive force will be f oimd to depend upon : (1) the number of turns
in the coil, (2) the rate at which the flux linked with the coil
changes.

The average electromotive force in volts is given by

N<t> 10-8

e =

(74)

where N is the number of turns in the coil, <t> is the total change of
flux in lines linked with the coil, and t is the time in seconds re-

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186 DIRECT CURRENTS

quired to insert or withdraw this fliix from the coil. 10~* reduces
the flux 0 to practical units so that e becomes volts. The minus
sign indicates that the induced emf . is in opposition to the force
which produces it.

J is the average rate of change of flux, so that the induced

electromotive force may be said to be proportional to the number
of turns and the rate of change of flux.

Example. — A flux of 1,600,000 lines links a coil having 360 turns. This
flux through the coil is decreased at a uniform rate to zero in 0.2 second.
What is the induced electromotive force during the time of withdrawal?

e=350i^^l0-.
= 26.25 volts. Ans,

The fact that the currents produced by induction oppose the
motion producing them should be carefully noted, for this prin-
ciple is manifest in practically all types of electric machinery.
This principle was first formulated by Lenz, in a form known as
LfCnz's Law which says:

^^ In all cases of electromagnetic indiLction, the induced currents
have such a direction that their reaction tethds to stop the motion
which produces them.'^

This law is also based upon the law of the conservation of
energy. That is, the induced currents, which represent energy,
are produced at the expense of the mechanical energy required
to push the magnet in the coil against their opposition, or the
energy required to withdraw the magnet against the opposition
of the induced currents, which try to prevent this withdrawal

146. Electromotive Force of Self-induction. — If a coil be con-
nected to a battery and a switch S closed (Fig. 161), current will
begin to flow in the coil. This current produces a flux linking the
coil. As this flux increases it must induce an emf. in the coil,
the magnitude of which depends on the number of turns in the
coil and the rate at which the flux increases. By Lenz's Law,
and also from a consideration of Fig. 160(a), the electromotive
force thus induced must have such a direction as to oppose the
increase in the flux linking the coil and hence must oppose any
increase of current. Therefore this current cannot reach its

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THE MAGNETIC CIRCUIT

187

maximum value at once, but is retarded in its rise by the
opposing electromotive force.

In Fig. 162 is shown the rise of current in a circuit containing
resistance only, the impressed voltage being 10 volts and the re-

Flux

Fig. 161. — Relation of emf. of self-induction to current.

sistance 20 ohms. When the switch S is closed the current
reaches its maximum or Ohm*s Law value of 0.5 ampere at once.
In the case of the inductive circuit, the current approaches
its Ohm's Law value more or less slowly. To be exact, it takes
an infinite time for the current to reach its Ohm's Law value, al-

o.e

0.4

^0.8

<0.2

0.1

^=10V.
i?=20JV
L= 0

8i
01

litch

.02 .04 .06 .08 .10 .12 .14
Second
Fig. 162. — Rise of current in a non-inductive circuit.

though in a comparatively short time it reaches substantially
this value. An idea of the time required to build up a current
in an inductive circuit may be obtained from the inductance and
the resistance of the circuit. The ratio of the inductance in
henry s to the resistance in ohms, L/R^ is called the time constant
of the circuit. This is the time in seconds required for the current
to reach 63.2 per cent, of its final value. It is a measure of the

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188

DIRECT CURRENTS

rapidity with which the current in a circuit rises to its ultimate
value. Fig. 163^ shows the rise of current in a circuit whose
impressed voltage is 10 volts, resistance 20 ohms and inductance
0.6 henry. The time constant of this circuit is 0.6/20 = 0.03
seconds and is shown on the diagram. This curve should be
compared with Fig. 162, in which the circuit has the same inr^
pressed voltage and the same resistance but has no inductance.

°~6.

0 6

0.4

£0.8

O0.2

0.1
Switch rloBed>«

10 V

«=
h =

20 n'

0. e Henn

b/a

t^i/R

.02 .01 .06 .06 aO .12 .14 .16 .18
Second

FiQ. 163. — Rise of current, in an inductive circuit.

Example. — ^A relay having a resistance of 400 ohms and an inductance
of 0.4 henry is connected across a 110-volt circuit. What is the time con-
stant of the relay? To what value does the current in the relay rise in this
time?

04
The time constant = jj^n ~ ^-^^ second.

i = 0.632^ = 0.1738 amp.

This delayed rise of current in a circuit due to self inductance
should be carefully kept in mind, since it accounts for some of
the time lag observed in relays, trip coils, etc. When a short-
circuit takes place there may be considerable delay between the
time at which the short-circuit occurs and the opening of the
breaker or switch controlled by the relay. The effect of induc-
tance is also one of the controlling factors in the initial current-
rush on short-circuit.

If an inductive circuit carrying current be short-circuited, the
current does not cease immediately, as it does in a non-inductive

^ The equation of the curve showing the rise of current is*="D(l"~*ir)

where E is the impressed voltage, i == current at time, t seconds after closing
switch and e the Napierian logarithmic base. The current increases at a
rate of E/L amp. per second at the instant when the switch is closed. <

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THE MAGNETIC CIRCUIT

189

circuit under similar conditions, but continues to flow and does
not become zero until an appreciable time after the instant of
the short-circuit. This is due to the electromotive force of self-
induction. The flux linking the coil is due to the current, and
when the current decreases, this flux also decreases. In decreas-
ing, the flux induces an electromotive force in the coil. In the
same way that the current due to the induced electromotive
force tended to prevent the flux being withdrawn in Fig. 160(6),
so now the electromotive force of self-induction tends to prevent
the decrease of the current.

0.6

A A

Swil

«hcl^

wed

r—

u n 9

E-

10 V

|0^
5 0-2

20fi
0.8£«nry

0.1

0 .02 .04 .06 .08 40 .12 .14 .16
Second
FiQ. 164. — Decay of ctirrent in an inductive circuit.

A curve ^ showing the decrease of the current with time is given in
Fig. 164. The circuit has the same constants as the circuit shown
in Fig. 163. It is usually advisable to fuse the battery so that it
will not be injured, since short-circuiting the inductive circuit also
short-circuits the battery, as is shown in Fig. 164.

It thus appears that the effect of inductance is always to oppose
any change in circuit conditions. If the current tends to in-
crease, inductance opposes it; if it tends to decrease, inductance
tends to oppose this decrease. Inductance corresponds to inertia
in mechanics. A body having inertia opposes any force tending
to set it in motion when the body is at rest, and if the body is in
motion, inertia opposes any force tending to bring the body to
rest.

^ The equation of this curve,

- Rt

i = he ^

where i is the value of the current at a time, t seconds after the closing of
the switch, and /o is the initial value of current.

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190

DIRECT CURRENTS

If, after having established the current in the circuit of Fig. 163,
the switch S be opened, a noticeable arc will appear at the switch
blades. This arc will be much greater in magnitude than that
formed at the contacts of the switch in the circuit of Fig. 162,
with resistance only in the circuit, although the current and cir-
cuit voltage are the same in each case. This arc is due to the
electromotive force of self-induction and in some circuits may
have such a value as to cause severe al*cing at the switch contacts.
In fact this voltage has been known to reach such values in
alternator fields as to puncture their insulation when the field

AWVWVVV

OQQQQQQQOQQQQOOQ.

.Field DiM:harge

Field

Fig. 165. — Field-discharge switch with connections.'

circuit is opened. To protect the field from puncture, a field
discharge switch shown in Fig. 165 is often used. At the instant
of opening the switch the field (and the hne temporarily) is
paralleled by the field discharge resistance. The energy of the
field is dissipated partly in this resistance rather than at the
switch contacts. Contact with switches opening inductive cir-
cuits, even in the case of very low voltages, should be carefully
avoided. Not only is there the danger of being burned by the
arc, but of being injured from the high induced voltages as well.
Calculation of the Electromotive Force of Self-induction. — From
equation (74) page 185, the electromotive force induced in a coil
due to a change in the flux hnking the coil is

e =-Ar|lO-»

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THE MAGNETIC CIRCUIT

191

where N is the number of turns, and <t>/t the rate at which the
flux changes.

Remembering that

L = -^ 10-» or N<l> 10-8 = jr,7 (equation 73, page 184),

and also that the electromotive force of self-induction opposes the
change in current, its value may be writtenic^

The electromotive force ol self induction is proportional to the
product of the inductance and the rate of change of current with
respect to time. The minus sign indicates that this electro-
motive force opposes the change of current.

If the inductance varies as well as the flux, equation (75) may-
be written:

--i.L\±l\',

(76)

the additional term accounting for the electromotive force due to
any change in the inductance.

Example, — The field circuit of a
generator has an inductance of 6
henrys. If the field current of 12
amp. is interrupted in 0.06 second,
what is the average induced electro-
motive force in the field winding?

e = 6 g^ = 1,440 volts.

Ans.

FiQ. 166.-

a suspended

147. Energy of the Magnetic
Field. — To establish a magnetic
field energy must be expended.
To maintain a constant field
does not require an expending
of energy even in electromagnets.
The energy lost in the exciting
coils of electromagnets is ac-
counted for as heat in the copper and is not concerned with the
energy of the magnetic field itself. The energy of the magnetic
field is stored or potential energy and is similar to the energy of
a raised weight, Fig. 166. Work is performed in raising the

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-Energy of
weight.

192 DIRECT CURRENTS

weight to its position, but no expenditure of energy is required
to maintain the weight in this position. The energy of the weight
due to its position is Wh foot-pounds, where W is the weight in
pounds and h the height in feet through which the weight has
been raised. This energy is available and can be utilized in
many ways.

In the same way the energy stored in the magnetic field is
available and may make itself manifest in many ways, as, for
example, the arc at the switch contacts. In an alternating cur-
rent circuit this energy may all be returned to the circuit.

The energy of the field in joules, or watt-seconds, is

W = 1/2LP (77)

where L is the circuit inductance in henrys and / the current
flowing.

Example, — ^In a circuit having an inductance of 4 henrys, the current is
10 amp. What is the energy of the magnetic field.' If this circuit is
iatemipted in 0.2 second, what is the average value of the power expended
by the magnetic field during this time?

W = 1/2 X 4 X 102 = 200 watt-seconds. Ans,

200
P = ^ = 1,000 watts = 1 kilowatt. Arts.

Equation (77) shows that the energy of the magnetic field
is proportional to the square of the current. Therefore if
the current can be reduced by a suitable resistance to one-half
its initial value before opening a highly inductive circuit, the
energy of the arc at the switch contacts can be reduced to one-
fourth of its initial value. This fact should be remembered when
opening the field circuit of a dynamo.

A very common use of the electromotive force of self-induction
occurs in the so-called spark coil used for gas lighting. This
coil consists of a considerable number of turns of wire wound on a
laminated iron core. The core is usually made of iron wires as
shown in Fig. 167. This coil is connected between the bell-
ringing battery B and the grounded gas pipe. The other
terminal of the battery is connected directly to the insulated
contact on the gas burner. When the two contacts on the burner
meet, the circuit is closed, and a magnetic field is established in
the laminated core of the spark coil. As the two contacts of the

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THE MAGNETIC CIRCUIT

193

burner separatCj they snap apart and the circuit is broken sud-
denly^ Consequently, a high electromotive force of self-induction
is produced in the spark coil. This causes a hot arc at the con-
tacts, which ignites the gas, the gas being turned on simultaneously
with the closing of the contact points and by the same mechanism.

Gas Cock

'^Ground Connection
FiQ. 167. — Electric gas ignition.

The spark coil may be considered as having a magnetic field
which is built up as the two contacts at the gas jet wipe by each
other. Energy is thus stored in the magnetic field. When
this energy is released suddenly by the contacts snapping open,
considerable power is developed resulting in a hot spark at the
contact points.

148, Mutual Inductance. — In Fig. 168 are shown two coils,
A and B. Coil A is connected to a battery through a switch >S.
Coil B is not connected to any source of voltage, but to a gal-
vanometer. Coil B is placed so that its axis is nearly coincident
with that of A and the two coils are close together. When the
switch S is closed, current flows in coil A, building up a field
which links the coil. The position of B with regard to A results
in a considerable part of the magnetic flux produced by A linking
B, Therefore, if the current in A be interrupted by opening
the switch S, or if it be altered in magnitude, a change of flux
simuUaneovsly occurs in B inducing an emf. in S. This emf.
is detected by the galvanometer connected across the terminals
of B. Upon closing the switch S the galvanometer will deflect
momentarily, and upon opening the switch S its deflection will
reverse, showing that the induced voltage on opening the circuit
is opposite in direction to the induced voltage on closing the

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194

DIRECT CURRENTS

circuit. Because coil 5 is in such a relation to A that an emf .
is induced in B due to the change of flux in A, these two coils
are said to possess mulvxiL inductance. The induced emf. is an

Fio. 168. — Mutual inductance between two coils.

electromotive force of mutual induction and its magnitude,
equation-(73), page 184, is

62 =^ N^ ^ 10-8 volts

where N2 is the number of turns in coil B, 02 the change in mag-
netic flux from coil A which links coil B, and t the time in seconds
required to change the flux by 02 lines.

Even though coils A and B be brought close together, all the
flux, 01, produced by coil A does not link coil B. Only a certain
proportion, if, of <t>i links B, K being less than unity. That is:

62 = iV^2 ^ 10-8 volts

(78)

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THE MAGNETIC CIRCUIT 195

K is often called the coefficient of coupling of the circuits A and B,
As N2 and K are constants for any particular circuit and position
and 01 may be assumed proportional to /i, the current in coil
Ay equation (78), may be written

62 = Af^ 10-8 volts (79)

where Af is the miUual inductance or coefficient of mutual in-
duction in henrys between coil A and coil B.

M = :^^ (80)

The mutual inductance of two circuits may be defined as that
factor which when multiplied into the time rate of change of
current in one circuit, gives the induced voltage in the other
circuit.

Example,— CoiX A (Fig. 168) has 400 turns and coil 5 has 600 turns.
When 5 amp. flow in coil A, a flux of 600,000 lines links with A, and 200,000
of these lines link coil B, What is the self inductance of coil A with B
open-circuited, and what is the mutual inductance of the two coils?

J N4>i 400 X 500,000,^., ^., ,

Li = —J- = = — r^ 10 ' - 0.4 henry. Ans,

I o

The induced voltage in B due to the current in A rising to 5 amp. in
1 second will be

e, = JVj Y = 600 X 200,000 X lO"' = 1.2 volts.

as a change of 5 amp. in coil A changes the flux in coil B by 200,000 Unes.
Therefore:

1.2 =3f Xj

M = 0.24 henry. Ana,

or using equation (80)

,, 0.4 X 600 X 500,000 ,^_. ^ ^ . ,

M - = 10 ' = 0.24 henry.

The mutual inductance of two circuits may be materially in-
creased by linking the circuits with an iron core. Thus, if the
two coils, similar to those shown in Fig. 168, be placed upon an

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196

DIRECT CURRENTS

iron core (Fig. 169) the coefficient of coupling, Ky may be made
very nearly unity. That is, practically all the flux linking coil A
also links coil B.

FiQ. 169. — Effect of iron core upon mutual inductance.

A very common example of mutual inductance occurs in the
induction coil (Fig. 170). A primary winding, P, of compara-
tively coarse wire and few turns, is wound on a laminated iron
core C. This winding is connected to a battery B, The primary
current is interrupted by passing through the contact D, against
which the iron armature A is held by a spring. When the core C

Spark>^

FiQ. 170. — Induction coil.

is magnetized by the primary current, the armature A is drawn
toward it and away from D, opening the circuit and causing the
flux in the core to drop practically to zero. The spring then
pulls the armature A against the contact D again, and the cycle
is repeated. By this process the flux in the core C is continually
being established and then destroyed.

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THE MAGNETIC CIRCUIT , 197

On the same core is placed a secondary winding, /S, consisting of
many turns of fine wire. This winding is thoroughly insulated
from the primary winding, but as it is wound on the same core
as P, the two coils have a high value of mutual inductance.
Because of the change of flux in the core, due to the interruptions
of the primary current, a high alternating emf. is induced in
the secondary. This induced electromotive force may be con-
sidered as due to the mutual inductance existing between the
primary and the secondary coils. The induction coil has many
practical applications. Its wide use in automobile and gas en-
gine ignition systems is important.

149. Magnetic Pull. — It has been shown that a force exists
between magnetized surfaces. This force can be accurately
calculated if the surfaces are parallel and quite close together,
being given by

where / is the force in dynes, A the area of each of the two sur-
faces in square centimeters, and B the flux density in gausses.
This becomes :

B^A
P = 24:64 kilograms

if B is expressed in kilolines per sq. cm.

F = -^-^^^lb
72,130,000

if B is in lines per sq. in. and A in square inches.

Example. — The core of a solenoid is 2 in. in diameter and a total flux of
200,000 lines passes from the end of the core into an iron armature of
• equal area. What is the pull on the armature in pounds?

A = ^ (2)2 = 3.14 sq. in..

B = 200,000/3.14 - 63,800 lines per sq. in.
„ 63,8002 X 3.14

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CHAPTER IX
ELECTROSTATICS: CAPACITANCE

So far, electric currents, or electricity in motion, has only been
considered. Electricity when in motion is called dynamic
electricity. Electricity may, however, be stationary or at rest.
Under these conditions the electricity is called static electricity.
There is no difference in the nature of static and dynamic elec-
tricity. The static electricity usually appears different because
of its extremely high potential and small quantity.

160. Electrostatic Charges. — If the terminals of an electro-
static induction machine be connected to two equal ellipsoids.

FiQ. 171. — Electrostatic charges on insulated ellipsoids.

which are conducting and are insulated. Fig. 171, the ellipsoid
connected to the positive terminal will be charged with positive
electricity and that connected to the negative terminal will
be charged with an equal amount of negative electricity. The
charges will distribute themselves over the entire surface of the
ellipsoids, but the density of the charges will be greatest on the
ends of the ellipsoids which are adjacent. This is due to the
fact that the positive and negative charges attract each other.

If the two wires from the electrostatic machine be disconnected
the two charges will not be sensibly affected. In time they
will leak away through the insulating supports.

198

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ELECTROSTATICS: CAPACITANCE

199

If the two ellipsoids were free to move they would come to-
gether. If they were comiected together with a wire a spark
would be observed at the instant that contact was made, showing
that current flows for an instant from one ellipsoid to the other.
Both of the above effects are due to the fact that the positive
and negative charges attract each other.

161. Electrostatic Induction. — If a positively charged el-
lipsoid A (Fig. 172(a)) be brought near another insulated el-
lipsoid Bj which initially had no charge, a minus charge will be
found on the end of B nearest A. As B did not hold any charge
initially, and it is assumed to be perfectly insulated, no electricity
can have gone out from B and none can have reached it from

c3 c::^) c>

ia) (h)

Fig. 172. — Electrostatic induction.

external sources, so that the net charge on B must still be zero.
Therefore, a positive charge 6' must also appear on B at the
outer end farthest from A. This charge must be equal to, 6,
and as the two are of opposite sign the net charge on B is
still zero. It will be noted that the minus charge 6 is as near
as possible to the positive inducing charge a, whereas the positive
charge 6' is as far away as possible from the positive charge a.
This is due to the fact that the unlike charges attract each
other and that like charges repel each other.

Also charges a and h are called bound charges, and charge V
is a free charge. This may be proved by connecting B to ground
(Fig. 172(6)). The charge V will be found to have escaped to
ground, whereas the two charges a and h remain. Charge V
will seek a position as far away from a as possible.

If a were a negative charge, h would be a positive charge.

The above experiments are all illustrative of the following
laws of electrostatics.

Charges of unlike sign attract each other and charges of like sign
repel each other.

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200 DIRECT CURRENTS

A positive charge will indiice a negative charge on a body near it,
or

A negative charge will induce a positive charge on a body near it.

This is similar to magnetic induction, where a north pole
induces a south pole, etc. (See Par. 16.)

162. Electrostatic Lines. — Unit electrostatic charge is defined
as that charge which, if placed 1 cm. distant from an equal
charge in air, will be repelled with a force of 1 dyne.

If a unit positive charge, P, which can move freely, be placed
at various points in the field near two oppositely charged bodies,
it will be found to move along certain well defined paths, the
path in each case being determined by the point at which the

Fig. 173. — Electrostatic field between charged conductors.

unit charge starts. The unit charge starting from the posi-
tively-charged body will always move along a definite path until
it reaches the negatively-charged body. The several paths
which such a charge may follow are shown in Fig. 173. This is
similar to the behavior of a unit north pole when placed in a
magnetic field. When a difference of potential is produced
between two conductors an electrostatic^^ld also results. The
intensity of this field at any point is equal to the force which is
exerted on a unit positive charge at that point. Such a field
may be represented by lines just as with the magnetic field.
The density of the Unes represents the field intensity. The
field between two irregular bodies is sketched in Fig. 173, the
lines of force being represented by the paths which the unit
charge would follow if allowed to move freely.

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ELECTROSTATICS: CAPACITANCE

201

PfT^^^

An electrostatic line of force begins at a positively-charged
conductor and ends at a negatively-charged conductor. In this
respect it resembles a magnetic line of force which begins at a
north pole and ends at a south pole. The electrostatic line of
force is not like a magnetic line of induction which is always a
closed curve. (See Par. 11.)

Electrostatic lines of force distribute themselves exactly as do
the flow lines or stream lines of an electric current, or the magnetic
lines in a magnetic field.

There is one difference, however, between electrostatic lines,
on the one hand, and magnetic lines and electric current lines on
the other. No matter how much current
flows in a conductor, the conductor is not
injured mechanically, provided it can be
kept cool. Neither is a magnetic con-
ductor injured, no matter how many
magnetic hnes exist in it. But there is a
limit to the number of electrostatic lines
which may exist in a medium. If the
lines become too concentrated the medium
cannot withstand the stresses which result
and it is ruptured or "breaks down.''
This break-down may be followed by a
dynamic arc, which increases the injury to
the medium by burning.

In a gaseous medium it is possible for a partial break-down
to occur. Let a needle point in air, Fig. 174, be raised to a
high potential above a plate. The electrostatic lines will be con-
centrated at the needle point but will be spread out over the plate.
As the stress is most highly concentrated at the needle point,
the air will obviously break down at this point first. This break-
down can be detected by the blue glow or corona^ which appears
around the needle point, and at the same time an odor of ozone is
evident. Complete rupture cannot occur between the point
and the plate, at least at first, because the air beyond a certain
region aa is still not stressed to the break-down point.

As the potential is raised, however, the boundary of the
disrupted region will advance to 66, and will continue to

1 See Chap. XII, Vol. II.

Fig. 174. — Electro-
static stress lines be-
tween a needle-point
and a plate.

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202 DIRECT CURRENTS

advance with increasing potential until the remaining air can
no longer support the stress, when complete break-down
takes place.

Dielectrics. — If electrostatic phenomena are being considered,
the medimn between two conductors is called a dielectric. This is
in distinction to the properties of the same medium, as an in-
sulator which relates to electrical conduction. For example, air
is not a particularly good dielectric, its dielectric strength being
only about 75,000 volts to the inch, but it is one of the best in-
sulators known.

The ability of a substance to resist electrostatic break-down
is called its dielectric strength. This is expressed in volts per
unit thickness when the substance is placed between flat electrodes
having rounded corners. For example, the dielectric strength
of air is approximately 3,000 volts per mm. Rubber and var-
nished cambric have a much greater dielectric strength than
air, that of rubber being in the neighborhood of 16,000 volts
per mm., or 400,000 volts per in., and that of cambric being
about twice as great as the value for rubber.

The volts per unit thickness impressed across a dielectric is
called the voUage gradient. For example, if 24,000 volts are im-
pressed across 30 mils of insulation, the gradient is 24,000/30 or
800 volts per mil.

163. Capacitance. — Two conductors separated by a dielectric
is called a condenser.

Battery

-y^

Condenser

Fio. 175. — Charging and discharging a condenser.

Fig. 175 shows two conducting plates connected to a battery,
the plates being separated by a dielectric. There is also a single-
pole, double-throw (S.-P.D.-T.) switch S and a galvanometer G in
the circuit. If the switch S be closed to the left, the galvanometer
vvill deflect momentarily, and then come back to zero. This
indicates that when the switch is closed, a quantity of electricity

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ELECTROSTATICS: CAPACITANCE

203

passes through the galvanometer, but that the current ceases
to flow ahnost inunediately. This current flows for a time
sufiicient to charge the condenser. After the condenser has
become fully charged, the current ceases because the emf. of
the condenser is equal and opposite to that of the battery. As
this condenser emf. opposes the current entering the condenser
it may be considered as a back emf. Any current which may
flow after the condenser has become fully charged is a leakage
current flowing through the insulation. If the switch S be opened
for a short time, and then closed again, no deflection of the gal-
vanometer will be noted unless there is leakage through the
insulation.

This phenomenon of charging a condenser from a battery is
not unUke the filling of a tank T from a reservoir R, Fig. 176.

Reservoir

FiQ. 176. — Reservoir and connected tank.

When the valve V is first opened, water will rush through the
connecting pipe and will continue to flow at a diminishing rate
until the level f?, of the water in the tank T, is equal to the level
of the water in the reservoir. If the tank does not leak, no water
flows through the pipe after the water levels have become equal.
In the same way the condenser. Fig. 175, takes current until
its potential is equal to that of the battery, after which current
ceases to flow.

To prove that electricity has actually been stored in the con-
denser. Fig. 175, the switch S may be closed to the right. This
short-circuits the condenser through the galvanometer. The
galvanometer now deflects momentarily in a direction oppo-
site to that on charge, showing that the current now flows
ovt of the positive plate. The condenser now becomes com-
pletely discharged, as is shown by there being no longer any
deflection of the galvanometer.

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204 DIRECT CURRENTS

If the voltage of the battery, Fig. 175, be increased, the gal-
vanometer deflection on charge and on discharge will increase
also. This is due to the fact that the charge given to the con-
denser is proportional to the voltage across its terminals, just as
the amount of water in the tank will be proportional to its
height H (Fig. 176). The relation between the voltage, and the
charge in a condenser may be expressed by the equation:

Q =^ CE (81)

That is, the quantity of electricity in a condenser is equal to the
voltage multiplied by a constant C. This constant C is called
the capacitance of the condenser. The practical imit of capaci-
tance is the farad. If C is in farads and E in volts, Q is in coul-
ombs or ampere-seconds.

The farad is too large a unit for practical purposes, as a con-
denser having a capacitance of 1 farad would be prohibitively
large. The capacitance of the earth as an isolated sphere is less
than one thousandth of a farad. The microfarad^ equal to one
millionth of a farad, is the unit of capacitance ordinarily used.

By transposition, equation (81) may be written as follows:

C = Q/E (82)

E = QIC (83)

As an example of the use of the above relations, consider the following
problem :

A condenser has a capacitance of 200 microfarads and is connected across
600-volt mains. If the current is maintained constant at 0.1 amp., how long
must it flow before the condenser is fully charged?

The quantity in the condenser, when fully charged, is Q = 0.000200 X
600 = 0.12 coulomb or ampere-second.

0.12 = O.U

< = 1.2 seconds. Arw.

154. Specific Inductive Capacity or Dielectric Constant.— A

parallel plate condenser (Fig. 177(a)), with air as a dielectric, has
a measured capacitance C\. If a slab of glass or of hard rubber
be inserted between the plates so as to fill the intervening
space completely (Fig. 177(6)), and the capacitance of the con-
denser again be measured, it will be found to.be greater than its
previous value. Let this new value be C2. The increase in

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ELECTROSTATICS: CAPACITANCE

205

capacitance obviously must be due to the presence of the glass
or rubber.

The ratio C2/C1 = ic is called the sped jic inductive capacity , or
dielectric constant or permittivity j of the material between the con-
denser plates. The specific inductive capacity of air is assumed
to be unity, just as the magnetic permeability of air is likewise
assumed to be unity.

Fio. 177.-

(a) (6)

-Plate condenser having air and then glass as a dielectric.

In the table are given the specific inductive capacities of some
of the more common dielectrics :

* Bakelite 4 . 1 to 8.8 Paraflan

Glass 5.5 to 10 Rubber compounds

Ice 86.4 Hard rubber

Mica 2.5 to 5.5 Transformer oils
Paper 1.7 to 2.6

166. Equivalent Capacitance of Condensers in Parallel. — Let it

be required to determine the capacitance, C, of a number of con-

1.9 to 2.3
3 to 6
1.5 to 3.5
2.3 to 2.6

JcT "1^

FiQ. 178. — Capacitances in parallel.

densers in parallel, the condensers having respective capacitances
of Ciy Ciy C3. This arrangement of condensers is shown in Fig.
178. Let the common voltage across the condensers be E and
the total resulting charge Q. Obviously,

Q = CE
and

Qi = CiE, O2 = C^y Qb = CzE

•For more complete data see "Standard Handbook," Section 4, Par.
238, et seq.

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206 DIRECT CURRENTS

The total charge

Q = Qi + Q2 + Q3 = C^

CE = CiE -|- C%E -|- Cs£f

CE = i?(Ci + C2 + Ca)

.-. C = Ci + Ca + Ca (84)

TJiat is, if condensers are connected in parallel, the restdting
capacitance is the sum of the individual capacitances.

This is analagous to the grouping of conductances in parallel
in the electric circuit.

Example, — Three condensers, having capacitances of 5, 10, and 12 micro-
farads, respectively, are connected across 600-volt mains, (a) What single
condenser would replace the combination? (&) What is the charge on
each condenser?

(a) C = 5 + 10 + 12 = 27 microfarads Ana,

(6) Oi = 5 X 600 = 3,000 microcoulombs

Qj = 10 X 600 = 6,000 microcoulombs
Q« = 12 X 600 = 7,200 microcoulombs. Ana.

Total charge = 16,200 m.c. = 27 X 600 m.c. (check).

166. EQUIVALENT CAPACITANCE OF CONDENSERS IN SERIES.—

In Fig. 179, three condensers, having capacitances of Ci, C2,
and Cz respectively, are connected in series across the voltage E,

It is desired to determine the
capacitance of an' equivalent
. single condenser. Let Si, E2, and

^1=

^ C-

Ej Ez be the potential differences

Z9. i. across the condensers Cu C2, and

■^^ E% C3,respectively. After the voltage

i. E is applied to the system, there

f3 will be + Q units of charge on
the positive plate of Ci, and by
the law of electrostatic induction

Fig.' 179. — Capacitances in series. r\ -x x i_ • j j

— Q umts must be mduced on
its negative plate.
Now consider the region a which consists of the negative plate
of Ci, the positive plate of C2, and the connecting lead. This
system is insulated from all external potentials, since it is as-
sumed that the condensers have perfect insulation. Before the
voltage was applied to the system of condensers, no charge ex-

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ELECTROSTATICS'. CAPACITANCE 207

isted in the region a. After the application of the voltage, the
net charge in this region must still be zero, as no charge can flow
through the insulation. Therefore, + Q units must come into
existence in order that the net charge in the region a may re-
main zero. (+Q +(— Q)) = 0. This charge of + Q units
will go the plate of C2 since it is repelled by the + charge on Ci
just as the charge 6', Fig. 172 (a), took a position on the end of
the ellipsoid as far as possible from the positive inducing charge
a. The same reasoning holds for the region 6, between C2 and
C3. Therefore, each of the three condensers in series has the
same charge Q. (This is analagous to resistances in series, each
of which must carry the same current if no leakage exists.)
Consider the voltages J?i, E^y E3.

El = ^-, Et = ^, ^3 = ^ from equation (83), page 204.
d C2 ^-3

The sum of the three condenser voltages must equal the line
voltage:

Ei -\- E2 ")~ Ez = E

E =-^+^+^
Ci C2 C3

Also E = y^, a.a by definition the equivalent condenser C must
have a charge Q.

Substituting this value for E,

C C 1 C2 (^3

. That is J the reciprocal of the equivalent capacitance of a number
of condensers in series is equal to the sum of the reciprocals of the
capacitances of the individual condensers.

In assuming for condensers connected in series that with direct
current the potentisd across each condenser is inversely propor-
tional to its capacitance, the factor of leakage is absolutely
neglected. If the condensers are even slightly leaky, however,

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208 DIRECT CURRENTS

a current flows through the series and eventually the potential
distributes itself according to Ohm's Law.

El = IRiy Ez = 7/22, and E3 = IRz

where I is the leakage current, and fii, /?2, and Rs are the respec-
tive ohmic resistances of the three condensers.

Example of condensers connected in series:

Consider that the three condensers of Par. 155, having capacitances of
6, 10, and 12 microfarads respectively, are connected in series across dOO-
volt mains. Determine (a) the equivalent capacitance of the combination;
(h) the charge on each condenser; (c) the potential across each condenser,
assuming no leakage.

(a) 1,= !+-^+^ =0.383

C = 1/0.383 = 2.61 microfarads. Ana.
(h) Q = 2.61 X 600 « 1566 microcoulombs, .

on each condenser.

^^ ^1 = 5X10-' -313 volts

« 1,566 X 10-' --^ ,.

^^ ^ 10 X 10-^ r 157 volts

^»^ ^'y2^X^10-^' =130 volts. Ana.

157. Energy Stored in Condensers. — As a certain quantity of
electricity is stored in a condenser and a difference of potential
exists J^etween the positive and negative plates, energy must
be stored in the condenser. The existence of this energy is
shown by the spark resulting from short-circuiting the condenser
plates. The energy in joules or wattnseconds is

W = 1/2 QE (86)

This may also be written :

W = 1/2 CE^ (87)

W = 1/2 QyC (88)

The similarity in form of (87) to the equation for the energy
stored in the magnetic field should be noted. (See equation (77)
page 192, Par. 147.) The energy stored in the electrostatic field
is proportional to the square of the voltage^ whereas the energy
stored in the electro-magnetic field is proportional to the square
of the current.

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ELECTROSTATICS: CAPACITANCE 209

Example, — Determine the stored energy in each of the condensers in
series of Par. 156 and the total stored energy.
Using equation (88), '

^ -/ (1,566 X 10-«)« no>iw 1

Wi^}i 5 X 10-^ " ^ ^'^'

W.-H ^^fg^^^y' = 0.1225 joule Ans,

TF, = H ^^^^>f y = 0.1020 joule Ans.

The total energy TT = K (1,566 X 10-« X 600) = 0.4698 joule. Ana.

168. Calctilation of Capacitance. — ^As a rule it is impossible to
calculate the capacitance of a condenser, or the mutual capacitance
of conducting bodies, because of their
complex geometry and also because
the dielectric constants of the inter-
vening media are not always ac-
curately known. There are some
simple cases, however, where accurate C= ^rdtdxio*

calculations are possible. ^

mi_ 1 X J ■!-!• ^o^ • jt_ -"Q* 180. — Capacitance of a

The plate condenser, Fig. 180, is the pute condenser.

simplest form of condenser. Let A

be the area of one side of each plate in square centimeters, d
the distance between plates in centimeters and k, the dielectric
constant of the medium between the plates. The capacitance is

^ = MxtxW °^i«rof*^ads. (89)

In this equation it is assumed that the electrostatic lines between
the two plates are parallel.

The total capacitance of a simple plate condenser of this type
cannot be accurately calculated for the following reason. All
the electrostatic lines do not lie between the plates as certain
lines pass from the back of the positive plate to the back of the
negative as shown in Fig. 181 (a). This results in the actual
capacitance being greater than the value as just calculated. This
error may be avoided by using one more plate in one group than
in the other. Fig. 181 (6). In this case the area A, equation (89),
includes both sides of all the plates with the exception of the two
outside ones. As the charge on both outer plates is of the same

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210

DIRECT CURRENTS

sign and the plates have the same potential, no electrostatic
lines can pass between them. An error due to the bulging or
"fringing" of the lines near the edges of the plates may occur
unless the plate area is large compared with the distance between
plates.

(a) Electrostatic leakage lines ot a plate
condenser.

Fig. 181.

(h) Multi-plate condensers.

Example of Condenser Design, — It is desired to construct a plate condenser
having a total capacitance of 8 microfarads. The plates are of tin-foil
6 in. X 8 in. and 1 mil thick. The dielectric is of paper 7 in. X 9 in. and 2
mils thick and having a dielectric constant of 3. How many sheets of paper
and of tin-foil are necessary? What will be the dimensions of the condenser?
The area of each plate is:

6 X 8 X (2.54) « = 309.6 sq. cm.
The distance between plates:

d = 0.002 X 2.54 = 0.00508 cm.

The capacitance between two plates (from equation 89) :
3 X 309.6

Therefore:

4t X 0.00508 X 9 X 10«

= 0.01616 mf .

= 495 sections are needed.

0.01616

These sections are indicated at d, Fig. 181 (6). This means that 496
plates and 495 sheets of paper are necessary.
Thickness :

Tin-foil = 496 X 0.001 = 0.496 in.
Paper = 495 X 0.002 = 0.990 in.

1.486 in.

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ELECTROSTATICS: CAPACITANCE

211

Volume of condenser proper = 7 in. X 9 in. X 1.49 in. Ans,

Of course additional outside insulation and a protective covering arc

necessary.

The capacitance in microfarads of two co-axial cylinders, the outer of

which has a radius of Ri cm. and the inner a radius of Rt cm., is

^ 0.0388ic - .,

C =» 5" mf . per mile

This equation is applicable to single-conductor underground cables.

159. Measurement of Capacitance. — There are two common
methods of measuring capacitance, the direct-current or ballistic
method and the alternating-current or bridge method.

The direct-current method employs a galvanometer which is
used ballistically. It can be shown that if the moving coil of

Galv.

standard
Condenaer

FiQ. 182. — Ballistic method of measuring capacitance.

the ordinary galvanometer have considerable inertia and be
properly damped, its maximum throw, due to the impulse pro-
duced by the sudden passage of a current through the coil, is
proportional to the total quantity of electricity passing through
the galvanometer. This assumes that the entire charge passes
through the coil before the coil begins to move. Let D be the
maximum galvanometer throw in centimeters. Then:

Q = KD (90)

Where Q is the quantity and K the galvanometer constant.

To make the measurement, the apparatus is connected as
shown in Fig. 182. A battery B supplies the current for the
apparatus. The measurement may be made on either the
charge or the discharge of the condenser or check measurements
may be made using both the charge and the discharge. If the
condenser is at all leaky, the discharge method is preferable.

When the switch S is closed to the left the condenser Ci is

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212 DIRECT CURRENTS

charged through the galvanometer and the maximum throw of the
galvanometer is read. Several check readings should be taken.
The galvanometer should return immediately to zero. If it
shows a steady deflection it indicates a leaky condenser. In a
corresponding manner the ballistic throw of the galvanometer
may be read on discharge by closing switch S to the right after
charging. Let Di be the deflection of the galvanometer when Ci
is connected, Qi the quantity going into the condenser, and E the
voltage across the condenser. Then by equation (90)

Qi = KD,

Also Qi = CiE

where Ci is the unknown capacitance.

.\CiE = KDy (a)

If now the standard capacitance C^ be substituted for the
unknown condenser and another set of readings taken,

or C2E = KD, ^ ^

Dividing (o) by (b),

CtE KD2

j^ is the galvanometer constant.

It is often desirable to use an Ayrton shunt in such measure-
ments as it gives the apparatus greater range. When such a
shunt is used, proper correction must be made for its multiplying
power.

In the bridge method two capacitances form adjacent arms of
a Wheatstone Bridge and two resistances form the other two arms,
Fig. 183 (a). An alternating-current supply is preferable. The
secondary of an induction coil may be used as the source of power
or a battery with a key may be made to charge and discharge
the system as shown in Fig. 183 (6). A telephone is used as a
detector except in (6). Let C, be the unknown capacitance and
C2 a standard which may or may not be adjustable. Ri and R2
are two known resistances, one of which should be adjustable
unless Ci is so.

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ELECTROSTATICS: CAPACITANCE

213

Either C2 or one of the resistances is adjusted, until there is
no sound in the telephone, showing that the bridge is in balance.
Under these conditions:

Cx -K2

C2 -Ki

Cx " ^ip~

(91)

FiQ. 183. — Bridge methods of measuring capacitance.

When a battery is used, a double contact key K is necessary.
K is pressed and released, and until the bridge is balanced, the
galvanometer will deflect both upon the charge of the system, when
the key is pressed, and upon the discharge, when the key is
released. The bridge is balanced when the galvanometer does
not deflect on either charge or discharge. Equation (91) is
then applicable.

In the above measurements, it is assumed that there is little
if any leakage through the condensers.

Galv.
Ayrton Shont

Perfect Cable

— "&i"<=r

-i-x-

Faialt
FiQ. 184. — ^Locating an open in a cable.

160. Cable Testing — Location of a Total Disconnection. — In
Chap. VII, it was shown that a grounded fault in a cable
could be located by suitable resistance measurements, such as
the Murray and Varley loop tests. If a cable be totally discon-
nected and its broken ends remain insulated these loop tests are
impossibl^r The distance to the fault may now be determined

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214 DIRECT CURRENTS

by capacitance measurements. The connections are shown in
Fig. 184. The capacitance Ci of the length, x, to the fault is
first measured by the ballistic method. If a similar perfect cable
parallels the faulty cable, the two are looped at the far end and
the capacitance C2 of a length I of the perfect cable plus the length
I — X — 21 — X oi the faulty cable is measured.

Let c be the capacitance per ft. of each cable, assxmied to
be the same for each :

Ci = xc = KDi

where K is the galvanometer constant and Di the deflection
corresponding to Ci.
Likewise,

C2 = (2Z - x)c = JRlDs

Dividing one equation by the other,
X ^ Di
21- X D2

The capacitance per unit length and the total capacitance do
not enter into the equation, so that it is not necessary to use a
standard condenser for the calibration of the galvanometer. The
capacitances of the various lengths are proportional to the gal-
vanometer deflections when corrected for the setting of the Ayr-
ton shunt.

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CHAPTER X
THE GENERATOR

161. Definition. — A generator is a machine which converts me-
chanical energy into electrical energy. This is accomplished by
means of an armatm'e carrying conductors upon its surface,
acting in conjunction with a magnetic field. Electrical power is
generated by the relative motion of the armature conductors
and the magnetic field.

In the direct-current generator the field is usually stationary
and the armature rotates. In most types of alternating-
current generators the armature is stationary and the field
rotates. Either the armature or the field is driven by mechanical
power applied to its shaft.

162. Generated Electromotive Force. — It was shown in Chap.
VIII that if the flux linking a coil is varied in any way, an

(a) Maximum lines paasins throngh oo!l ^ ^ ) ^.^ lines passing through coll

Fig. 185. — Simple coil rotating in a magnetic field.

electromotive force is induced in the turns of the coil. The
action of the generator is based on this principle. The flux
linking the armature coils is varied by the relative motion of
the armature and field.

In Fig. 185 a coil revolves in a imiform magnetic field produced
by a north and a south pole. In (a) the coil is perpendicular to
the magnetic field and in this position the maximum possible
flux links the coil. Let this flux be 0.

If the coil be rotated counter-clockwise a quarter of a re-
volution, it will lie in the position shown in (6). As the plane of

215 ( \

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216

DIRECT CURRENTS

the coil is parallel to the flux no lines link the coil in this position.
Therefore, in a quarter revolution the flux which links the coil
has been decreased by 0 lines. The average voltage induced in
the coil during this period is, therefore,

e = iVr| 10-« (Chap. VIII, equation 74)

where N is the number of turns in the coil and t the time required

for a quarter revolution. But t = j^ where R = the revoliUions

per second. Therefore, the average voltage during a quarter
revolution is

e = 4iVr/j0lO-8 volts

The generation of electromotive force in a moving coil of this
type, which is similar to those used in dynamos, may also be

analyzed by considering the
total electromotive force as
being due to the sum of
the electromotive forces
generated in each side of
the coil. The electro-
motive force of one turn
is the sum of the electro-
motive forces in each con-
ductor forming the sides
of the turn, since these
conductors are connected
in series by the end con-
nections of the turn. The individual electromotive forces are
then considered as being generated in the conductor rather
than induced in the coil. This in no way conflicts with the fact
that the induced electromotive force is also due to the change
of flux linked with the coil. The same total emf. is obtained
under either assumption.

Consider the conductor a6. Fig. 186, free to slide along the two
metal rails cd and ef. The rails are connected at one end ce by a
voltmeter. A magnetic field having a density of B lines per sq.
cm. passes perpendicularly through the plane of the rails and
conductor.

FiQ. 186. — Conductor cutting a uniform
magnetic field.

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THE GENERATOR 217

Let the conductor ab move at a uniform velocity to the position
aV. While this movement is taking place, the voltmeter will
indicate a certain voltage. This voltage may be attributed to
either of two causes.

1. As conductor ab moves to position a'V the flux linking
the conducting loop formed by ce, the rails and oft, is increased,
because of the increasing area of this loop.

2. An electromotive force is generated in the conductor ab
since it cuts the magnetic field.

Similarly, the electromotive force developed by the coil in
in Fig. 185 may be attributed to the emf.'s generatied in the
conductors on opposite sides of the coil through their cutting
of magnetic lines. These conductors are connected in series
by the end conductors, or connectors, which in themselves
generate no electromotive force. The direction of the electro-
motive forces developed in the coil sides are such that these
emf.'s are additive.

The electromotive force in volts generated by a single conductor
which cuts a magnetic field is

e = BlvlO-^ (93)

where B, I and v are mutually perpendicular.

B is the flux density of the field in gausses, I the length of
conductor in centimeters^ and v the velocity of the conductor
in. centimeters per second.

That the electromotive force induced by a change of the flux
linked with a coil is the same as that obtained by considering
the emf.'s generated by the cutting of magnetic lines by the con-
ductor which make up the coil may be illustrated by a concrete
example. Let the flux have a density of 100 lines per sq. cm..
Fig. 186. The distance ab is 30 cm. and aa' is 20 cm. The con-
ductor ah moves at a uniform velocity to position a'b' in 0.1
second. What is the electromotive force across C6?

The change of flux linking the coil is:

0 = 30 X 20 X 100 = 60,000 lines.

This change occurs in 0.1 second.
Then by equation (74), page 185

e = 1^^10-« = 0.006 volt.

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218

DIRECT CURRENTS

Applying equation (93) ,

V = ^ = 200 cm./sec.

e = 100 X 30 X 200 X lO"* = 0.006 volt.

It will be seen that the same result ia obtained whether the
electromotive force is considered as being generated by the con-
conductor itself cutting the field or whether it is considered as
being induced by the change in flux linking the coil.

163. Direction of Induced Electromotive Force. Fleming's
Right-hand Rule. — A definite relation exists among the direc-
tion of the flux, the direction of motion of the conductor and the

Fore finger along lines of force. Thumb in direction of motion,
gives direction of induced emf.
FiQ. 187. — Fleming's right-hand rule.

Middle finger

direction of the electromotive force in the conductor just as
a definite relation exists between the direction of current and
of the flux which it produces.

A very convenient method for determining this relation is the
Fleming right-hand rule. In this rule the fingers of the rigM hand
are utilized as follows:

Set the fore-finger, the thumb, and the middle finger of the
right hand at right angles to one another (Fig. 187). If the
fore-finger points along the lines of flux and the thumb in the direc-
tion of motion of the conductor, the middle finger will point in the
direction of the induced electromotive force.

This rule is illustrated by Fig. 187.

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THE GENERATOR

219

164. Voltage Generated by the Revolution of a Coil. — A

coil of a single turn is shown in Fig. 188 (a). The coil rotates
in a counter-clockwise direction at a uniform speed in a uniform
magnetic field. As the coil assumes successive positions, the
electromotive force induced in it changes. When it is in position
(1) the electromotive force generated is zero, for in this posi-
tion neither conductor is cutting magnetic Unes, but rather is
moving parallel to these lines. When the coil reaches position
(2), (shown dotted) its conductors are cutting across the lines
obliquely and the electromotive force has a value indicated

Fig. 188. — Emf. induced in a coil rotating at constant speed in a uniform mag-
netic field.

at (2) in Fig. 188 (6). When the coil reaches position (3) the
conductors are cutting the lines perpendicularly and are therefore
cutting at the maximum possible rate. Hence the electromotive
force is a maximum when the coil is in this position. At position
(4) the electromotive force is less, due to a lesser rate of cutting.
At position (5) no lines are being cut and as in (1) there is no
electromotive force. In position (6) the direction of the electro-
motive force in the conductors will have reversed as each con-
ductor is under a pole of opposite sign to that for positions (1)
to (5). The electromotive force increases to a negative maximuiri
at (7) and then decreases until the coil again reaches position
(1). After this the coil merely repeats the cycle.

This induced electromotive force is alternating and an emf.
varying in the manner shown is called a sine wave of electro-
motive force. This alternating electromotive force may be
impressed on an external circuit by means of two slip-rings,

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220

DIRECT CURRENTS

Fig. 189. Each ring is continuous and insulated from the other
ring and from the shaft. A metal or a carbon brush rests on
each ring and conducts the current from the coil to the external
circuit. (See Vol. II, Chap. I.)

If a direct current is desired, that is, one whose direction is
always the same, such rings cannot be used. A direct current
must always flow into the external circuit in the same direction.

Fig. 189.-

To External Circuit
-Current taken from rotating coil by means of slip-rings.

As the coil current, must necessarily be alternating, since the
emf. which produces it is alternating as has just been shown,
this current must be rectified before it is allowed to enter the
external circuit. This rectification can be accomphshed by
using a split ring such as is shown in Fig. 190. Instead of using
two rings, as in Fig. 189, one ring only is used. This is split

(6)

Fig. 190. — Rectifying effect of a split ring or commutater.

by saw cuts at two points diametrically opposite each other.
The two ends of the coil are connected one to each of the sections
or segments so produced.

A careful consideration of Fig. 190 will show that, as the direc-
tion of the current in the coil reverses, its connections to the
external circuit are simultaneously reversed. Therefore, the

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THE GENERATOR

221

direction of flow of the current in the external circuit is not
changed. The brushes pass over the cuts in the ring when the
coil is perpendicular to the magnetic field or when it is in the so-
called neutral plane and is generating no voltage, as shown in
Fig. 188. These neutral points are marked 0-0-0 in Fig. 190 (6).

By comparing Fig. 188 (6) with Fig. 190 (6) it will be seen that
the negative half of the wave has been reversed and so made
positive.

A voltage with a zero value twice in each cycle, as shown
in Fig. 190, could not be used commercially for direct current
service. Also a single-coil machine would have a small output

Reaultant
Electromotivefoi

Fig. 191. — Effect of two coils and four commutator segments upon the electro^

motive force wave.

for its size and weight. The electromotive force wave of Fig.

190 may be improved upon by the use of two coils and four com-
mutator segments. Fig. 191. This gives an open circuit type of
winding, since it is impossible to start at any one commutator
segment and return to this segment again by following through
the entire winding. In this particular arrangement the full
electromotive force generated in each coil is not utilized, as one
coil passes out of contact with the brushes at points a, a, a, Fig.

191 (6), and the voltage shown by the dotted lines is not utiUzed.

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222

DIRECT CURRENTS

165. Gramme-ring Winding. — This type of winding in its
elementary form, Fig. 192, consists of insulated wire wound
spirally around a ring (or hollow cylinder of iron) with taps
taken from the wire at regular intervals and connected to com-
mutator segments. This winding is simple, and has the advan-
tage that a single winding is adapted to any number of poles,
if the voltage limitations do not prevent. The portions of the
conductors which lie inside the ring cut no flux and act merely
as connectors for the active portions of the conductors. Because
of the small proportion of active conductors a relatively large
amount of copper is required in such a winding. In small

Fig. 192. — Gramme«ring winding.

machines, there is not sufficient room to carry these inactive
conductors back through the armature core. In a gramme-ring
winding formed coils cannot be used and this makes the winding
expensive. This type of winding has a high inductance, which
renders good commutation difficult.

It will be noted that the electromotive force between brushes
in a gramme-ring winding is the sum of the electromotive forces
of all the coils that lie between brushes. When one coil passes
a brush another moves forward to take its place. Fig. 193 shows
the electromotive force between brushes due to four coils, it
being assumed that the voltage curve for each is a sine wave.
The electromotive force of each coil is plotted separately. These
electromotive forces do not all have their zero value at the same
time nor do they reach their maximum value at the same time

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THE GENERATOR

223

owing to the positions of the individual coils. The resultant
electromotive force at any point is the sum at this point of
these individual electromotive forces. This voltage should be
compared with the electromotive force obtained with the open-
coil winding shown in Fig. 191, in which the resultant electromo-
tive force does not equal the siun of the individual electromotive
forces but is made up of the successive tops of the individual
waves. It will be noted that a fairly smooth resultant electro-
motive force is obtained with four coils, the "ripples" being
noticeable but comparatively small in magnitude.

Fig. 193.-

-Resultant electromotive force due to four series-connected coils
between brushes.

A gramme-ring winding is called a dosed winding, since it is
possible to start at any one point in the winding and return to
the same point again by passing continuously through the
winding.

166. Drum Winding. — The objections to the ring winding are
overcome by the use of the drum winding. The conductors
of this winding all lie upon the surface of the armature and are
connected to one another by front and back connections or coil
ends {ad and 6c, Fig. 194, are coil ends). With the exception
of these end connections, all the armature copper is "active,"
that is, it cuts flux and so is active in generating electromotive
force.

The sides of each coil should be about one pole pitch (the
distance between centers of adjacent poles) apart; If one con-
ductor is under a north pole the other is then under a south pole,
and as both move in the same direction, but under different poles,
the electromotive forces of these two conductors will be in op-
posite directions, Fig. 194. Due to the manner in which these
conductors are connected at their ends, the electromotive forces
in the individual coils are additive.

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224 DIRECT CURRENTS

In most gramme-ring windings, and in the earlier drum-wound
machines, the surface of the armature core was smooth. The
conductors were held in position partly by projecting pins, and
were prevented by binding wires from flying out under the action
of centrifugal force. The smooth core construction has been
superseded by the "iron-clad" type where the conductors are
embedded in slots as indicated in Fig. 197. The slots are lined
with insulation and the conductors are held in firmly by wooden
or non-conducting wedges in the larger machines (see Kg. 224),
and by binding wires in the smaller types (see Fig. 214). These
constructions are much better mechanically than the smooth

Commutator;

Fig. 194. — Two coils in place on a 4-pole, drum-wound armature.

core armature type and they also permit a much shorter air-
gap. On the other hand, as the coils are embedded in iron, they
have a high inductance. This makes commutation more difficult
and the flux pulsations due to the armature teeth give pole-face
and tooth losses.

167. Lap Winding. — Direct-current armatures are usually
wound with former-made coils, Fig. 195. These coils are usually
wound on machines with the necessary number of turns, and
are then wound with cotton or mica tape. They are then bent
into proper shape by another machine. The two ends are left
bare so that later they may be soldered to the commutator bars.
The span of the coil, called the coil pitch, should be equal or nearly
equal to the pole pitch, so that when one side of a coil is under a
north pole the other is under a south pole. This span may be as
low as nine-tenths of the pole pitch, in which case a fractional
pitch winding results.

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THE GENERATOR

225

Usually two coil sides occupy one slot, one coil side lying at the
top and the other at the bottom of the slot. That is, if the side
of one coil is in the bottom of a slot, its opposite side lies in the
top of some other slot. This allows the end connections to be
easily made as the coil ends can be bent around one another in
a systematic manner, passing from the bottom to the top layer by
means of the peculiar twist in the ends of the coils.

"ID

Fig. 195. — Formed armature coils.

The bundle of wires constituting one side, ob (Figs. 194 and
196), of a coil will be termed a winding element. This may
consist of one or of several conductors taped together. Even
when there are several conductors, they will be shown as a single
conductor in the wiring diagram, as indicated in Fig. 196. Obvi-
ously there will be twice as many of these elements as there are
coils. The number of elements that the coil advances on the back
of the armature is the hack 'pitch of the winding and will be
denoted by 2/5. This back pitch is obtained by the connection 6c,
Fig. 194. The number of elements spanned on the commutator

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226

mRECT CURRENTS

end of the armature is called the front pitch and will be designated
by y/. This may be greater or less than the back pitch bid not
equal to it. If it be greater, the winding is retrogressive, and if it
be less, the winding is progressive. This is illustrated in Figs. 197
and 198. Conductor 1 is connected on the back of the armature
to conductor 10. Therefore, the back pitch yh = 9. Conductor
10 is then connected back to 3 on the front of the armature,
the connection being made at the commutator segment.

winding y
Elements

.^X

winding/
Elements

-r-TT

'^y^

:^'^

EZE

FiQ. 196. — Single coil representing a 3-turn coil of an armature winding.

Therefore, the front pitch yf == 7. This winding is therefore
progressive.

As most windings are now made in two layers, only two-layer
windings will be considered. The conductors or elements lying
in the top of the slots will be given odd numbers and those in tlie
bottom of the slots even numbers. Fig. 197. As one side of a
coil lies in the bottom of a slot and the other side in the top of a
slot, obviously yh and y/ must both be odd. Further, if they were
both even, all the conductors could lie only in either the odd or
the even slots but could not lie in t)oth. In a simplex lap winding
having two elements per slot, the return connection cannot be
made back to the original slot but it must always lead back to
a slot which is next to the original slot. Thus, in Fig. 197, the
connection is from the top conductor 1 to conductor 10, thence
back to 3, i.e., to the top of the next slot. Therefore, the front
and back pitches can only differ from each other by 2.

That is,

2/^ = 2// ± 2 (94)

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THE GENERATOR

The average pitch

y =

_ Vh + Vt

227

(95)

Fig. 197. — Simplex lap winding having back pitch of 9 and front pitch of 7.

Fig. 198. — Development of a 4-pole lap winding.

The + sign in (94) indicates that the winding is progressive,
that is, progresses in a clockwise direction when viewed from the
commutator end. The — sign indicates a retrogressive winding

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228 DIRECT CURRENTS

whose advance is in a counter-clockwise direction when viewed
from the commutator end.

It will be seen that for every coil one commutator segment is
necessary. Therefore, the number of commutator segments

where Z is the total number of winding elements on the surface
ot the armature and N is the number of coils.

From Figs. 197 and 198 it will be seen that the winding ad-
vances one commutator segment for each complete turn.

In designing a winding it is necessary that the opposite sides of
each coil lie under different poles so that the two electromotive
forces generated in the coil sides may be additive. Hence the
average pitch should be nearly equal to the number of elements
per pole.

The three fundamental conditions to be fulfilled by a lap
winding are:

(1) The pitch must be such that the opposite sides of the coil
lie under unlike poles.

(2) The winding must include each element once and only
once.

(3) The winding must be re-entrant or must close on itself.

Example, — Assume that the armature of a 4-pole machine has 18 slots.
Design a two-layer lap winding having two elements per slot.

There are 36 elements. The average pitch should be nearly equal to
^ — 9. The back pitch can be made equal to 9.

2/6 = 9 y/ = 7

Starting at 1, the winding will progress as follows:

l-10-3-12-5-14r-7-16^9-18-l 1-20-13-22-15-24^17-26^19

28-21-30-23-32-25-34^27-36^29-2-31-4-33-6-35-8-1.

The above is called a winding table. It is very useful in check-
ing the winding. By proper checking it may be seen that each
conductor is included once and only once and that the winding
closes at the same conductor, 1 in this case, at which it began. The
winding is shown in Fig. 198 as if it were split axially and rolled
out flat. It will be noted that the brushes rest on segments to

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THE GENERATOR 229

which are connected elements which lie midway between the
poles, as 11 and 19, for example.*

168. Lap Windings — Several Cofl Sides per Slot — In the larger
sizes of machines it is often necessary to place several coil sides
or elements in one slot, usually 4, 6, or 8. More than eight coil
sides per slot are rarely used. The reason for placing several
coil sides in a slot is as follows: If two elements per slot were
used, one in the top layer and one in the bottom layer, a large
number of slots would be necessary. This would reduce the
size of the slots and make the space factor (ratio of the copper
cross-section to the slot cross-section) low. Also the tooth roots
would be so narrow that the teeth would be mechanically weak.
By placing several elements in each slot the number of slots is
reduced and larger slots result. This also reduces the cost of
winding.

Coils made up from several individual coils are shown in Fig.
195. The two or three coils are taped as one and are placed
in the slots as a unit. A careful examination of the armature
of Fig. 214, page 246 shows four wires running from each coil side
to the commutator, indicating a quadruple coil.

The numbering and connections of the conductors are in no
way different from those already described in the case of but two
coil sides per slot.

The selection of the pitch, where several coil sides per slot
are used, is more restricted than it is with two elements per slot.

Assume that a 6-pole machine has 72 slots and six elements
per slot. The total number of elements on the armature surface:

Z = 72 X 6 = 432

The pitch should be approximately

Let Vh = 71

Vf = 69

If this back pitch is used a coil must reach from conductor 1
to conductor 72 (Fig. 199). Then the next coil will obviously
reach from conductor 3 to conductor 74* These two coils, there-
fore, span different distances on the armature and accordingly

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230 DIRECT CURRENTS

must have different spans, as a study of Fig. 199 will show. In
practice it is desirable that the coils be all the same when pos-
sible, and further it should be possible to tape all three coils to-
gether and place them in the slots as a imit.

If in the above case yb = 73 and y/ = 71, the coil containing
conductor 1 will reach from the upper left-hand side of slot A to
the lower kft-hand side of slot JS, that is, from conductor 1 to
conductor 74. Conductor 3 will reach from the center and top
of slot A to the center and bottom of slot B, and conductor 5 wiU
reach from the upper right-hand side of slot A to the lower right-

FiG. 199. — Method of connecting the conductors of a triple coil.

hand side of slot B. As all three coils now span the same dis-
tance on the armature, they will be equal in size, form, etc. More-
over, the three single coils can be taped together to form a triple
coil and placed in the two slots as a unit. Therefore, if three coils
have their adjacent sides in the top of one slot, their other sides
should lie together in the bottom of some other slot. This condi-
tion is obtained by making the back pitch one greater than a
multiple of the number of coil sides or elements per slot. For
example, in the illustration just given, yt is equal to 73, one
greater than 72, 72 being a multiple of 6.

169. Paths through an Armature. — If foiu* batteries, each hav-
ing an electromotive force of 2 volts and a current capacity of
10 amp. be connected in parallel, Fig. 200 (a), there will be
four paths for the current to follow in going through the batteries.
The voltage of the combination will be 2 and the ampere capac-

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ity 40, making a total power capacity of 80 watts. If now these
same batteries be arranged in two groups of two in series, Fig.
200 (6), there result but two paths for the current to follow, but
the voltage is now 4 volts. The current capacity is now 20 amp.,
and the power capacity is 4 X 20 == 80 watts, its previous value.
Similarly the conductors in an armature may be so connected
that certain groups of conductors are in series. These groups
may then be so connected that there are two or more paths in
parallel. To determine the number of such parallel paths, start

(a) (6)

Fio. 200. — Parallel and series-parallel arrangement of batteries.

at one of the machine terminals, as for example, the negative,
and see how many different paths through the armature it is
possible to follow in order to reach the positive terminal.

The simplest arrangement of conductors occurs in the gramme-
ring winding. Pig. 201 (a) shows a winding for a 4-pole
niachine.

Starting at the (— ) terminal, one path may be followed by
going to brush (a), through the winding at (1) to brush {d) and then
to the (+) terminal.

A second path is obtained by going to brush (a), then through
path (2) to brush (6) and then to the (+) terminal.

A third path is obtained by going to brush (c), through path (3),
then through brush (6) to the (+) terminal.

A fourth path is obtained by going to brush (c), through path
(4) to brush {d) and then to the (+) terminal.

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DIRECT CURRENTS

This makes four separate paths between the ( — ) and (+) termi-
nals, these paths being in parallel.

Assume that there are 10 amperes per path and 20 volts be-
tween bushes. The armature may be considered as being equiva-
lent to four batteries connected as shown in Fig. 201 (6), each
battery delivering 10 amperes at 20 volts. Battery 1 corresponds
to path 1, battery 2 to path 2, etc.

Fig. 201. — Four paths in parallel through an armature.

It will be seen that the four batteries are connected in parallel
because their four positive terminals and their four negative ter-
minals are respectively connected together. The total current
delivered will be 40 amperes at 20 volts. In a similar manner
each path in the ring winding will deliver 10 amperes, making
20 amperes per brush or 40 amperes per terminal. The potential
difference between brushes will be 20 volts.

The paths through a drum winding are not as easy to follow as
those through a ring winding. Fig. 202 shows the 18-slot
drum winding of Fig. 198 developed in circular form. For the
sake of simplicity two paths are shown with heavy lines, one

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from brush a to brush b, and the other from brush c to brush d.
These constitute two paths. By tracing through the lighter
lines, two more paths may be found, one between brushes c and
b and the other between brushes a and d, making four paths in all.
In all simplex lap windings there are as many paths through the
armature as there are poles.

Fig. 202. — Heavy lines show two of the four parallel paths of a lap winding.

170. Multiplex Windings.— Fig. 203 shows a 36-slot, 4-
pole winding, in which every alternate slot is filled. There are
two conductors per slot. The back pitch, 2/&,is 17, and conductor
1 connects to conductor 18 on the back of the armature. Con-
ductor 18 then connects to 5 on the front of the armature, making
the front pitch y/ = 13. Instead of returning to the conductor
differing by ^ from the initial conductor, the return is made to
a conductor differing by 4 from the initial conductor. Conduc-
tors S and 4 are not connected to, this winding. Furthermore,
only alternate commutator segments are utilized. It will be'

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DIRECT CURRENTS

seen that this winding closes on itself after going once around the
armature; that is, this winding is re-entrant and is in itself com-
plete in the same manner as any simplex 18-slot winding. (See
Fig. 202.)

Fia. 203. — Duplex doubly-re-entrant lap winding — one winding only being

shown.

As this winding uses only alternate slots, and alternate commu-
tator segments, another winding, the dupUcate of this one, can be
placed in the vacant slots, this new winding having the same front
and back pitch as the other, and being connected to the commuta-
tor segments not utilized by the other. This winding will also
close on itself, and is, therefore, re-entrant.

These two windings are separate and are insulated from each
other on the armature, but are connected together electrically by

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the span of the carbon brushes on the commutator. This condi-
tion is perhaps more clearly shown in the simple gramme-ring
winding of Fig. 204 (a), where one winding is in solid lines and the
other in dotted lines. These two windings are in parallel, so that
the number of paths is now twice what it would be in a simplex lap
winding. As each of the two windings closes on itself, the wind-
ing is said to be doubly re-entrant. It is necessary with this
type of winding that the brush span at least two commutator
segments.

(o) Duplex doubly re-entrant gramme- (h) Duplex singly re-entrant gramme-
ring winding. ring winding.
Fig. 204.

When there are two such windings in parallel, the winding
is said to be duplex. Therefore, this is a doubly re-entrant
duplex winding. Obviously, three or more such windings can
be placed on an armature, making the winding triplex, quad-
ruplex, etc., the number of such windings being called the
multiplicity of the winding.

Let m = the multiplicity of the winding.

The number of paths p' in a lap winding is

p' = mp (96)

where p is the number of poles.

The relation of the back and the front pitch becomes

yb^ yf ± 2m (97)

This should be compared with equation (94) where m = 1.

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236 DIRECT CURRENTS

If the number of coils, Fig. 203, be odd, that is, if there axe
35 or 37 coils and commutator segments, the winding will not
close after having gone once around the armature, but will return
one slot, or two conductors, to the right or to the left of the one
at which it started. (If there are more than four elements
per slot the winding may return to the same slot at which it
started, but removed by two conductors from the conductor at
which it started.) Therefore, this winding does not close or
become re-entrant, after having passed once around the armature,
but must pass around once again before closing. This is il-
lustrated in Fig. 204 (6). The initial winding starts at a. After

(a) Duplex doubly re-entrant winding. (b) Duplex singly re-entrant winding.

Fig. 205. — Duplex windings in diagrammatic form.

passing once around the ring armature it does not close at a as
does the winding in Fig. 204 (a), but terminates at 6, one con-
ductor removed from a. The second winding, shown dotted,
starts at h and after passing once around the armature, closes at a.
Although this winding passes around the armature twice, it
only closes once, so is said to be singly re-entrant. Therefore,
this constitutes a singly re-entrant duplex winding. The two
windings are the same electrically. Their difference is best
illustrated by the two simple diagrams of Fig. 205.

171. Equalizing Connections in Lap Windings. — ^Lap windings
may consist of several paths in parallel, the parallel connections
being made through the brushes. If several batteries are con-
nected in parallel and their emf .'g are not equal, ciu'rents circulate
among the batteries, even when no external load is being sup-
plied. This means a constant loss of energy which heats the
batteries.

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This same condition exists in generator armatures. Because
of very slight inequalities in the air gap, due to the wearing
of the bearings, lack of mechanical alignment, etc., there may be
slight differences of electromotive force in the different paths
through the armature. These differences of emf. will cause
currents to flow between different points in the armature,
and these currents must flow through the brushes even when no
current is being delivered by the generator. To relieve the
brushes of this extra current, several points in the armature

1 I I I im ' I ri rii I i

f r [ ] 11 r fVi I ] ivi II I) I i^:i f r'-rrrrri

FiG. 206. — Simplex lap winding with equalizing connections.

which are simultaneously at equal potentials are connected
together by heavy copper bars. This allows these circulating
currents to flow from one point in the armature to another with-
out . passing through the brushes. To make these equalizer
connections, the number of coils should be a multiple of the
number of poles, and the coils per pole should be divisible by
some small number as 2 or 3.

As an example, assume an 8-pole generator having 12 slots
per pole and two coil sides per slot. There will be 96 slots and
192 coil sides. The number of coil sides per pole will be 24.
Let t/b = 25 and t// = 23. A portion of this winding is shown in
Fig. 206. It will be noted that every fourth coil is connected
to an equalizing connection. The coils that are connected to
the same equalizing connection occupy the same positions rela-
tive to the poles. (See the two half -coils drawn with heavy lines.)
This is necessary as such coils should be generating the same

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DIRECT CURRENTS

voltage at every instant. It will be noted in Fig. 206 that the
two segments under the two positive brushes are connected
together by an equalizing connection.

Theoretically, every coil should be connected to an equalizing

connection, but as this would
require an undue number of
such connections, it is suf-
ficient, practically, to connect
every third or fourth coil.
This is the reason that the
number of coils per pole should
be divisible by a small number
as 2, 3 or 4. Fig. 207 shows
a large direct-current arma-
ture with the equalizer con-
nections at the back of the
armature.

172. Wave Winding.— It
has been shown that in the

F.0. 207.-General Electric Co. direct-cur. «»»« of the Up winding a COn-
rent armature with equalizer rings. ductor under one pole is Con-
nected directly to a conductor
which occupies a nearly corresponding position imder the
next pole. This second conductor is then connected hack
again to a conductor imder the original pole, but removed
two or more conductors from the initial conductor. This is

(a) Lap Winding: (b) Wave Winding

Fig. 208. — Lap and wave windings.

Vf >\

shown in Fig. 208 (a), where conductor ab under a north pole
is connected to conductor cd having a corresponding positicm
under the next south pole. Conductor cd is then connected
to cf which is adjacent to ab under the original north pole.

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Obviously it would make no difference as far as the direction and
magnitude of the induced emf . in the winding is concerned if the
connection, instead of returning back to the same north pole,
advanced /ort/;ard to the next north pole, as shown in Fig. 208 (6).
When' the connection is so made, the winding passes successively
every north and south pole before it returns again to the original
pole, as shown at a'V in Fig. 208 (6). The winding after passing
once around the armature reaches conductor a'h' lying under the
same pole as the initial conductor ah. When a winding advances
from pole to pole in this manner, it is called a wave winding. The
number of units* spanned by the end connections on the back of
the armature is called the hack pitch and is denoted by yh in
Fig. 208 (6). This is similar to the corresponding term in the
lap winding shown in Fig. 208 (a). The number of elements
which the end connections span on the commutator end of the
armatiu*e is the front pitch and is denoted by i//. This should
also be compared with Figi 208 (a). As in the lap winding, y/
and yh must both be odd in order that one side of a coil may lie
in the top of a slot and the other side in the bottom of a slot.
Unlike the lap winding, yf may equal yh in the wave winding.

The above is illustrated as follows:

A certain wave winding may have a back pitch of 23 and a
front pitch of 19. The average pitch

Likewise both the front and the back pitch may each be 21
making the average pitch 21.
In any event, the average pitch

y-'-^ (98)

y may be either even or odd.

When the winding viewed from the commutator end falls
in a slot to the left of its starting point as a'6'. Figs. 208(6) and
209 (a), after passing once around the armature, the winding
is retrogressive. If, on the other hand, it falls to the right of its
starting point, as shown in Fig. 209 (6), it is progressive.

The wave winding is much more restricted in its relation to the

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DIRECT CURRENTS

number of slots and coils than is the lap winding, for the following
reason. In a simplex wave winding, after having passed once
around the armature, the winding must fall two conductors either
to the right or to the left of the conductor at which it started.
Thus in Fig. 209 (a), if there are two conductors per slot and
conductor ab lies in the bottom of one slot, conductor a'6'must
lie in the bottom of the slot next to ab. As there are two coil
sides in each slot, this means that conductors oft and a^V will
differ from each other by 2.

(a) RetrogreBsive wave winding-
commutator segments.

-34
Fig. 209.

(6) Progressive wave winding — 32
conmiutator segments.

Let y be the average .pitch. Assume that the winding closes
after passing once aroimd the armature, which, of course, it should
not do as this would constitute a short-circuit. Then:

py = Z
where p is the number ot poles and Z the number of coil sides or
elements. But the winding must not close after passing once
around. In fact, it must not close until every slot is filled.
Therefore, after passing once around the armature, the product
py cannot equal Z but must be Z ± 2. That is:

py = Z ±2
or

Z ±2

y =

The + sign indicates a progressive winding and the
gressive winding.

(99)
sign a retro-

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As an illustration, assume that a 4-pole armature has 63 slots
and four conductors per slot, making 252 winding elements. Let
the average pitch be 63, the front and back pitch both being
63. As in thie lap winding diagrams, a single-turn coil will be
used to represent a coil having several turns, as indicated in Fig.
210. Starting at conductor 1, the winding will advance as follows :

l-64-127-190-(253 or 1)

That is, the winding will close on itself after going once around
the armature, which condition constitutes a short-circuit and

Windiog;
Elements

jjjT i 1 1 I I 1 1 iTrr

Fig. 210. — Single-turn coil representing a 3-turn coil for winding diagram.

makes the winding impossible. (The method by which a winding
may be placed in these slots will be shown later.) Therefore, a
wave winding is impossible in a 4-pole machine if 252 winding
elements are to be included.

Let N-c be the number of commutator segments, which is also
the number of coils.

Nc = Z/2, Z = 2Nc

Let j>i = pairs of poles = p/2 p = 2pi

Substituting in equation (99)

2Nc ± 2

^= 2v.

Nc = v^y ±

(100)

If Pi is odd and y is odd, the product piy is odd, as the product of
two odd numbers is always odd. Adding or substracting unity
makes Ne even.

Therefore, with a wave winding whose average pitch is odd and
having^, 10, 14 goles, or^ 5^^ pairs of poles, the number of com-
mutator segments and coils must each be even. If the average
pitch is even the number of commutator segments and coils
must each be odd.

16 . '

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DIRECT CURRENTS

On the other hand, if j>i is even, corresponding to 4, 8, or 12
poles, the product piy is even, so that A^c must be odd. The ap-
pUcation of equation (100) is illustrated in Fig. 209. There are
6 poles and the average pitch, y, is 11. Applying equation (100),
iNTc = 3 X 11 + 1 = 34; AT, = 3 X 11 - 1 = 32. The 34 seg-
ments are shown in (a). Fig. 208, which gives a retrogressive
winding, and the 32 are shown in (6), which gives a progressive
winding. Nc is even in either case.

This is another limitation of the wave winding and shows
why the 252-conductor (12&-coil) winding just considered is im-
possible. The number of coils must be odd in a 4-pole winding.
However, if one coil were omitted, making 250 elements, the
winding would progress as follows: _ -

l--64-127-190-(253 or 3)
-«6^129-192-5, etc.

That is, the winding would advance by two conductors after each
passage around the armature, which condition makes the winding
possible. This,of course, reduces the number of commutator seg-

Commutajtor

Fig. 211. — Dummy coil and "creeping" in a forced wave winding.

ments and coils from 126 to 125, an odd number. If, in this case,
the armature stampings were standard, having 126 slots, the
winding would be possible by omitting one coil. This coil
would be inserted in the slots just the same as the otfter coils,
except that its ends would not be connected to the commutator
segments but would be taped and thus insulated from the main

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winding. The coil would serve only as a filler, and is called a
"dummy coil." In this case there would be a slight ''creeping"
of the winding with respect to the commutator, as shown in Fig.
211. This is called a forced winding.

If the coils used in a wave winding consist of more than one
turn, they will have the ends brought out and connected in the
manner shown in Fig. 195 (&) and in Fig. 210.

173. Number of Brushes. — Fig. 212 shows the beginning of a
wave winding, which begins at positive brush a and advances

Fig. 212. — Wave winding — 3 positive brushes connected by the winding itself.

once around the armature. This is a 6-pole winding in a machine
having 44 commutator segments. The pitch is found from
equation (100).

44 = 3i/ ± 1

y =

44 ± 1

= 15, using the + sign.

This is also equal to the number of commutator segments by
which the winding advances per pair of poles, or each time that
it is connected to the commutator. Therefore, the segment
connections, starting with 1, are 1-16-31-2-17, etc., as shown
in Fig. 212. The winding ends- one segment beyond the

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244 DIRECT CURRENTS

starting point for each complete passage around the armature,
showing that the correct pitch has been chosen.

There are three positive brush sets, a, 6, and c, and also three
negative brush sets, the same number that would be used in a
lap winding. It should be noted that the three + brushes, a, 6,
and c are all connected together direcUy by the winding. More-
over, the conductors which connect these three brushes all he
between the poles in the neutral plane, where they are not cutting
any magnetic lines and are for the instant, therefore, dead
conductors. Hence, if brushes 6 and c were removed, the current
could easily pass through the armature to brush a and thence to
the external circuit. In like manner, two of the negative brushes
could be removed, without serious disturbance. It is desirable
to utilize all six brush sets, as two brush sets would mean a
commutator three times as long in order to obtain the necessary
brush area.

In a wave winding only two brushes are necessary y regardless of
the number of poles, althoiigh it is usvAdly desirable to use the same
number of brushes as poles.

There are cases, however, where it is desirable to use only two
brushes. The best example is in railway motors where it would
be difficult to obtain access to foiu* or six brushes. By means
of a small hand hole in the motor casing, it is a comparatively
simple matter to reach two brushes located on the top of the
commutator.

174. Paths through a Wave Winding. — In a simplex wave
winding there are always two parallel paths, regardless of the
number of poles. Fig. 213 shows a 4-pole, 17-slot, simplex wave
winding, having two coil sides per slot. One of the paths is shown
by the heavy lines. Approximately half the winding is shown
heavy, the other half constituting the other path. (The coils
short-circuited by the brushes are not included.) A wave wind-
ing may be duplex, triplex, or have any degree of multiplicity
just as the lap winding may.

The paths through the armature depend only on the degree of
multiplicity and not on the number of poles. A simplex wave
winding always has two paths, a duplex winding four paths, etc.

It is interesting to compare the current and voltage of an
armature for the various ways of connection. Consider a 6-i)ole

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machine. When connected as a simplex lap winding let its emf .
be 300 volts and the armature current per terminal be 120 amp.

Fig. 213. — 17-slot, 4-pole, simplex wave winding; back pitch = 9, front pitch =
7; one of two parallel paths shown heavy.

The following table gives the values of current and emf. obtain-
able when the winding is changed, the total number of armature
conductors remaining fixed.

Simplex lap . .
Dupjex lap . . .
Triplex lap. . .
Simplex wavp
Duplex wave .
Triplex wave .

Paths

Volts

300-
150
100
900
450
300

Amperes

Kw.

12(^
240
360
40
80
120

36
36
36
36
36
.36

It will be noted that in this particular machine the triplex
wave winding gives the same result as the simplex lap winding.
The kilowatt capacity is not affected by the connection used.
The above relations should be kept in mind when it is desired

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DIRECT CURRENTS

to change a machine from one current and voltage rating to
another. This may often be done by merely changing the
conmiutator connections.

176. Uses of the Two Types of Winding. — A wave winding
has an advantage in that it gives a higher voltage with a given
number of poles and armature conductors. It is used, therefore,
in small machines, especially those designed for 600-volt circuits.

Armature Core
Shaft

Cam mutator

Armature On Is

Armattire

(a) 25 H.P. wave-wound Westinghouse generator armature.

(6) End view of

an armature showing open construction — Westinghouse commu-
tating-pole D. C. motor.

Fig. 214.

In this case a lap winding would result in a very large number of
small conductors. This in turn means a higher winding cost and
less efficient utilization of the space in the slots.

The wave winding has the additional advantage that the elec-
tromotive force in each path is produced by series-connected
conductors, which lie under successive north and south poles.

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Any magnetic unbalancing, therefore, due to such causes as air-
gap variation and difference in pole strength, does not produce
cross currents, because the corresponding conductors of each and

Fio. 215. — Low- voltage, high-speed G. E. armature for electrolytic work.
(Note double commutator and shrink rings.)

every path are moving by the same poles and the effect of such
unbalancing will be the same in each path. Hence no equalizer
connections are necessary.

Leeos ro be aitached
to zoft\miAtor bars.

Treated duck strips protect
coils from rubbing.

Ffsh paper cells prorpct
coils in core slots.

Ventitsrion
holes.

Coi'is fit compaciiy Willi
«et sides lo^ether

End-plate
riveted to core.

•Armature keyed to shaft which may be
removed without dislurbinj windings.

Fig. 216. — Partly wound armature showing method of assembling coils (Westing-
house).

The possibility of using only two brushes with a wave winding,
and the corresponding advantage in railway motors, have already
been mentioned.

When large currents are required, the lap winding is more
satisfactory, since it gives a large number of paths. As 200

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DIRECT CURRENTS

Fig. 217. — Frame rings — Westinghouse type S. K. motor.

£f tff

i^k

^dwi

?K^t ^

BgfB

^^m

v^^^^

J^^^^^^oi fjJ' L 0 rr 1 p 1 1 ic

Fig 218.— Westinghouse 230-volt, 35 H.P., 850 R.P.M., shunt motor.

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amp. per path is practically the limit, a large number of paths
must be used where heavy current output is desired. This is
particularly true of large engine-driven multipolar generators.

Figs. 214 and 215 show two different types of armature and
Fig. 216 shows an armature in the process of being wound.

DYNAMO CONSTRUCTION

176. Frame and Cores. — The frame or yoke of a dynamo has
two functions. It is a portion of the magnetic circuit (see Figs.

nnnnml^if-^ Single SUmplngt

^sssmmnj

Bhant FMd

Pole

Fig. 219. — Construction of a 12-pole, direct-connected, engine-driven generator.

38, 39 and 40) and it acts as a mechanical support for the machine
as a whole. In small machines, where weight is of Uttle import-
ance, the yoke is often made of cast iron. The feet almost always
form a part of the casting. In another type of construction a
steel plate is rolled around a cyUndrical mandrel and then welded,
Fig. 217. The feet in this case are made of steel stampings and
are riveted on, Fig. 218. In larger machines the yoke is made
of cast steel and is usually more or less oval in cross-section.
Figs. 219 and 220. The feet are a part of the yoke casting.
The yoke for the larger machine is usually cast in two pieces

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DIRECT CURRENTS

which are bolted together. This facilitates the shipment of
large machines and allows the armature to be removed easily.
177. Field Cores and Shoes. — The field cores are made of
forged steel, cast steel and steel laminations. When made of
cast or forged steel they are usually circular in cross-section, as
such a section allows the minimum length of turn for a given

Fig. 220. — Westinghouse engine-driven, 300 K.W., 500- volt, 150 R.P.M. generator.

core section. These cores are held to the yoke by bolts, Figs. 219
and 220. The laminated cores are built of sheet steel stampings,
Fig. 221. They are stacked so that the pole tip comes alter-
nately on one side and the other. This results in there being
but half the iron in a pole tip cross-section and so producing a
saturated pole tip, which assists commutation. When stacked
to the proper thickness, they are riveted together and dove-tailed
to the yoke. In this case a separate pole shoe is not necessary.
A laminated or a solid steel pole shoe may be bolted to the solid

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cores, the laminated type being used on the larger machines
to reduce pole face losses. (See Par. 228, page 358.)

178. The Armature. — The armature is made of sheet steel
discs (14 to 25 mils thick) punched out by a die. The slots may
be cut by the die or they may be cut out afterward with a

Fig. 221. — Field core lamination and pole piece assembled — Westinghouse

D. C. motor.

slotting machine. In small motors these stampings are keyed
directly to the shaft, Fig. 222, After every 2 or 3 inches of lami-
nations a suitable spacer is inserted to form a ventilating duct.
Fig. 223. The laminations are clamped together by end plates.
Fig. 222, which are in turn held by nuts on the shaft or by bolts

Fig. 222. — Armature construction of a small motor.

passing through the laminations. The laminations are perfor-
ated to allow air to pass through the armature axially and out
radially through the ducts. Frequently a blower is attached to
the end of the end plate. Fig. 222, to facilitate ventilation.

In machines of medium size, the stampings are assembled
and keyed to an armature spider, which is in turn keyed to the

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DIRECT CURRENTS

shaft, Fig. 223. This reduces the amount of sheet steel neces-
sary and at the same time permits a free passage of air through
the center of the armature. This air is then thrown out through
the ventilating ducts by centrifugal action, as indicated by the
arrows. The stampings in Fig. 223 are clamped together by

Fig. 223. — Cross-section of a moderate size generator; armature stamping.

end plates held by bolts. These end plates may also serve as
supports for the overhang of the armature coils.

When the armature becomes greater than 30 in. in diameter,
it is not economical to stamp out a complete ring. Such arma-
tures are made up of segments similar to those shown in Fig. 219.
These are dove-tailed to the armature spider, each segment lap-
ping the joint in the next layer.

The slots may be straight sided. Fig. 223, in which case the

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THE GENERATOR

253

conductors are held in the slots by binding wires. In the larger
machines the conductors are held in the slots by wooden wedges,
Fig. 224. The slots must be well insulated, as grounds are trouble-
some and are expensive to repair. A layer of a hard substance
such as fish paper, fiber or press board should be placed next to
the laminations. This in turn should be lined with varnished
cambric or empire cloth. The conductors themselves are usu-
ally covered with cotton insulation, except in the heavy bar wind-
ings. The groups of conductors are bound together in one coil
by cotton tape. (See Fig. 216.)

— ^Wooawi Wedge

Ptpm board or paper
EiDpIn Cloth
-Tape

-— D.c.a

(a) (6)

^a) Open slot containing triple coil

Open slot containing two coil si

Semi-closed slot and "mush" winding.

Fig. 224.— Types of slot,

I}) Open slot containing two coil sides, 12 turns per coil.

To reduce the flux irregularities in the air gap, due to the teeth,
a semi-closed slot. Fig. 224 (c), is used occasionally. In this case
the individual conductors must be placed in the slot one by one,
so the coil ends must be taped after the coils are placed in the
slots. Such a winding is called a ''mush'' winding. The ex-
pense of winding prevents the general use of this type of slot in
direct-current machines.

179* The Commutator. — The commutator is made of wedge-
shaped segments of hard-drawn or drop-forged copper, insulated
from one another by thin layers of mica. The segments are held
together by clamping flanges {DD, Fig. 225), which pull the
segments inward when the flanges are drawn together by through-
bolts. These flanges are prevented from short-circuiting the
segments by two cones of built-up mica {F, Fig. 225). This
construction is illustrated by the commutator of the machine
shown in Fig. 223.

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254

DIRECT CURRENTS

The leads from the armature coils may be soldered into small
longitudinal slits in the ends of the segments or the segments

(il) Assembled commutator.

{B) Commutator bar.
(C) Mica commutator insulating strip.
(DD) Clamping flanges.
{E) Drawn steel tube.
(F) Insulation used between clamping flanges and commutator bars.

Fig. 225. — Crocker- Wheeler commutator and details.

may have risers (Fig. 223) to which these leads are soldered.
(Also see Fig. 214 (6).)

FiQ. 226. — Shunt field coil and edgewise series winding.

180. Field Coils. — The field coils are usually wound with
double-cotton-covered (d.c.c.) wire. The coils are dried in a
vacuum and then impregnated with an insulating compound.

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THE GENERATOR

255

The outer cotton insulation is often protected by tape or cord on
the outside. In the larger machines an air space is often left
between layers for ventilating purposes. The coils are also
wound on metal spools, Fig. 226. An edgewise series winding,
set some distance from the shunt winding, also is shown here.

Fig. 227. — Rocker ring and brush holder.

181. The Brushes. — The function of the brushes is to carry
the current from the commutator to the external circuit. They
are usually made of carbon, although in very low-voltage machines

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256 DIRECT CURRENTS

they may be made of copper gauze, or patented metal compounds.
The brush holder, Fig. 227, is fastened to the brush stud and holds
the brush in its proper position on the commutator. The brush
should be free to shde in its holder in order that it may follow
any irregularities in the commutator. The brush is made to
bear down on the commutator by a spring, Fig. 227. The pres-
sure should be from 1 to 2 lb. per sq. in. To decrease the elec-
trical resistance the upper portion of the brush is copper plated
and this plating is connected to the brush holder by a pig-tail
made of copper ribbon. A rocker ring with cross connections
is also shown in Pig. 227.

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CHAPTER XI

GENERATOR CHARACTERISTICS

182. Electromotive Force in an Armature. — The path of the
magnetic flux from the poles of a generator into the armature, and
a curve showing the flux distribution are given in Fig. 228. The
ordinate at each point is proportional to the flux density in the
air-gap at that point. The maximum flux density is given by
the ordinate -Bmax- The positive ordinates of the distribution
curve are north pole flux entering the armature and the negative
ordinates are flux leaving the armature and entering a south pole.

^/ffliiini[iftw\^TO^> ^/ff/M/iiiiiiiiiii\\\^

^\ /77f/tMniiiinn\\

FiQ. 228.-^Flux distribution at no load of a D. C. generator.

The total flux leaving a north pole is given by the area under one
of the positive parts of the distribution curve. Similarly, the
total flux leaving the armature and entering a south pole is given
by the area of one of the negative parts of the distribution curve.
Each positive part and each negative part of the curve may be
replaced by a rectangle having the same area, as shown by the
dotted line, Fig. 228. The height of this rectangle will be B
maxwells per square centimeter, which is equal to the average
value of the flux density under an entire pole pitch.

Let it be required to determine the average electromotive
force induced in a single conductor as it passes through the flux
of successive poles.

Let the total flux leaving a north pole or entering a south pole
be <l> maxwells. Let A be the pole area in sq. cm., I the active
length of the conductor in cm., s the speed of the armature in
revolutions per second, and P the number of poles.
17 257

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258 DIRECT CURRENTS

When the conductor passes through the distance ab, or one
pole pitch, the average induced voltage, by equation (93), page
217, is

e = Blv 10-8
where B is the average flux density, I the active length of the
conductor in cm., and v the velocity of the conductor in cm. per
second.

where i is the time required for the conductor to traverse the
distance ab.

e = -^^fll0-=|l0-

since Bl (ab) gives the total flux between the points a and
b as cut by the conductor and is therefore equal to <l>.

The time

•-i

Therefore,

the

average voltage per

conductor

If there are Z such conductors and p paths through the arma-
ture, there must be Z/p such conductors in series. (See Par. 169.)
Hence the total voltage generated between brushes is

- - f^ (>»»

Example, — A 900 r.p.m., 6-pole generator has a simplex lap winding.
There are 300 conductors on the armature.

The poles are 10 in. square and the average flux density is 60,000 lines
per sq. in. What is the voltage induced between brushes?

0 = 10 X 10 X 50,000 = 5,000,000 Unes
8 = 900/60 = 15 r.p.s.
P = 6

p = 6 (see Par. 169)
. „ 5,000,000 X 15 X 6 X 300 „_ ,^ .
E = -' -- —Q-^^iQ-s = 225 volts. Am.

183. The Saturation Curve. — Equation (101) may be written
as follows:

where S = r.p.m.

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GENERATOR C/. ^ , ^ miSTICS

259

The quantity within the parenthesis is constant for a given
machine and may be denoted by K,
Therefore:

E = Kit>S (102)

The induced emf. in a machine, therefore, is directly propor-
tional to the flux and to the speed.

If the speed be kept constant, the induced voltage is directly
proportional to the flux, ^.

The flux is produced by the field ampere-turns, and as the
turns on the field remain constant, the flux is a function of the
field current. It is not directly
proportional to the field current
because of the varying permea-
bility of the magnetic circuit.

Fig. 229 shows the relation
existing between the field am-
pere-turns and the flux per pole. ^
The flux does not start at zero |
ordinarily but at some value |
sUghtly greater, owing to the g
residual magnetism in the ma- S
chine. At first the line is prac-
tically straight, as most of the
reluctance of the magnetic
circuit is in the air-gap. At
the point q the iron begins to
be saturated arid the curve falls away from the straight line.

The number of field ampere-turns for the air-gap and for the
iron can be approximately determined for any point on the curve.

Let it be required to determine the ampere-turns for the gap
and for the iron at the point c. From the origin draw oh tan-
gent to the saturation curve and also draw thfe horizontal line ac.
The line oh is the magnetization curve of the air gap, if the
reluctance of the iron at low saturation be neglected. Therefore,
the ampere-turns required by the gap are equal to ah and those
required by the iron are equal to 6c.

From equation (102) the induced voltage is proportional to
the flux, if the speed is maintained constant. Therefore, if the
induced voltage be plotted against field current as abscissas.

Field Ampere-Turns '^NIj
Fig. 229. — Saturation curve.

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260

r \f RECTI ^ ^cENTS

a curve similar to that of Fig. 229 is obtained. This is shown in
Fig. 230 and differs from the curve of Fig. 229 only by a constant
quantity {KS). Two curves are shown in Fig. 230, one plotted
for 1,200 r.p.m. and the other for 900 r.p.m. The curves are

similar, any ordinate of the
lower curve being 900/1,200
of the value of the correspond-
ing ordinate of the upper
curve. Thus, at ordinate ac,
ab ^ 900
ac " 1,200

Also at ordinate a'</
a^ 900
aV 1,200

If the saturation curve of

a generator for one speed has

been determined, saturation

curves for other speeds may be readily found by the method

just indicated.

184. Hysteresis. — The saturation curve oab, Fig. 231 (a), is
determined for increasing values of the field current. If when

Field Current
Fig. 230. — Saturation curves
different speeds.

for two

(a)-

Fig. 231. — Hysteresis loops.

point 6 is reached the field current be decreased, the curve
will not retrace its path along the curve bao. For any given
field current, the corresponding induced voltage will now be
greater than it was for increasing field currents. This is shown
by the curve bed. This is due to hysteresis in the iron. (See
page 181, Par. 142.)

Fig. 231 (b) shows the effects obtained when the curve is

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GENERATOR CHARACTERISTICS

261

carried up aloDg the path oah, back to c, and at c the field
current is again increased, the curve ultimately coming back to
oab at the point a.

It is evident that for any given value of field current, there
is no single value of flux. The value of flux for any given field
current depends upon whether the field current was increased
until it reached the value in question or whether it was decreased.
This characteristic of the magnetic circuit should be carefully
borne in mind, for the operating characteristics of both gen-
erators and motors are affected to a considerable degree by
hysteresis in the magnetic circuit.

185. Determmation of the Saturation Curve. — To determine
,the saturation curve experimentally, connect the field, in series

Am.
O O

r^'

D.C.

Supply

Field

F:g. 202. — Connections for obtaining saturation curve.

with an ammeter, across a direct current source of power. A
voltmeter should be connected across the armature tei^minals.
Obviously the ammeter measures the field current, values of which
are plotted as abscissas; the voltmeter reads the values of induced
armature voltage, which are plotted as ordinafes. These con-
nections are shown in Fig. 232. As the voltage drop within the
armature due to the voltmeter current is negligible, the terminal
volts and the induced volts under these conditions are identical.
During the experiment the speed should be determined each time
that the other readings are taken. If the speed cannot be mam-
tained constant, corrections can be easily made for any variation,
by the method described in Par. 183.

When the saturation curve of a shunt generator is determined

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262

DIRECT CURRENTS

it may be difficult to obtain a sufficiently high resistance to
reduce the field current to its loT^er values. A drop wire con-
nection, Fig. 233, allows field currents as low as zero to be ob-
tained without the use of excessive resistance. Such a connection
is easily made with the well-known ''3-point" type of field
rheostat, shown in Fig. 233.

oEieU

Fig. 233. — Drop- wire connections for obtaining field current.

In determining the saturation curve experimentally, the field
current should be varied continuously in one direction, either up
or down, as shown in Fig. 231 (a). Otherwise minor hysteresis
loops, such as shown in Fig. 231 (6), will be introduced.

The field current in this experiment should be obtained from
a supply other than the generator itself, for two reasons: H
the generator excited its own field, the voltage and field current
would be inter-dependent and it would be difficult to adjust
the field current without the voltage in turn changing this ad-
justment. Also a voltage drop would exist in the armature due
to the field current. The voltmeter would not then be read-
ing the true induced voltage, although the error from this cause
would be sUght.

186. Field Resistance Line. — By Ohm's Law the current in a
circuit is proportional to the voltage, for a constant resistance.
If the current be plotted against volts. Fig. 234, a straight line
passing through the origin results. For example, if the resistance
of a field circuit be 50 ohms, the current will be 2 amperes when
the voltage is 100 volts; 1.5 amperes when the voltage is 75 volts,
and 1 ampere when the voltage is 50 volts. This relation is
shown in Curve II, Fig. 234. Curve I shows the resistance line
for 80 ohms field resistance. It will be noted that at 80 volts the
current is 1.0 ampere, at 40 volts it is 0.5 ampere, etc. Curve III
shows the same relation for a field resistance of 40 ohms.

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GENERATOR CHARACTERISTICS

263

It will be noted that the higher the resistance the greater the
slope of the resistance line. In fact the slope of the line is equal

120
110
100
00

300

50
40
SO
20
10

.sS*^

0.5 1.0 1.5 2.0

Amperes
FiQ. 234. — Field resistance lines.

2.5

to the field resistance in ohms since the tangent, of the angle which

the line makes with the axis of abscissas is -i.

Load

Fig. 235. — Shunt generator connections.

187. Types of Generators. — There are three general types of
generator in common use, the shunt, the compound and the series.
In the shunt type the field circuit is connected across the arma-
ture terminals, usually in series with a rheostat. Fig. 235. The

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264

DIRECT CURRENTS

shunt field, therefore, must have a comparatively high resistance
in order that it may not take too great a proportion of the gener-
ator current. The compound generator is similar to the shunt, but
has an additional field winding connected in series wth the arma-
ture or load. Fig. 270, page 296. The series generator is excited
entirely by a winding of comparatively few turns connected in
series with the armature and load. (See page 301.)

100

Qfl

^',

— i-

1
1

— r

^60

—

5)A

9t\

1^.

■fx r*

Field Current
FiQ. 236. — Method of shunt generator building up.

188. The Shunt Generator. — Fig. 236 shows the saturation
curve of a shunt generator and its shunt field resistance line
drawn on the same plot. This field, it will be observed, has a
resistance of 24 ohms, so that at 120 volts it takes 5 amp.; at
60 volts 2.5 amp. ; etc.

At the instant of starting a generator the induced voltage is
zero. The generator may come up to voltage in the following
manner: As the generator is brought up to speed there will be a

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GENERATOR CHARACTERISTICS 265

small voltage cho, in this instance about 4 volts, induced in the
armature due to the residual magnetism of the machine. This
4 volts also exists across the field, because the field is connected
across the armature terminals. The value of field current which
flows in virtue of this 4 volts can be obtained by drawing a hori-
zontal hne from a imtil it meets the field resistance line at 6. The
current in this particular case is o6'or about 0.2 ampere. By con-
sulting the saturation curve it will be seen that for this field
current the induced voltage, 6'c, is about 8 volts. The 8 volts
produces about 0.33 ampere in the field, as may be seen by pro-
jecting across to the field resistance line at d. This field current
od' produces a voltage d'e, which in turn produces a higher value
of field current. Thus it will be seen that each value of field
current produces a voltage in excess of its previous value and this
increased voltage in turn increases the field current, that is, the
action is cumulative. The machine will continue to build up
until point / is reached, where the field resistance line crosses the
saturation curve. The machine cannot build up beyond this
point for the following reasons:

Consider a point h on the field resistance line, above /. This
point represents a field current og' of about 5.3 amperes. To pro-
duce this field current requires a voltage g'h of about 128 volts.
But this field current of 5.3 amperes produces an induced voltage
g'g of only 122 volts. If 128 volts are required to produce the
field current of 5.3 amperes and the machine can only produce
122 volts at this field current, it is obvious that the machine can-
not build up to the point h.

It is evident that a machine would build up indefinitely if its
iron did not become saturated.

189. Critical Field Resistance. — If the resistance of the field
be increased to 60 ohms, the field resistance line will be repre-
sented by oa, Fig. 237. This line crosses the saturation curve at
point a', corresponding to about 6 volts. Therefore, with this
value of field resistance, the generator will not build up beyond
a'. If the field resistance be slowly decreased until the field resist-
ance line "reaches oh, the generator will be observed to start
building up rapidly. It will of course stop building up at the
point V, The value of the field resistance corresponding to oh is
called the critical field resistance. In this particular case the

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266

DIRECT CURRENTS

resistance is 120/3.25 or 36. 1 ohms. If the field resistance exceeds
the critical value, the generator cannot build up.

190. Generator FaUs to Build Up. — There are three conunon
reasons for a generator failing to build up. (1) The shunt field
may be connected in such a way that the current sent through
it on starting is in such a direction as to ''buck*' or reduce the
residual magnetism instead of increasing it. Under these condi-
tions, the generator cannot of course build up. To test for this,

120
110
100

.-^

Orit

ical
tance

•\

^^60

30
20
10

.0 .

Field Current
Fig. 237. — Critical field resistance.

open the field circuit. If the voltage rises when the field is opened,
the current is bucking the residual magnetism and the field should
be reversed. If opening and closing the field produces no effect
upon the voltmeter it may be assumed that the field circuit is
open.

(2) The field resistance may be greater than the critical field
resistance. In this case, the procedure is to reduce the field
resistance until the machine builds up.

(3) There may be no residual magnetism in the machine, due
to jarring or to too long a period of idleness. If the armature cir-

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GENERATOR CHARACTERISTICS 267

cuit is not open and the voltmeter is known to be all right, the
absence of residual magnetism will be indicated by the voltmeter's
not reading. To remedy the difficulty it may be necessary to
connect the field terminals temporarily across a separate supply
circuit in order to build up the residual magnetism. This is
called ''flashing'' the generator. If the generator has a series
field, a convenient method is to connect a low voltage source,
such as a storage battery or even a dry cell, across the series field.
This may produce enough magnetism to cause the machine to
begin to build up. One or two trials may be necessary in order
to secure the proper polarity.

191. Armature Reaction. — Fig. 238 (a) shows the flux pass-
ing from the field poles through an armature when there is no
current in the armature conductors. This flux is produced
entirely by the ampere-turns of the field. The neutral plane,
which is a plane perpendicular to the flux, coincides with the
geometrical neutral of the system. At the right is shown a
vector F which represents the mmf . producing this flux, in magni-
tude and direction. At right angles to this vector F is the neutral
plane.

In Fig. 238 (6) there is no current in the field coils, but the
armature conductors are shown as carrying current. This cur-
rent is in the same direction in the armature conductors as it
would be were the generator under load. The current obviously
flows in the same direction in all the conductors that lie under
one pole. The current is shown as flowing into the paper on the
left-hand side of the armature. (This current direction may
be checked by Fleming's right-hand rule, Par.* 163.) These
conductors combine their mmf.'s to send a flux downward
through the armature, as shown in the diagram, this direction
bfeing determined by the corkscre^ rule. The conductors on
the right-hand side of tl^e armature are shown as carrying current
coming out of the paper. They also combine their mmf.'s to
send a flux downward through the armature. That is, the conduc-
tors on both sides of the armature combine their magnetomotive
forces in such a manner as to send flux down through the arma-
ture. The direction of this flux is perpendicular to the polar
axis. To the right of the figure the armature mmf. is represented
in direction and magnitude by the vector Fa- f

Fig. 238 (c) shows the result obtained when the field current

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268

DIRECT CURRENTS

and the armature current are acting simultaneously, which occurs
when the generator is under load. The armature magnetomotive

(a) Ourrent in Field Ck>il only

nn^n^xr.

-nugoono

■Brash Axla

Cc) Current in both the
Armature and Field
Fig. 238. — Effect of armature reaction upon the field of a generator.

force crowds the symmetrical field flux shown in (a) into the upper
pole tip in the north pole and into the lower pole tip in the south
pole. As the generator armature is shown rotating in a clockwise

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GENERATOR CHARACTERISTICS

269

• direction, it will be noted that the flux is crowded into the trailing
pole tip in each case. On the other hand, the flux is weakened
in the two leading pole tips.

The effect of the armature current is to displace the field in the
direction of rotation of the generator. It should be borne firmly
in mind that the flux is not pulled around by the mechanical
rotation of the armature.

To the right of Fig. 238 (c) the armature reaction is shown by
vectors. The field vector F and the armature vector Fa combine
at right angles to form the resultant field vector Fo* The direc-
tion of Fo is downward and to the right, which corresponds td the
general direction of the resultant flux in the drawing. The
neutral plane must be at right angles to Foy provided the direc-
tion of the resultant flux is the same as that of the resultant
magnetomotive force.

As the neutral plane is perpendicular to the resultant field,
it will be observed that it too has been advanced. It was shown

X>0xnagiieiIziiiff Pomponenti
of the Armatare limf.

OroM Magnetizmg

Component of the

Armatare Mmf.

Fig. 239. — Relation of armature field to brush axis.

in Chap. X that the brushes should be set so that they short-cir-
cuit the coil undergoing commutation as it is passing through the
neutral plane. When the generator delivers current the brushes
should be set a little ahead of this neutral plane, as will be shown
later. If the brushes are advanced to correspond to the advance
of the neutral plane, all the conductors to the left of the two
brushes must still carry current into the paper, and those to the
right must carry current out of the paper. The result is shown

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270

DIRECT CURRENTS

in Fig. 239. The direction of the armature field moves with the
brushes. Its axis always lies along the brush axis. Therefore
Fa) instead of pointing vertically downward, now points down-
ward and to the left, as is shown by the vectors. Fa may be
resolved into two components, Fd parallel to the polar axis and Fc
perpendicular to this axis.

It will be noted that Fd acts in direct opposition to F, the
main field. Fig. 238. Therefore, it tends to reduce the total flux

nnnnnn

< ti ) Demii£7ie ti zing Armature
Conductors

(b) Cr^i^ Ma^etizing

■ja-^jL^^-p^ \ Armfltnre Conductors

Fio. 240. — Demagnetizing and cross-magnetizing components of armature

reaction.

and so is called the demagnetizing component of armature re-
action. Fc acts at right angles to F and produces distortion,
Therefore, it is called the cross-magnetizing component of arma-
ture reaction.

The exact conductors which produce these two effects are
shown in Fig. 240. In (a) the brushes are shown as advanced
by an angle /3 to correspond to the advance in the neutral plane.
All the conductors within the angle 2/8, both at the top and at

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GENERATOR CHARACTERISTICS 271

the bottom of the armature, carry current in such a direction
as to send a flux through the armature from right to left. This
may be checked by the corkscrew rule. These conductors thus
act in direct opposition to the main field and are therefore called
the demagnetizing armature conductors. Their magnetomotive
force is represented by the component Fd, Fig. 239.

Fig. 240 (6) shows the flux produced by the conductors not
included within twice the angle of brush advance. The direc-
tion of this flux is downward and perpendicular to the polar
axis. These conductors cross-magnetize the field. The mmf.
producing this flux is represented by the component Fc, Fig. 239.
The resultant of Fd and Fc is F^.

It should be remembered that the sum of both the demag-
netizing and cross-magnetizing ampere-turns is equal to one-half
the number of ampere-conductors.

Example, — A 4-pole dynamo has ^8S surface conductors. The machine
is lap wound and delivers 120 amp. to the external circuit. The brushes
are advanced 15 mechanical de-
grees. How many demagnetizing Brush Axi8>^^ /
and how many cross-magnetizing y
armature ampere-tums are there? /

Twice the angle of brush lead Fd / ir- Field

is 30**. There are four brushes, so
that the total number of degrees
covered by the demagnetizing con-
ductors is 120°. Therefore ^i the
conductors on the armature, or
96 conductors, are demagnetizing

conductors. Fio 241.— Resultant effect of armature

As the machine is lap wound and field mmf's.

there are four paths through the
armature. The current per path = 120/4 = 30 amp.

Demagnetizing ampere-conductors — 30 X 96 = 2,880.

Demagnetizing ampere-turns — 2,880/2 = 1,440. Ans.

The number of cross-magnetizing conductors must be % of the con-
ductors on the armature. Therefore, the number of cross-magnetizing
ampere-tums is

192 X 30 „ ^„^ ,
2 "= 2,880. Ans.

Fig. 241 shows the method of finding the resultant magneto-
motive force acting on the armature. F is the field magneto-
motive force and F^ is the armature magnetomotive force, acting
along the brush axis after the brushes have been advanced. Fo is
the resultant of the two, being less than F due to the demagnetiz-

272

DIRECT CURRENTS

ing component of F^- Fa can be resolved into two components
at right angles to each other, Fd the demagnetizing component
of the armature mmf . and Fc the cross-magnetizing component
of the armature mmf.

192. Armature Reaction in Multipolar Maclunes. — Reactions
occur in multipolar machines in the same manner as in the bipolar
machines that have just been described. The picture to the eye
may be a little different, however. In Fig. 242 the armature and

FiQ. 242. — Field flux of a multipolar generator.

the field poles of a multipolar machine are shown, the armature
being shown as a flat surface, for convenience.

In (a) are shown the alternate north and south poles, together
with the magnetic flux entering the armature. There is no cur-
rent flowing in the armature conductors. In (6) the flux dis-
tribution is shown. It will be observed that it is symmetrical
about the polar axis. It is substantially constant under the pole
shoe and drops off gradually at the edges, due to fringing. It
falls to zero and reverses in the interpolar spaces. The neutral
plane is the region where the flux is zero and under no-load
conditions is midway between the poles.

Fig. 243 (a) shows the armature conductors carrying current,
the field current being zero. These armature conductors
produce a flux in a manner similar to that shown in Fig. 238 (b).
The magnetomotive force of the armature is not uniform, but
varies uniformly from zero at the pole axis to a maximum in the

GENERATOR CHARACTERISTICS

273

center of the interpolar spaces. The armature conductors be-
tween the lines qr and st may be considered as constituting a pan-
cake coil, the current flowing into the paper in the conductors
on the left and out of the paper in the conductors on the right.

_ I -j L^ I «

Armature*
Magnetomotive force
Flax
I .

(a)

Fig. 243. — Flux due to armature reaction in a multipolar generator.

Midway between gr and st the magnetomotive force will be a
maximum, as the magnetomotive forces of all the conductors on
both sides are acting together at this point. The magnetomotive
force directly under the pole centers is zero since for every ampere-

Resultant FJux-.^

!/" .^ A 1 Bieucrai nane

FiadFlu:

Fig. 244. — Resultant flux found by combining field flux (Fig. 242) and arma-
ture flux (Fig. 243).

conductor on one side of any pole axis there is a symmetrically-
spaced ampere-conductor on the other side carrying an equal cur-
rent on the same direction. The net magnetomotive force at the
pole-centers due to all such ampere-conductors is obviously zero.
The magnetomotive force distribution along the air-gap is shown

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Google

274 DIRECT CURRENTS

by the dotted line, Fig. 243 (6). Owing to the high reluctance
of the interpolar spaces, the^iix curve has not the same shape as
the magnetomotive force curve but droops in the interpolar
spaces as shown in Fig. 243 (6).

The resultant flux is found by adding the two flux curves of
Figs. 242 and 243, as is done in Fig. 244. (This assumes constant
permeability in the iron.) It will be noted that the flux peaks
on the trailing pole tip, Fig. 244, as in the case of the bi-polar
generator. Also the neutral plane has advanced in the direction
of the rotation. In order to keep the brushes in the neutral plane
they should be advanced as this neutral plane advances. Fig. 245

Fio. 245. — Field distortion in a 4-pole generator.

shows the crowding of flux into the trailing pole tips in a 4-pole
generator.

193. Compensating Armature Reaction. — As the cross-mag-
netizing effect of the armature usually necessitates the shifting
of the brushes with load, it is desirable to minimize armature reac-
tion if this can be done conveniently. One practical method,
when laminated pole cores are used, is to use a stamping having
but one pole tip, as shown in Fig. 221, page 251. These are
alternately reversed when the core is built up. This leaves
spaces between the pole tip laminations, resulting in the pole tips
having but one-half the cross-section of iron along their lengths.
Therefore, the pole tip becomes highly saturated and its per-

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GENERATOR CHARACTERISTICS

275

meabiKty greatly reduced. This tends to prevent the flux from
crowding into the traiHng tip.

Another method is to introduce longitudinal
slots in the pole faces, as shown in Fig. 246.
These slots introduce high reluctance in the
path of the armature flux but have Kttle
effect on the field flux.

The Thompson-Ryan method is to com-
pensate armature reaction by magnetomotive
forces equal and opposite to those of the
armature. In order to be effective, these
compensating magnetomotive forces should
be equal and opposite to those of the armature at every point.
This principle is illustrated by Fig. 247 (a), which shows

Fig. 246. — Longi-
tudinal alots in pole-
face for reducing
armature reaction.

(a)
Fio. 247a. — Compensation of armature reaction with pole-face conductors.

j"H"a~H"rH~iri^~H^"S^rH~airr¥^

(6)
Fia. 2476. — Turns in pole faces to compensate armature reaction.

conductors embedded in the pole faces close to the armature.
Each conductor carries a current opposite to that of its corre-

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• 276 DIRECT CURRENTS

spending armature conductor. This winding is connected in
series with the armature so that the magnetomotive forces are
opposite and equal at all loads. These windings allow the use
of a very short air-gap, with the accompanying reduction in field
copper and in field loss.

The Thompson-Ryan principle has been applied to many
modern machines where commutation difiiculties are unusually
great, as in alternating-current series motors (see Vol. II, page
281) and in large rolling-mill motors. The conductors are
installed in the pole faces in the manner indicated in Fig. 247 (6).
This type of construction is used in the Ridgeway dynamo.

The conductors are connected in series with the armature and
are so adjusted that their ampere-turns are in almost exact op-
position to the armature ampere-tmns at each point. They do
not as a rule have the same number of conductors as the armature
because the armature current at each point ii^ that of one armature
path, whereas the current through this compensating winding is
the sum of the currents in the various parallel paths through the
armature. The small poles between the main poles. Fig. 247 (6),
^re saturated for any flux tending to leak between the main poles.

Armature reaction is also reduced by increasing the length of
the air gap, thus oflFering higher reluctance to the armature flux.
A longer air-gap means more field copper and a greater field
current, however.

194. Commutation. — It has been shown that the electromotive
force induced in any single coil of a direct cm*rent generator is
alternating, and in order that the current may flow always in the
same direction to the external circuit, a commutator is necessary.
Fig. 248 shows the changes of current in an armature coil as it
approaches and recedes from the brushes. It is assumed that
ideal commutation is being realized. The load is such that 20
amp. flow in each path of the armature, making 40 amp.
leaving the machine by this one brush. The current distribution
throughout the brush is also assumed to be uniform.

When in positions (1), (2) and (3) each coil (and, therefore,
successive positions of any one particular coil) carries 20 amp.
As the brush covers four segments and the current distribution
is uniform, 10 amp. must flow into the brush from each seg-
ment. Therefore, when passing from position (3) to (4), the

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GENERATOR CHARACTERISTICS

277

coil must lose the 10 amperes which pass from this segment into
the brush. Hence, in position (4) the coil carries only 10 amperes.
Before reaching position (5) the coil gives up another 10
amperes so that the current is zero when the coil reaches position
(5). When the coil reaches position (6) the current flows through
the coil in the reverse direction, due to current entering the
brush from' another armature path. The current reaches 20
amperes in position (7) and remains 20 amperes in the further
positions (8), (9) and (10).

+ Brush

\\ \ \ \N

h-* 80— ao-u- io-^ 0 *-io ■»^ja6->-y ♦-ao ••-»

Fig. 248. — Current in coil undergoing commutation — ideal conditions.

Therefore, commutation consists of two parts:

1. Reversing the current in any coil from its full positive value
to an equal negative value. This reversal must take place in the
short time interval required for a segment to pass under the
brush.

2. The current supplied by the two paths meeting at the brush
must be conducted to the external circuit.

Part (1) is illustrated by Fig. 248 (6). The current in the
coil is +20 amperes until the brush is reached, when it reverses
at a uniform rate to a value of —20 amperes. This is perfect
commutation.

The -foregoing ideal commutation is only approximated in
practice. There are two causes preventing its realization.

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278

DIRECT CURRENTS

It will be noted that when the coil is in positions (4), (5) and
(6) it is short-circuited by the brush. If any voltage is being
induced in the coil when it is in these positions, a large current
will necessarily flow, since the resistance of the short-circuited

16 6 ^

Fig. 249. — Short-circuit currents through brush.

coil is very low. This resistance consists merely of the resistance
of the coil plus the contact resistance of the brush. This contact
resistance constitutes the major portion of the total resistance.
Fig. 249 shows assumed currents of 15 and 5 amperes flowing

Fio. 250. — Change of current in coil when brushes are too far back of the neutral

plane.

in coils (4) and (5) respectively, due to voltages induced in them
while they are being short-circuited by the brush.

If the local short-circuit currents of Fig. 249 be superposed
upon those of Fig. 248 (a) the current distributes itself over the

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GENERATOR CHARACTERISTICS

279

brush in the manner shown in Fig. 250 (a). There are now 45
amperes entering the brush and 5 amperes leaving it. Therefore,
the brush has to handle 50 amperes instead of 40, and in one
place there are 20 amperes per segment, or twice that which
occurred under the ideal conditions of Fig. 248. This will tend
to produce heating and undue spaxJdng under the heel of the
brush.

Fig. 250 (6) shows the manner in which the current in the coil
varies under these new conditions. Instead of dropping uni-

-I- Brush

\\\\\

T-^Hriizrnaancziiz:

aa-»- 20-> 2(j-> lOo-*. -*-6o -^26o *-f20o *-fiOa *^20a -*— 80a

Fig. 251. — Commutation with the brushes too far ahead,
f ormly from 20 amperes it first rises to 25 amperes before starting
to reverse. It will be noted that the time for reversing from
+20 amperes to —20 amperes has been reduced from time t to
time hj which makes commutation more difficult. The curve of
Fig. 250 occurs when the brush is too far back of the neutral
plane. Voltages are then induced in the coils as they are under-
going commutation.

The curves of Fig. 248 (h) and 250 (h) are called commutation
curves.

If the brushes are placed tod far ahead of the neutral plane,

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280

DIRECT CURRENTS

short-circuit currents flow under the toe of the brush, resulting
in the current distribution and commutation curve of Fig. 251.
This condition produces undue sparking under the toe of the
brush.

If the brushes are too wide, both the heel and the toe will
short-circuit coils in which voltages are induced, resulting in the
commutation curve of Fig. 252. Moving the brushes either
backward or forward does not assist matters in this case. The
only remedy is a narrower brush.

Fig. 252. — Commutation with too wide a brush.

196. The Electromotive Force of Self-induction. — Fig. 253 (a)
shows an armature coil just as it is entering the commutation
zone. The slot conductors are embedded in iron and, due to
the current flowing in the coil, considerable flux passes through
the coil, in this case upward. Let the value of the flux be <^i.
In Fig. 253 (6) the same coil is shown just after it has left the

v^wt^k"^;

V^^i^^"-

l\g/;bi^L

^'L

(a) Before commutation ( b ) After commutation

Fig. 253. — Change of flux through a coil undergoing commutation.

conmiutation zone. The current through the coil is the same

as its previous value, but it now flows in the reverse direction.

The flux is still <^i but it has been reversed in direction.

Therefore, in the time t seconds required for a segment to

pass the brush or commutating zone, the flux has changed by

2<t>i lines. This is shown in Fig. 254, where the ideal commutation

curve is assumed. This change of flux will induce a voltage,

2<^i
e = — N --- 10"^ volts (from equation 74, page 185)

N being the number of turns in the coil. °'^'^'^^' by (^OOglC

GENERATOR CHARACTERISTICS 281

This voltage, with its proper direction, is shown in Fig. 254.
It is called the electromotive force of self-induction.

Instead of looking upon this as a voltage phenomenon, it may
be considered as follows: The armature coil has self -inductance,
and mutual inductance with other armature coils. This induc-
tance tends to prevent the current reversing in the same manner
that inductance tends to prevent change of current in any circuit.

Therefore, even though the brushes are set exactly in the
neutral plane and the coils undergoing commutation are cutting
no magnetic Knes, there will be a voltage induced in the coil
due to its own self-inductance and to mutual inductance. To

ElectromotiTo force o(
SelMndactioD

Fig. 254. — Electromotive force of self-induction in a coil undergoing commuta-
tion.

ehminate this voltage it is necessary to set the brushes ahead of
the neutral plane in a generator. When the coil is undergoing
commutation it finds itself in a field of the same polarity as that
which the conductors leaving the commutation zone are about to
enter. Therefore, this field induces a voltage which assists the
current to reverse.

Another way of stating it is that the electromotive force in-
duced in the coil due to its cutting flux in the zone ahead of the
neutral plane is in exact opposition to the electromotive force of
self- and mutual induction, shown in Fig. 254, and so neutraUzes
it.

It is therefore necessary that the brushes be kept ahead of
the neutral plane in a generator y in. order to obtain satisfactory
commutation under load conditions.

196. Sparking at the Commutator. — The voltages induced
in a coil due to the shifting of the neutral plane and also due
to its own self-inductance are comparatively low in value,
being of the order of magnitude from a few tenths of a volt to
perhaps 4 or 5 volts. But they are acting in a circuit having
a very low resistance. The resistance of one coil is extremely
low so that most of the circuit resistance is at the brush contact.

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282 DIRECT CURRENTS

If the brush contact resistance is too low, these short-circuit
currents may reach such excessive values as to produce severe
sparking at the brushes. On the other hand, a low resistance
brush is desirable from the standpoint of carrying the current
out to the external circuit with minimum contact loss.

Copper brushes have a very low contact resistance, but the
short-circuit currents are excessive when they are used. There-
fore, their appUcation is limited to very low voltage, high
current machines. In this case copper gauze is often used.
Another disadvantage of using copper brushes is that they
*'cut" the conunutator mechanically.

Carbon brushes have a much higher contact resistance than
copper and therefore limit the short-circuit ciu*rents, giving much
more satisfactory results. In addition, they are more or less
graphitic in their composition and so lubricate the commutator
to a certain extent. Unusually hard carbon brushes may cut the
commutator. Different grades of carbon are required for differ-
ent machines.

The passage of the current from the commutator to the brush
is more of an arc phenomenon than it is one of pure conduction.
A careful examination will show myriads of minute arcs existing
between the brush surface and the commutator. The voltage
drop between the commutator and the brush, instead of
being proportional to the current (as it would be with conduc-
tion only) is substantially constant and is equal to about 1
volt per brush. Bits of copper may be found in the positive
brush due to the arcing. The voltage drop across the negative
brush is different from that across the positive brush, due to the
copper being positive in one case and negative in the other.
These facts all substantiate the arcing theory.

Another proof is the so-called "high mica." After a machine
has been in operation for a considerable time, it often happens
that the mica insulation between the commutator segments pro-
trudes above the surface of the commutator, resulting in so-called
"high mica," Fig. 255. It was long supposed that this was
due to the mica being harder than the copper, which resulted in
the wearing away of the copper more readily than the mica. The
fallacy of this supposition is of course evident. Even though the
mica is much harder than the copper, the two must always wear
evenly for the brush cannot grind the copper until it conaes in

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GENERATOR CHARACTERISTICS

283

contact with it. Hence the brush must grind down the mica
before it can touch the copper if ''high mica" is due to mech-
anical abrasion alone.

The rational explanation of high mica is dwelt upon in some
detail by B. G. Lamme in a paper presented before the American
Institute of Electrical Engineers.^ The copper is not worn away
as generally supposed, but is carried away by the minute arcs
that exist between the brush and the commutator, as shown in Fig.
255. This may 'be proved by nmning two similar machines for
the same periods of time, one of the machines delivering current
and the other having no current at all in the brushes and com-

FiG. 255. — Commutator with high Fig. 256. — Undercut mica com-
mica. mutator.

mutator. High mica will ultimately appear on the commutator
which carries current, if conditions warrant, whereas it will be
found impossible to produce high mica on the machine which
carries no current.

High mica may be reduced by the use of fairly hard brushes
which grind the mica down. In modem practice the mica is
under-cut by many manufacturers, that is, the top of the mica is
below the commutator surface, as is shown in Fig. 256. There
is some disadvantage in this construction, in that small bits
of copper, carbon and dirt Collect in the groove and may ulti-
mately short-circuit the segments. These grooves can be easily
cleaned out, however.

The result of any arcing under the brush is to pit the commu-
tator. As irregularities and depressions in the commutator
tend to prevent the brush making intimate contact with the

* "Physical Limitations in D.C. Commutating Machinery," by B. G.
Lamme, A. I. E. E. Trans.y Vol. XXXIV, Part II (1915), page 1739.

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284

DIRECT CURRENTS

commutator, arcs of increasing magnitude will be formed. The
deeper the depressions, or the higher the mica, the larger and
more vigorous these arcs become. Hence, any condition which
produces sparking and so roughens the commutator only in-
creases the sparking and roughening, or, these actions are cumu-
lative. If a commutator is sparking badly and the cause of the

Fig. 257. — Proper method of fitting brushes.

sparking is not corrected, the commutator will deteriorate very
rapidly and soon become inoperative.

The brushes should be fitted very carefully to the commu-
tator surface by grinding with sandpaper in the manner shown
in Fig. 257. Carbon on the surface of the commutator should
be removed with an oily cloth. Do not use waste. A slightly
roughened commutator may be partially smoothed with fine
sandpaper. Do not use emery, as the particles of emery are
conducting and may short-circuit the commutator bars. If the

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GENERATOR CHARACTERISTICS 285

commutator is grooved by the brushes, or is otherwise in poor
condition, it should be turned down in a lathe.

Other difficulties, such as loose mica and loose segments, are
more serious in character. It is often possible to rectify these
difficulties by tightening up the commutator clamp bolts.

197. Commutating Poles (Interpoles). — Fig. 258 shows the
geometrical neutral or no-load neutral plane and the neutral
plane under load. It will be noted that this is merely Fig. 244
reproduced. If the brushes remained in the no-load neutral
plane, there would be severe sparking under load conditions, be-
cause of the very appreciable flux, <^2, due to armature reaction,
now existing in the neutral zone. The brushes will not commu-

Load Neutral^ ^^'V^^

Fig. 258. — Brush advance to proper commutating plane.

tate properly even if advanced to the load neutral plane. This
is due to the fact that the electromotive force of self-induction
still exists in the coils undergoing short-circuit, even if the voltage
due to the pole flux is zero. The brushes must be advanced so
that the short-circuited coils are cutting the flux <t>i of the next
pole, as shown in Fig. 258, in order that an electromotive force
may be generated which will balance the electromotive force of
self-induction. It will be noted that this position is in the fringe
of the next pole flux. A very slight movement of the brushes in
either direction makes a very marked change in the flux so it is
difficult to obtain good commutation under these conditions. In
fact, it may be impossible to obtain satisfactory commutation
because of the steepness of the flux-distribution curve. When the
best position of the brushes is obtained, the trailing tip of each
brush may be in too strong a field and the leading tip in too
weak a field.

If a flux having the same value as <^2, but opposite to it in
direction, could be produced in the geometrical neutral, it is
obvious that the flux in the neutral plane could be brought to zero
in spite of armature reaction. If a flux having a value <^2 + <l>\

286

DIRECT CURRENTS

were produced, satisfactory commutation would be obtained
without moving the brushes. It is the function of commutating
poles to produce just this flux.

Commutating poles consist of narrow poles located between the
main poles. They send a flux into the armature which is of the
proper magnitude to produce satisfactory commutation. For
example, in Fig. 258, the commutating pole must first produce a
flux equal to </>2 so as to neutralize, in the neutral zone, the in-

"^ — n^

L^«

FiQ. 259. — Flux produced by commutating pole alone.

crease of flux due to armature reaction. It must also produce an
additional flux <t>i to balance the electromotive force of self-
induction in the coil undergoing commutation. This commu-
tating pole flux is shown in Fig. 259. The pole producing it at
this point must be a south pole. Fig. 260 shows the resultant
flux obtained by combining Figs. 258 and 259.

Fio. 26O.-7- Resultant of main flux and commutating-pole flux — machine loaded.

As the armature reaction and the electromotive force of self-
induction in the coils undergoing commutation are both propor-
tional to the armature current, the compensating flux produced
by the commutating poles must also be proportional to the
armature current. The commutating poles are wound, therefore,
with a few turns of comparatively heavy wire and are connected
in series with the armature, as shown in Fig. 261. The air-gap

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GENERATOR CHARACTERISTICS

287

between these poles and the armature is large, so that the commu-
tating pole flux is nearly proportional to the armature current at
all loads.

It should be noted that the sequence of poles in the direction
of rotation in a generator is Ns and Sn, where the capitals refer
to the main poles and the small letters refer to the commutating
poles. Fig. 262 shows an interpole separate from the machine.
Fig. 263 shows the frame and field coils for a commutating pole
motor. It will be noted that only two commutating poles are

Fig. 261. — Connections of shunt field and commutating poles.

necessary in this 4-pole machine. Each pole has twice the
strength that it would have if four such poles were used. There-
fore, the proper commutating voltage is induced in but one side
of the coil undergoing commutation.

In practice, commutating poles are so designed that they pro-
duce a flux of greater magnitude than is necessary. The entire
commutating pole circuit is then shunted by a low resistance
shunt, this shunt being adjusted until the best condition of com-
mutation is obtained. The commutating pole shunt is shown in
Fig. 261. This shunt is sometimes made inductive so that the
proper proportion of current will flow to the commutating poles
on sudden changes of load, such as occur in railway generators.

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288

DIRECT CURRENTS

198. The Shunt Generator: Characteristics. — If a shunt
generator, after building up to voltage, be loaded, the terminal
voltage will drop. This drop in voltage will increase with in-

PiQ. 262. — Commutating pole and winding.

crease of load. Such a drop in terminal voltage is undesirable,
especially when it occurs in generators which supply power to
incandescent lamps.

Fiu.

263. — Frame and field coil for Westinghouse 30 H.P., direct-current, interpole

motor.

It is very important to know the voltage at the terminals of a
generator for each value of current that it delivers, because the
ability to maintain its voltage under load conditions determines

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GENERATOR CHARACTERISTICS

289

in a large measure the suitability of a generator for certain
specified service.

To test a generator, in order to determine the relation of
terminal volts to current, it is connected as shown in Fig. 235,
page 263. The machine is self excited and a voltmeter is con-
nected across its terminals to indicate the terminal volts. An
ammeter is connected in the Une to measure the load current. In
performing this test it is often desirable to connect an ammeter in
the field circuit so as to be able to follow the change in the field
current as the load is applied.

In starting the test, rated load should first be applied and
the field current adjusted until rated voltage is obtained. The

Load Current—/
Fig. 264. — Shunt generator characteristic.

load should then be thrown off and the no-load volts read on the
voltmeter. The load should then be gradually applied, reading
the volts and the current for each load. The speed of the gen-
erator should be maintained constant throughout. If the read-
ings be plotted as shown in Fig. 264, the so-called shunt char-
acteristic results. If, in a small generator, the load be carried
far enough, a rapid decrease of voltage will occur, as shown
in Fig. 264. This is called the break-down point of the generator.
Further application of load results in a very rapid decrease of
voltage and beyond a certain point any attempt at increase of
load results in a decrease of current rather than an increase. The
load may even be carried to short-circuit conditions and yet the
current will actually decrease as short-circuit is approached.
This is due to the fact that the field is short-circuited and any

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290

DIRECT CURRENTS

current flowing at short-circuit is due to the residual magnetism
of the machine only.

If the external resistance be now increased, the voltage will
rise slowly and will ultimately reach a value not far below that at
which it started. The fact that the voltage follows a different
curve when the short-circuit is removed is primarily due to
hysteresis. When the load is being appUed, the voltage is drop-
ping and the iron is on the part of the cycle represented by c, Fig.
231 (a). When the voltage starts to increase, it returns along the
path a, Fig. 231 (a), page 260. There is less flux for a given field

280

2S0

—

ftd

T<\

—

2U0

J5160

LOAD CUR

0.C GENERATOR

40*

60 100 150 200 250 800 350 400 460 600
Amperes

Fig. 265. — Typical shunt characteristic.

current and consequently less voltage is induced in the machine
upon the return curve. This, together with a lesser field current
resulting from the lower voltage, accounts for the return curve
lying below the other.

In practice, machines are operated only on the portion ab
(Fig. 264) of the characteristic. Fig. 265 shows this portion of
the curve for a 100-kw., 230-volt generator. The rated current
is 100,000/230 = 435 amperes. The generator field rheostat is
set so that the generator terminal voltage is 230 when it is
delivering this load of 435 amperes.

There are three reasons for the drop in voltage under load of a
shunt generator:

(1) The terminal voltage is less than the induced voltage by
the resistance drop in the armature. That is, the terminal
voltage

V = E - laRa (103)

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GENERATOR CHARACTERISTICS

291

where E is the induced volts, /« the armature current and Ra
the armature resistance.

Example, — ^The voltage induced within the armature of a shunt generator
is 600 volts. The armature resistance is 0.1 ohm. What is the terminal
voltage when the machine delivers 200 amp. 7

Applying equation (103)»

V « 600 - (200 X 0.1) « 600 - 20 « 680 volts

(2) Armature reaction weakens the field and so reduces the
induced voltage.

(3) The drop in terminal voltage due to (1) and (2) results
in a decreased field current. This in turn results in a lesser
induced voltage.

Drop due to Annatnre

BeaotioB

Dr9P due to decreaMd

Field Current

PlO. 266.-

Amperes d
-Voltage drops in shunt generator.

The effect of each of these three factors is shown in Fig. 266.

It might appear that the voltage of the generator would drop
to zero, or practically so, of its own accord when the load first is
applied, because the foregoing cycle is cumulative. That is, a
lesser terminal voltage results in a weaker field, and a weaker
field results in a lesser induced voltage and therefore a lower
terminal voltage which still further weakens the field, etc. The
above cycle would result in the terminal volts reaching zero, if
the iron were not in some measure saturated. If a 10 per cent,
drop in terminal voltage resulted in a 10 per cent, drop in flux,
the generator would be unable to supply any appreciable load.
However, a 10 per cent, drop in terminal voltage, hence in field
current, probably results only in a 1 or 2 per cent, drop in

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292

DIRECT CURRENTS

flux, due to saturation and also hysteresis, as is illustrated in
Fig. 231 (a), page 260. Therefore, a generator when operating at
hi^h saturation maintains its voltage better than when running at
low saturation.

This is illustrated by Pig. 267, which shows two saturation
curves for a 230-volt generator, one at 900 r.p.m. and the other
at 1,200 r.p.m. If the no-load voltage of the generator in each
case is 230 volts, the generator will be operating at point (a)
on the 1,200 r.p.m. curve and at point (6) on the 900 r.p.m.
curve. As point (6) corresponds to a much higher saturation

Field Currentt Load Current

Pig. 267. — Relation of shunt characteristics to speed.

of the armature and field iron than (a), the generator will main-
tain its voltage better at 900 r.p.m. than at 1,200 r.p.m., as
shown by the characteristics in Fig. 267.

199. Generator Regulation. — The ability of a generator to
maintain its voltage under load is a measiu*e of its suitability for
constant potential service. The regulation shows quantitatively
the amount the voltage varies from rated load to no load.

The definition of regulation according to the A. I. E. E. Stand-
ardization Rules is the rise in voltage between rated load and
no load. This is usually expressed as a percentage. Regulation
may be more specifically defined as follows:

Regulation = 100 i~d^ — d volts (per cent.) (104)

As an example, in Fig. 266, the rise in voltage from b to a = be

be
Per cent, regulation = 100 -jr.

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GENERATOR CHARACTERISTICS

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In the 100-kw. generator whose characteristic is shown in
Fig. 265, the no-load voltage is 252 volts. The rated load
voltage is 230.

Per cent, regulation = 100 ^^ = 100 ^^ = 9.6 per cent.

230

200. Total Characteristic. — Reference is often made to the total
characteristic of a shunt generator. The shunt characteristic,
to which reference has already been made, is the relation existing
between load current and terminal volts. The total characteristic
is the relation between armature current and indu^ced volts.

The armature current diflFers from the load current by the
amount of current flowing in the field.

Total Characterlrtle

(104)

o "^"^

Fio. 268. — Total characteristic of shunt generator.

The armature current .

/a = / + //
when / is the load ciu*rent and // the shunt field current.
The induced volts

E = V + laRa

where V is the terminal voltage and Ra the armature resistance,
including brush and brush contact resistance. The total char-
acteristic is the curve showing the relation of la and E. It may
be foimd graphically from the shunt characteristic as follows:

Let qr, Fig. 268, be the shunt characteristic. Draw the field
resistance line Oa, as was done in Figs. 236 and 237. The line
will have the appearance of being nearly vertical, owing to the
fact that the abscissas are plotted to armature current scale.

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294 DIRECT CURRENTS

The horizontal distances from the OY axis, Oq to Oa^ give the
value of field current for each value of voltage. By adding these
distances horizontally to the shunt characteristic, the total cur-
rent is given by the resulting characteristic qe. For example,
at point c on the shunt characteristic the distance cfdf is added
horizontally at cd, giving point d on the characteristic qe.

The armature resistance drop line Oh is then plotted, assuming
that the brush contact resistance is constant. The voltage drop
in the armature is then proportional to the current. It is only
necessary to determine the drop e'f at some value of current
Oe', That is, the voltage drop

e'f = {Oe')Ra

Draw the Une Of, The vertical distances from the OX axis,
Oe' to 0/', give the armature drop for each value of current.
Adding these drops to the characteristic qe, as ef = e'f is added at
the point e, the total characteristic qf is obtained.

It should be borne in mind that the total induced voltage
multiplied by the total cmrent gives the total power developed
within the armature. All of this power is not available, however,
for two reasons:

(1) Some of thi^ power is lost in the armature itself, appearing
as la^Ra loss iu the armature copper.

(2) Some of the armature output is consumed in heating the
shunt field.

Example, — A 20-kw., 220-volt, shunt generator has an armature re-
sistance of 0.07 ohm and a shunt afield resistance of 100 ohms. What
power is developed in the armature when it delivers its rated output?

Rated current

T 220 _ _

^^ =100= 2-2 ^"^P-

la = 90.9 + 2.2 = 93.1 amp.

Field current

Armature current

Induced volts

^ = 220 + (93.1 X 0.07) = 226.5 volts.
Power developed in armature

P = 226.5 X 93.1 = 21.1 kw. Ans,

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GENERATOR CHARACTERISTICS 295

The same result may be obtained by adding power losses as follows:
Field loss

Pf = ^^* = 484 watts.

Armature loss Pa ^ (93.1)*0.07 = 607 watts.

Power developed in armature

P = 20,000 + 484 + 607 = 21,091 watts = 21.1 kw. Ana.

201. The Compound Generator. — ^The drop in voltage with
load, which is characteristic of the shunt generator, makes this
type of generator undesirable where constancy of voltage is essen-
tial. This applies particularly to lighting circuits, where a very

Shunt-field
Rheostat

Pig. 269. — Connections of a compound generator (short shunt).

slight change of voltage makes a material change in the candle-
power of incandescent lamps. A generator may be made to
produce a substantially constant voltage, or even a rise in
voltage as the load increases, by placing on the field core a few
turns which are connected in series with the load. These turns
are connected' so as to aid the shunt turns when the generator
delivers current, Fig. 269. As the load increases, the ciurent
through the series turns also increases and, therefore, the flux
through the armature increases. The effect of this increased flux
is to increase the induced voltage. By proper adjustment of
the series ampere-turns, this increase in armature voltage may be
made to balance the drop in voltage due to armature reaction

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296

DIRECT CURRENTS

and that due to the resistance drop in the armature. If the ter-
minal voltage is maintained substantially constant, the field
current will not drop as the load increases. Therefore, the three
causes of voltage drop, namely, armature reaction, laRa drop,

Shmit
: Field

(a) ShoEtShunt (6) Long Shunt

Fig. 270. — Compound generator connections.

and drop in field current (Fig. 266), are neutralized more or less
completely by the effect of the series ampere-turns.

The shunt field may be connected directly across the armature
terminals. Fig. 270 (a), in which case the machine is called
short shimt. If the shimt field be connected across the machine

Over
ComponndAd

_E!lat

impounded
Under
Compounded

Fio.

Current
271. — Compound generator characteristics.

terminals outside the series field, Fig. 270 (6), the machine is long
shunt. The operating characteristic is about the same in either
case.

If the effect of the series turns is to produce the same voltage
at rated load as at no load, the machine is said to be flat com-
pounded. (See Fig. 271.) It is seldom possible to maintain a

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297

constant voltage for all values of current from no load to rated
load. The tendency is for the voltage first to rise and then to
drop again, reaching the same voltage at rated load as was
obtained at no load. The particular shape of the characteristic
is due to the iron becoming saturated, so that the added series
ampere-turns do not increase the flux at full load as much as they
do at Kght load. When the rated-load voltage is greater than the
no-load voltage, the machine is said to be over compounded.
When the rated-load voltage is less than the no-load voltage, the
machine is said to be under compounded. Generators are
seldom under compounded.

lOO 200

(a) (b)

Fig. 272. — Over-compounded generator maintaining constant voltage at the end

of a feeder.

Flat-compounded generators are used principally in isolated
plants, such as hotels and office buildings. The size of the con-
ductors in the distribution system of such plants is determined
almost entirely by imderwriters' requirements as to carrying
capacity. Wires conformng to these requirements are usually
of such size that only a very small voltage drop takes place be-
tween the generator and the various loads.

Over-compounded generators are used where the load is
located at some distance from the generator. As the load in-
creases, the voltage at the load tends to decrease, due to the
voltage drop in the feeder. If, however, the generator voltage
rises just enough to offset this feeder drop, the voltage at the load
remains constant.

Example. — Consider the conditions shown in Fig. 272 (o). A certain
load is 4,000 ft. distant from the generator. The load is supplied over a
500,000 CM. feeder. The no-load voltage of the generator is 500 volts.
It is desired to maintain the load voltage at a substantially constant value

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298

DIRECT CURRENTS

Diverter

/VNAAA

of 500 volts from no load to the maximum demand of 300 amp. What
must be the characteristic of the generator?

If the cables were operated at the "normal" density the current would
be 500 amp. or 0.001 amp. per cir. mil (Par. 68), and the drop would be
0.01 volt per foot, making a total drop of 80 volts.

The actual drop is

1^ X 80 = 48 volts.

The generator terminal voltage should rise from a no-load value of 500
volts to 548 volts when 300 amp. are being delivered to the load. Fig.
272 ih).

Compound generators are usually wound so as to be somewhat
over compounded. The degree of compoimding can then be

regulated by shunting more
or less current away from the
series field. To do this a low
resistance shimt, called a di-
verter, is used, Fig. 273.

Compound generators
which supply 3-wire distribu-
tion systems usually have two
series field windings, one con-
nected to each side of the
armature. There are two
separate series windings on
each pole, one winding being connected to the positive terminal
and the other to the negative terminal of the machine. (See
Fig. 338, page 376.)

In a compound generator the induced voltage in the armature
is:

. S = 7 + I,R. + laRa (105)

where V is the terminal voltage, /« the series field current, la
the armature current, and R^ and Ra the series field and armature
resistance respectively. (jB« is the equivalent parallel resistance
of the series field and diverter, if a diverter is used. /, then
equals the combined current in the diverter and series field.)
In a long shunt generator J, = /o.

Example. — A compound generator, connected short shunt, has a terminal
voltage of 230 volts when it is delivering a current of 150 amp. The
shunt field current is 4 amp., the armature resistance 0.03 ohm and the

Fig. 273. — Series-field diverter.

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GENERATOR CHARACTERISTICS

299

series field resistance 0.01 ohm. Determine the induced voltage in the
armature, the total power generated in the armature and the disposition
of this power.

The series field current /, = 150 amp., and the armature current /« =
154 amp.

^ = 230 + (160 X 0.01) + (154 X 0.03) = 236.1 volto.
Total power generated

Pa = 236.1 X 154 = 36,400 watts = 36.4 kw.

Armature loss
Series field loss
Shunt field loss
Power delivered

P'o = 154* X 0.03 = 711 watts.

P. = 150* X 0.01 = 225 watts.

P.A = (230 + 1.5)4 = 926 watts.

P = 230 X 150 = 34,500 watts.
Total

y- ,1'

36,362 watta (check).

202. Effect of Speed. — ^Fig. 274 shows the saturation curve
of a 230-volt, compound generator, taken at 900 r.p.m. The

Cnrrent

<ShantElel

'^^\^

Shimt Held >i< i>l i v„ . «. ,.

Series Eleld ^ , *8«rle« Field

(a) (h) (c)

Fig. 274. — Effect of speed upon compound characteristic.

shunt field rheostat is so adjusted that the machine builds up to
a no-load voltage of 230 volts. To produce this result a certain
number of shunt field ampere-turns are necessary, as indicated
by the distance oa. When load is applied to the machine a
certain number of series ampere-turns are added . Let the number
of series ampere-tums be represented by the distance db. Neg-
lecting armature reaction, the induced voltage will be increased
by a value cd shown in heavy lines.

Let thiet same machine be speeded up to 1,200 r.p.m., Fig.
274 (6), and let the no-load terminal voltage still be 230 volts.

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300 DIRECT CURRENTS

The distance oa will now be less than it was in Fig. 274 (a),
owing to the increased speed. But the distance ah will be
the same in each case, as the increase of series turns depends
solely on the load. The increase of voUage cd is much greater in
(6) than in (a), owing to the lesser saturation of the iron. There-
fore, the higher speed machine will have the more rising char-
.acteristic, as is shown in Fig. 274 (c). It will be noted that
the effect of speed upon the compoimd characteristic is just
opposite to the effect of speed upon the shunt characteristic.
(See Fig. 267.) This is due to the fact that saturation opposes
change of the flux in each case.

203. Deteimination of Series Turns: Armature Character-
istic.— It is often desired to determine the number of series
turns which it is necessary to place upon the poles of a shunt
generator in order to make it either flat-compounded or to give
it any desired degree of compoimding.

To make the determination, adjust the no-load voltage to its
proper value.' Let this value of shunt field current be /i. Load
the generator to its rated load and by means of the field rheo-
stat bring the terminal volts to the desired value. Let the cor-
responding value of field ciurent be 1%. The necessary in-
crease of field ampere-tiu'ns is

where Nah = shunt-field turns (either turns per pole or total
turns may be used).

Let / be the rated- load current of the machine, and iV, the
necessary series turns.

Then NJ^ (h-I^N^

The number of series turns for flat-compounding may also be
obtained by means of the armature characteristic. The load is
appUed to the armatiu'e in the usual way. It is preferable to
excite the field separately, as shown in Fig. 275. Load is applied
and the terminal voltage is maintained constant by means of the
shunt-field rheostat. Corresponding values of field current and
armature current are noted. When the two are plotted (as

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GENERATOR CHARACTERISTICS

301

shown in Fig. 276) the resulting curve is the armature character-
istic. The field current increases more rapidly than the armature
current owing to saturation.

To determine the number of series turns necessary, multiply
the increase of field current fee by the shunt turns and divide by
the current Oa,

D.C.

Supply

FiQ. 275. — Connections for obtaining armature characteristic.

Series-field turns for flat compounding

N. = N.K

where N^h is the number of turns of the shunt field.

'^A

FiQ. 276. — Armature characteristic.

204. The Series Generator. — In the series generator the field
winding is connected in series with the armature and the external
circuit. It must consist necessarily of a comparatively few turns
of wire having a sufficiently large cross-section to carry the rated
current of the generator.

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DIRECT CURRENTS

The series generator in most instances is used for constant cur^
rent work, in distinction to the shunt generator which maintains
constant potential. Fig. 277 shows the saturation curve of a
series generator and also its characteristic. The saturation curve
differs in no way from that of the shimt generator. The external
characteristic is similar in shape to the saturation curve for low
saturation. The voltage at each point is less than that shown

yr ^ Drop due to armatuw
X ^ *^ reaction

Critical External /
Resistance LincV^

7/^"^

y/^ \ External 0

Armature and Eield \
Resistance Linej^i^J — -— *•

_ — ■ —

la(R.

+ iJ.)\

Amperes
Fig. 277. — Series generator characteristic.

by the saturation curve by the amount due to the drop through
the armature and field, la{Ra + i2«), and the drop due to arma-
ture reaction. The curve reaches a maximum beyond which
armature reaction becomes so great as to cause the curve to droop
sharply and the voltage drops rapidly to zero. These machines
are designed to have a very high value of armature reaction.

The machine builds up as follows:

If the series field is connected in such a manner that the current
due to the residual magnetism aids this residual magnetism, the
generator will build up, provided the external resistance equals or

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GENERATOR CHARACTERISTICS 303

is less than that indicated by the external resistance line Oa. The
line Oa is therefore called the critical external resistance line. As
the external resistance decreases, the external resistance line
swings down to the right, as has already been discussed for the
shunt generator, Par. 189. The line 06 is such a line. It would be
practically impossible to operate with an external resistance cor-
responding to the line Oa^ or to any line cutting the curve to
the left of d, as a small increase in external resistance would swing
the resistance line away from the curve resulting in the genera-
tor's dropping its load. The machine is designed to operate
along the portion 6c of the curve, which corresponds to substan-
tially constant current. The current is not affected by a con-
siderable change in external resistance, corresponding to the line
06 swinging up or down. To obtain close regulation the series
field is shimted by a rheostat. The resistance of this rheostat
is controlled by a solenoid connected in series with the line. In
this way the current delivered by the generator may be held
substantially constant.

In the past, the series generator has been much used in series
arc lighting. The Brush Arc machiuQ and the Thomson-Houston
generator are common examples of such machines. Both of these
have open-circuit armatures. (See Par. 164.) As the voltage on
the commutator ranges from 2,000 volts to 10,000 volts, the com-
mutators have wide gaps between segments. In the Brush Arc
generator there are as many as two or three separate commutators
connected in series so as to reduce the voltage per commutator
and also to smooth out the ripples in the voltage wave. (See Fig.
191.) There are but four segments per commutator. (For a
more complete description see "Dynamo Electric Machinery,"
S. P. Thompson, Vol. I.)

In Europe, power is transmitted by direct currents at poten-
tials as high as 50,000 volts, in the Thury System.^ This high
voltage is obtained by connecting several series generators in
series and transmitting at constant current. The voltage
increases with the load. The generators have two commutators,
one at each end of the armature. The potential may run as high
as 5,000 volts per commutator. Regulation is obtained by

iSee "Standard Handbook," Fourth Edition, Chap. XI, McGraw-
Hill Co.

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304

DIRECT CURRENTS

shunting the fields. The power is utiUzed by series motors con-
nected at the desired points in series with the Hne.

Series generators are often used as boosters on direct current
feeders. When a drop on a particular feeder becomes excessive,
it may be cheaper to install a booster, and utilize it at the peak
load, than to invest in more copper. The booster is a series
generator operating on the straight portion of the magnetization

Motor

Booster

I Volts drop te
Feedw

Fia. 278. — The series booster.

curve, the terminal voltage being proportional to the current
flowing through the machine. Likewise the voltage drop in the
feeder is proportional to the current in the feeder. If the gene-
rator be connected in series with the feeder. Fig. 278 (a), and ad-
justed properly, its terminal volts may be made always equal to
the drop in the feeder, as shown in Fig. 278 (6). Therefore, the
voltage at the load may be maintained constant. The booster
is direct-connected to a shunt motor taking its power from the
bus-bars. If the driving power should in any way be removed,

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GENERATOR CHARACTERISTICS 305

the series generator will reverse and operate as a motor. The
speed of a series motor without load is practically unlimited, so
that it will run away and tear itself to pieces. Therefore, such a
booster should never be belt-driven and should have some pro-
tective device to prevent its running away.

205. Effect of Variable Speed upon Characteristics. — ^When a
generator is being tested to determine its characteristic or its
regulation, it is assumed that the generator speed is maintained
at a constant value, the rated speed of the generator. Any drop
in voltage resulting from a drop in speed of the prime mover or
driving motor is not chargeable to the generator.

In practice, a drop in speed with load in the case of the prime
mover is often unavoidable. Therefore, the regulation of the
generator is made to include the voltage drop due to this de-
creased speed. When making out specifications, the regulation
of the generator when driven by its prime mover should be
specified. Speed correction applied to characteristics of gener-
ators is somewhat involved, because of the many factors which
enter the computation. For a more complete discussion see ''A
Solution of an Acceptance Test Problem, '' by W. B. Kouwen-
hoven, Elect. Wld., Vol. 71, Jan. 19, 1918.

206. The Unipolar or Homopolar Generator. ^^ — In the ordinary
direct-current generator, the voltage as generated is alternating
and the current must be rectified or commutated. In the uni-
polar generator, however, a direct current is generated, and no
commutator is necessary.

The principle of 'the unipolar generator is that of Faraday's
disc dynamo. Fig. 279 (a). If a disc be rotated between the
poles of a magnet, an emf. is generated between the center and
the rim of the disc. A current can be taken from the disc by
placing a brush at the center and another at the rim. The
disc shown in Fig. 279 (o) would not be practicable because
the electromotive force is generated only at one portion, so that
current can flow back through the disc even when the external
circuit is open. If an annular pole be used. Fig. 279 (6), an
equal electromotive force is generated along each radius, so -that
the current has no return path in the disc itself.

1 For more complete discussion see the ''Standard Handbook," Fourth
Edition, Section 8, Par. 228.

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DIRECT CURRENTS

Fig. 279 (c) shows a crossnaection of a unipolar machine.
The brushes 66 are of one polarity and the brush a is of the
opposite polarity. A hole in the casting allows access to brush
a. Such generators are sometimes made with a rotating cyl-
inder and are said to be of the axial type.

The chief disadvantage of the unipolar type of generator is the
very low voltage generated, even at high speeds. It is necessary
to connect several discs in series in order to obtain working

Fig. 279. — The unipolar generator.

voltages. The generator in Fig. 179 (c), having an armature
diameter of about 20 in., and running at S,000 r.p.m., would
give only about 40 volts. Another disadvantage is the difficulty
of conducting the current from the disc at the high speeds at
which these machines are necessarily run.

Such generators are manufactured by both the General
Electric Co. and the Westinghouse Co. Their field of appUca-
tion is that of a high speed, turbo-driven generator, designed for
high currents at low voltages.

207. The Tirrill Regulator. — It has been pointed out that the
voltage of a generator varies with the load, speed, etc. By
means of a Tirrill regulator, the voltage of a generator can be
maintained constant even under rapid fluctuations of load.

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GENERATOR CHARACTERISTICS

307

In addition, compensation may be made for line drop. The
voltage is controlled by small relay contacts, which short-circuit
the shunt field rheostat, the duration of the short circuit de-
pending upon the amount of regulation required. The field

Compensating
Resistance Shunt

Generator may be
shant or componnd
woand \

Compensating
Winding

Main Control
Magnet

(a)

Fig. 280(a) and (6).— The Tirrill regulator.

rheostat is usually set so that the generator voltage is 35 per cent,
below normal when the regulator is disconnected.

The diagram of the apparatus is shown in Fig. 280. The relay
magnet is Unshaped and has two solenoids, differentially wound,
upon its core. One winding is directly across the line. The

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308 DIRECT CURRENTS

other is connected across the line through the main contacts.
The relay contacts intermittently short-circuit the generator
field rheostat.

The main control magnet can open the main contacts or allow
them to close. These contacts are normally held closed by a
spring. Assmne that the voltage rises. The potential winding
of the main control magnet strengthens this magnet and opens
the main contacts. This opens one of the windings on the relay
magnet and so nullifies the diflferential action. The relay con-
tacts are then pulled open and the short circuit removed from
the generator field rheostat. This immediately reduces the
generator voltage. The reverse action takes place when the
voltage drops.

As a matter of fact both relays are constantly vibrating so
that the changes in the generator voltage are very small.

The relay contacts are shunted by a condenser to reduce
sparking. Owing to the fact that these contacts can carry
only a very small current, it is usually necessary to have the
regulator act on an exciter field, and so maintain the bus-bar
voltage constant through the exciter.

A compensating winding on the main control magnet may be
connected across a series shunt to give the system a rising
\roltage characteristic and so compensate for line drop.

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CHAPTER XII
THE MOTOR

208, Definition. — It was stated in Chap. XI that a generator
is a machine for converting mechanical energy into electrical
energy.

In a similar way the motor is a machine for converting electri-
cal energy into mechanical energy. The same machine however,
may be used either as a motor or as a generator.

209. Principle. — Fig. 281(a) shows a magnetic field of con-
stant strength or intensity in which is placed a conductor that

(a)

(J) w

Fig. 281. — Force acting on a conductor carrying current in a magnetic field.

carries no current. In (6) the conductor is shown as carrying a
current into the paper, but the field due to the N and S poles
has been removed. A cylindrical magnetic field now exists
about the conductor due to the current in it. The direction of
this field, which may be determined by the corkscrew rule, is
clockwise.

Fig. 281(c) shows the resultant field obtained by combining
the main field and that due to the current. The field due to the
current in the conductor acts in conjunction with the main field
above the conductor, whereas it opposes the main field below the
conductor. The result is to crowd the flux above the conductor
and to reduce the flux density in the region below the conductor.

309 Digitized by (^OOgle

310 DIRECT CURRENTS

It will be found that a force acts on the conductor, trying to
push the conductor down, as shown by the arrow.

It is convenient to think of this phenomenon as due to the
crowding of the lines on one side of the conductor. Magnetic
lines of force may be considered as acting like elastic bands imder
tension. These lines always are endeavoring to contract so as
to be of minimum length. The tension in these lines on the
upper side of the conductor is tending to pull it down as shown
in the figure.

If the current in the conductor be reversed, the crowding of
the lines will occur helow the conductor, which will tend to move
it upward, as shown in Fig. 281(d).

The operation of the electric motor depends upon the principle
that a conductor carrying current in a magnetic field tends to
move at right angles to the field.

210. Force Developed with Conductor Carrying Current. —
The force acting on a conductor in a magnetic field is directly-
proportional to three quantities: the strength of the field, the
magnitude of the current, and the length of the conductor lying
in the field. The force in dynes is given by

F = 5ZJ/10 dynes. (106)

where B is the flux density in lines per sq. cm. or gausses, I the
active length of the conductor in centimeters and I the current id
amperes. The direction of the field, the conductor, and the
direction of the force are all mutually perpendicular to one
another.

Example, — ^A coil consisting of 20 turns lies with its plane parallel to a
magnetic field (see Fig. 286), the flux density in the field being 3,000 lines
per sq. cm. The axial length of the coil is 8 in. The current per con-
ductor is 30 amp. Determine the force in pounds which acts on each
side of the coil. (See arrows in Fig, 286a.)
B = 3,000
Z = 8 X 2.54 = 20.32 cm.
/ =30

Fi = 3,000 X 20.32 X 30/10 = 182,900 dynes.
As there are 20 turns

F = 20 X 182,900 = 3,658,000 dynes
3,658,000/981 = 3,730 grams
= 3.73 kg.
3.73 X 2.204 = 8.23 lb. Ans.

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THE MOTOR

311

211. Fleming's Left-hand Rule. — The relation between the
direction of a magnetic field, the direction of motion of a con-
ductor in that field and the direction of the induced electromotive
force is given by Fleming's Right-hand Rule.

In a similar manner, the relation between the direction of a
magnetic field, the direction of a cm-rent in that field and the

Fio. 282. — Fleming's left-hand rule.

direction of the resulting motion of the conductor can be deter-
mined by using Fleming's Left-hand Rule.

Fleming's Left-hand Rule:

Point the forefinger in the direction of the field or flux, the middle
finger in the direction of the current in the conductor, and the thumb
vnll point in the direction in which the conductor tends to move.
This is illustrated by Fig. 282.

(a) Mator ib) Generator

Fio. 283. — Motor and generator action.

"Cotton

Another convenient method for determining the above relation
is to make use of the fact that the crowding of the magnetic
lines behind the conductor tends to push it along. It is necessary
merely to sketch the main field and the Unes about the conductor,
as shown in Fig. 283(a). It is evident that the Unes will be

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DIRECT CURRENTS

crowded at the right of the conductor so that the direction of
motion is to the left.

In Fig. 283(6) is shown a similar condition for a generator.
In this case the conductor, as a generator, moves to the right.
Hence in a generator the conductor must move against a force
tending to oppose its motion, and so the conductor requires a
driving force to keep it in motion. This driving force is supplied
by the prime mover to which the generator is connected.

212. Torque. — When an armature, a fly wheel or any other
device is revolving about its center, a tangential force is necessary
to produce and maintain rotation. This force may be developed
within the machine itself as in a motor or steam engine, or it
may be applied to a driven device such as a pulley, a shaft, a

Force dae to
Belt Tension

Fia. 284. — Torque developed by a belt and by gears.

generator, the driving gears on the wheels otf a street car, etc.
Fig. 284. The total effect of the force is determined not only by
its magnitvde but also by its arnty or radial distance from the
center of the pulley or gear to the line of action of the force.

The product of this force and its perpendicular distance from
the axis is called torque.

Torque may also be considered as a mechanical couple tending
to produce rotation. It is expressed in units of force and dis-
tance.

In the English system, torque is usually expressed in pounds-
feet. (This distinguishes it from foot-pounds which represent
work.)

In the c.g.s. system the unit of torque is the dyne-centimeter
(a very small unit), and in the metric system the unit is the
kilogram-meter.

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Exam-pie, — ^A belt is driving a 36-in. pulley as shown in Fig. 286. The
tension in the tight side of the belt is 90 lb. and that in the loose side is 30
lb. Determine the torque applied 3^ Lb.

to the pulley. ^^■^ ' "^ O

The two sides of the belt are
acting in opposition so that the
net pull on the rim of the pulley is

90 - 30 = 60 lb.

This force is acting 18 in. or
1.5 ft. from the center of the
pulley. Therefore the torque

T = 60 X 1.5 = 90 Ib.-ft. Arm,

Fio.

90 Lb.
285. — Example of torque produced
upon a pulley by a belt.

213. Torque Developed by
a Motor. — Fig. 286 (a) shows
a coil of a single turn, whose plane lies parallel to a magnetic field.
Current flows into the paper in the left-hand side of the coil and
out of the paper in the right-hand side of the coil. Therefore,
the left-hand conductor tends to move downward with a force
Fi and the right-hand conductor tends to move upward with a

Fio. 286. — Torque developed at diflPerent positions of a coil.

force Fi. These two forces tend to rotate the coil about its axis.
Both act to turn it in a counter-clockwise direction and so
develop a torque. As the current in each of these conductors is
the same and they lie in magnetic fields of the same strength, force
Fi = F2. In (a) the coil is in the position of maximum torque
because the perpendicular distance from the coil axis to the forces
acting is a maximum.

When the coil reaches the position (6) neither conductor can
move any farther without the coil itself spreading. This is a

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DIRECT CURRENTS

position of zero torque because the perpendicular distance from the
coil axis to the forces is zero.

If, however, the current in the coil be reversed when the coil
reaches position (6) and the coil be carried slightly beyond the

EL^iojaafi

^^j^^^®

Fio. 287. — Torque developed by belt conductors in motor armatures.

dead center, as shown in (c), a torque is developed which still
tends to turn the coil in the coimter-clockwise direction.

To develop a continuous torque in a motor, the current in
each coil on the armature must be reversed just as it is passing
through the neutral plane or plane of zero torque and a commu-
tator is therefore necessary. This is analogous to using a com-

THE MOTOR 315

mutator in connection with a generator in order that the current
delivered to the external circuit may be uni-directional.

A single-coil motor, like that shown in Fig. 286, would be im-
practicable as it has dead centers and the torque which it de-
velops is pulsating. A two-coil armature would eliminate the
dead centers, but the torque developed would stiU'be more or
less pulsating in character.

The best conditions are obtained when a large number of coils
is used, just as in the armature of a generator. In fact there
is no difference in the construction of a motor armature and a
generator armature. In Fig. 287 (a) an armature and a field are
shown for a 2-pole machine and the torque developed by each
individual conductor is indicated. Fig. 287 (&) shows an arma-
ture and a field for a 4-pole machine. The direction of the
torque developed by each belt of conductors is indicated by the
arrow at that belt.

In armatures of this type a very small proportion of the total
number of coils is undergoing commutation at any one instant.
Therefore, the variation in the number of active conductors is so
slight that the torque developed is substantially constant, for
constant values of armature current and main flux.

From equation (106), the torque developed by any armature
can be shown to be

T = K't ZI^ (107)

where K't = a constant of proportionality, involving the diam-
eter of the armatm-e, the parallel paths through
the armature, the choice of units, etc.
Z = number of conductors on the surface of the arma-
ture.
/ = current supplied to the armature, in amperes.
$ = flux from one north pole entering the armature.

For any particular machine Z is a fixed quantity, so that the
torque

T = Ktl^ (108)

where Kt is a new constant of proportionality.

That 'is, in a given motor, the torque is proportional to the
armature current and to the strength of the magnetic field.

This is a very important relation to keep in mind, for by its
use the variation of torque with load in the various types of motors
can be readily determined. ^^^^^^ ,,(^OOgle

316

DIRECT CURRENTS

Example. — ^When a certain motor is taking 50 amp. from the line it
develops 60 lb .-ft. torque. If the field strength is reduced to 75 per. cent,
of its original value and the current increases to 80 amp., what is the
new value of the torque developed?

If the current remained constant the new value of torque, due to the weak-
ening of the field, would be

0.75 X 60 = 45 Ib.-ft.

Due to the increase in the value of the current, however, the final value of
torque will be

^ 45 = 72 Ib.-ft. Ana,
ou

It must be remembered that the torque expressed by the above
equations is the entire torque developed by the armature. The
torque available at the pulley will be slightly less than this, due
to the torque lost in overcoming friction and in supplying the
iron losses of the armature.

214. Counter Electromotive Force. — The resistance of the
armature of the ordinary 10-horsepower, 110- volt motor is
about 0.05 ohm. If this armature were connected directly
across 110-volt mains, the current, by Ohm's Law, would be

110

/ =

0.05

= 2,200 amp.

This value of current is not only excessive but unreasonable,

especially when one considers
that the rated cmrent of such a
motor is in the neighborhood
of 90 amp. When a motor is
in operation, the current through
the armature is evidently not de-
termined by its ohmic resistance
alone.

The armature of a motor is in
every way similar to that of a
generator. The conductors on
its surface, in addition to carrying
current and so developing torque,
are cutting flux. Therefore,

they must be generating an electromotive force.

If the right-hand rule be applied to determine the direction of

this induced electromotive force (see Fig. 288), it will be found

Ifotion Qf Conductor

Fig. 288. — Relation of the direction
of currents and voltages in a motor
conductor.

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THE MOTOR 317

that it is always in opposition to the current. That is, it opposes
the current entering the armature. This induced emf . is called the
counter electromotive force or back electromotive force. As
the counter electromotive force opposes the current it must also
oppose the line voltage. Therefore, the net electromotive force
acting in the armature circuit is the difference of the line voltage
and the back electromotive force. Let V equal the line voltage
and E the back electromotive force. The net voltage acting in
the armature circuit is

V - E
The armature current follows Ohm's Law and is

TT .. r»

/. =^ ■'V^ (109)

where Ra is the armature resistance.

This equation may be transposed and written

E =V - laRa (110)

This should be compared with equation (104), page 293, which
is the similar equation for a generator.

Li a generator the induced emf. is equal to the terminal voltage
plvs the armature resistance drop. Li a motor the induced
emf. is equal to the terminal voltage minus the armature re-
sistance drop. The counter electromotive force must always
be less than the terminal or impressed voltage if current is to
flow into the armature at the positive terminal.

Example, — Determine the back electromotive force of a 10-hp. motor
when the terminal voltage is 110 volts and its armature is taking 90 amp.
The armature resistance is 0.05 ohm. >

^ = 110 - (90 X 0.05) = 110 - 4.5 = 105.5 volts. Ans. '^

An interesting experiment for demonstrating the existence
of counter electromotive force is shown in Fig. 289. A lamp bank
is connected in series with the armature of a shunt motor.
First close switch S^ which closes the field circuit. Then close
Si. At the instant of closing Si the lamps will bum brightly,
being practically up to candle-power. As the armature speeds
up, these lamps will become dimmer and dimmer, showing that

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318

DIRECT CURRENTS

the armature is generating a counter electromotive force which
opposes the Une voltage and so leaves less voltage for the lamps.
When the armature is up to speed, the lamps will be very dim.
If, however, the field switch S2 now be opened, the flux and, there-
fore, the counter electromotive force will be immediately reduced
to zero practically, which will be shown by the lamps again

00 (^O

Fig. 289. — Demonstration of counter electromotive force.

coming up to full candle-power. (In practice when a motor is
in operation, the field circuit should not be opened under any
conditions whatsoever.)

Equation (101), page 258, for the induced electromotive force
in a generator will obviously apply to a motor. That is, the
counter electromotive force

E = -^-TTTT^ volts
p 10^

where 4> is the total flux entering the armature from one north
pole, s the speed of the armature in revolutions per second,
P the number of poles, Z the number of conductors on the surface
of the armature, and p the parallel paths through the armature.

As Z, P, 7>, and lO""* are all constant for any given motor,
the counter electromotive force becomes

E = Ki<t>S
which is identical with equation (102), page 259, S being given
in R. P. M.

Solving for speed

S ^K

(111)

where

K = 1/Ki

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THE MOTOR 319

The speed of a motor is directly proportional to the counter
electromotive force and inversely proportional to the field.

Substituting for E in (111) its value given in (110), the speed
becomes

>S ^ K^^ ^-^- (112)

This ij3 a very important equation for it shows the law of speed
variation of a motor with changes of load.

Example. — ^A certain motor has an" armature resistance of 0.1 ohm.
When connected across 110-volt mains and taking 20 amp. its speed
is 1,200 r.p.m. What is its speed when taking 50 amp. from these same
mains, with the field increased 10 per cent. ? .

Applying (112)

^110 -'50X0.1 105

Si 02 __ 0f __ 105 <f>i

Si ^ 110-20X0.1 108 4>i ' 108

Therefore:

01 4>i

Si = 1,200

^-I'^iSS

But 02 » 1.10 01

Therefore:

^^ = ^'200i5|.^=l,060r.p,m. Ans.

216. Annature Reaction and Brush Position in a Motor. —
Fig. 290(a) shows a motor armature carrying current. Due
to the armature ampere-turns, a magnetomotive force Fa is pro-
duced in the armature, and the direction of flux produced by this
mmf . is at right angles to the polar axis. Fig. 290 (6) shows the
vectors representing the magnitudes and directions of the arma-
ture mmf. Fa and the field mmf. F. By adding these two
vectorially, the resultant mmf. Fo is obtained. The total flux
produced by Fo is distorted as shown in Fig. 290 (c). It will be
noted that (1) the flux has been crowded into the leading pole
tips, and (2) tlie neutral plane perpendicular to the resultant
field has moved backward. Therefore in a motor it is necessary
to move the brushes backward with increase of load, whereas in a

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320

DIRECT CURRENTS

generator they are moved forward. Were it not for the electro-
motive force of self-induction (see Par. 195), the brush axis
would coincide with the neutral plane. Due, however, to the
necessity of counteracting this last electromotive force, the
brushes are set behind this load neutral plane, as is shown in Fig.
290 (c). That is, in both the motor and the generator it is neces-

onnoprn

Keuiml I'lancv \^

Brush Axis

FiQ. 290. — Armature reaction in a motor,

sary to set the brushes beyond the load neutral plane in order to
counteract this electromotive force of self-induction.

This backward movement of the brushes is accompanied by a
demagnetizing action of the armature upon the field, as indicated
in Fig. 290 (d), where F'^ is the demagnetizing component of

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THE MOTOR

321

Fa' Therefore, as the load is increased on a motor the armature
reaction tends to increase the motor speed. In fact instances
have been known where motors with short air-gaps (producing
high armature reaction) have run away when the load wasappUed.
Fig. 291 shows the armature conductors carrying current
and passing under successive north and south poles. It will be
noted that the armature reaction Fa in the first inter-polar
space is upward. (See Fig. 243.) Therefore, if a commutating
pole is to be used it must be a north pole, in order to oppose this

O O 0O O0O O 0

Rotation

$ O

ee®e®®®®®®000000000o

Fig. 291. — Relation of commutating poles to main poles in a motor.

magnetomotive force of the armature by tending to send a
flux down into the armature. F'a must then be opposed by a
south pole. Therefore in a motor, the relation of main poles and
commutating poles, in the direction of rotation, is Nn /Ss, or
opposite to the corresponding relation for a generator. (See
Fig. 261, page 287.)

The polarity of the interpoles should be carefully investigated
with a compass, if a motor happens to be sparking badly from
some unknown cause, as the sparking may be due to their being
incorrectly connected.

216. The Shtmt Motor. — ^The shunt motor is connected in the
same manner as a shunt generator, that is, its field is connected
directly across the line in parallel with the armature.

A field rheostat is usually connected in series with the field.

If load is applied to any motor it immediately tends to slow
down. In the case of the shunt motor this decrease of speed
lowers the back electromotive force, as the flux remains substan-
tially constant. If the back electromotive force is decreased,
more current flows into the armature (see equation 109, page
317). This continues until the increased armature current pro-
duces sufficient torque to meet the demands of the increased load.

The suitabiUty of a motor for any particular dutj^ is determined

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DIRECT CURRENTS

almost entirely by two factors, the variation of its torque with
load and the variation of its 8j>eed with load.

In the shunt motor the flux is substantially constant. There-
fore, from equation (108), the torque will vary almost directly
with the armature current. For example, in Fig. 292, when the
armature current is 30 amp. the motor develops 40 Ib.-ft.
torque, and when the current is 60 amp. the motor develops
80 Ib.-ft. torque. That is, when the current doubles the torque
doubles.

140

120

100

' /

/■

/•

0 10 20 30 40 60 eo 70 80

Amperes

Fio. 292. — Shunt and series motors; torque-current curves.

The speed of a motor varies according to equation (112), where

7- laRa

S = K

<t>

In the case of the shimt motor, K, V, Ra, and <^ are all sub-
stantially constant. Therefore, the only variable is 7o. As the
load on the motor increases, la increases and the numerator of
this equation decreases. As a rule the denominator changes only a
small amount. The speed of the motor will then drop with increase
of load, as shpwn in Fig. 293. As laRa is ordinarily from 2 to 6
per cent, of F, the percentage drop in speed of the motor is of this
order of magnitude. For this reason the shunt motor is con-
sidered a constant speed motor, even though its speed does drop
slightly with increase of load.

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THE MOTOR

323

Owing to armature reaction, <^ ordinarily decreases slightly
with increase of load and this tends to maintain the speed con-
stant. Occasionally the armature reaction is sufficiently great
to give a rising speed characteristic with increase of load.

Speed Regulation.— The speed regulation of a shunt motor is
almost identical with the voltage regulation of a shunt generator.
It is defined in the A. I. E. E. Standardization rules as being the
difference in the no-load and the rated-load speed divided by

Current « 800

Fio. 293. — Typical shuni motor characteristics.

the no-load speed. That is, in Fig. 293, the percentage speed
regulation is

^-^100 = ^100
ca ca

Example. — The speed of a shunt motor falls from 1,100 r.p.m. at no
load to 1,050 r.p.m. at rated load. What is its percentage speed regulation?

„ , ^. 1,100 - 1.050 ,_^ ... ^ ,

Regulation = TTon ~ 4.54 per cent. Ans.

The speed regulation is a measure of a motor's ability to main-
tain its speed when load is applied.

Fig. 293 shows the three essential characteristics of a shimt
motor, the torque, the ipeed, and the efficiency, each plotted
against current. The effect of the machine losses upon the effi-

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324 DIRECT CURRENTS

ciency will be discussed in the next chapter. It will be noted
that the shunt motor has a definite no-load speed. Therefore
it does not run away when the load is removed, provided the field
circuit remains intact.

Shunt motors are used where a substantially constant speed is
required, as in machine shop drives, spinning frames, blowers, etc.
There is an erroneous impression that shimt motors have a
low starting torque and therefore, should not be started under
load. Starting boxes are usually designed to allow 125 per cent,
of full-load current to flow through the armature on the first
notch. Therefore, the motor develops 125 per cent, of full-load
torque at starting. By decreasing the starting resistance, the
motor could be made to develop 150 per cent, of full-load torque
without trouble. Ordinary starting boxes however will overheat
under these conditions, if the starting period is too long.

217. The Series Motor. — In the series motor the field is con-
nected in series with the armature, as shown in Fig. 294. The

field has comparatively few turns of
wire and this wire must be of suf-
ficient cross-section to carry the rated
armature current of the motor.

In the series motor the flux, <^, de-
pends entirely on the armature cur-

FiG. 294.-Connection8 of a ^.^^^ jf ^j^^ ^^^^ ^f ^^ie motor is
series motor.

operated at moderate saturation, the
flux will be almost directly proportional to the armature
current. Therefore, in the expression for torque,

T = Ktl<l>
if <^ is assumed to be proportional to /, the expression becomes
T = K':P (113)

when K\is a, constant.

The torque is proportional to the square of the armature cur-
rent, as shown in Fig. 292. When the current is 30 amp. the
torque is 20 Ib.-ft.; at 60 amp. the torque is 80 Ib.-ft. That is,
the doubling of the armature current results in the quadrupling
of the torque. It will be noted thai as the current increases
above 60 amp., the torque rises very rapidly. This charac-
teristic of the series motor makes its use desirable where large
increases of torque are desired with moderate increases in cur-

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THE MOTOR 325

rent. In practice, saturation and armature reaction both tend
to prevent the torque increasing as rapidly as the square of the
current. '

When equation (112) is applied to the series motor, the speed

S = K ^ " ^"^^" + ^'^ (114)

where if is a constant, V the terminal voltage, /« the motor cmv
rent, Ra the armature resistance including brushes, i2« the series
field resistance and <l> the flux entering the armature from a
north pole. /?„ the resistance of the series field, is now added to
the armature resistance in order to obtain the total motor resist-
ance. Both la and 0 vary with the load.

As the load increases, the voltage drop in the field resistance
and the armature resistance increases because this voltage drop is
proportional to the current. Therefore, the back emf . becomes
less, which causes the motor to run more slowly, although this
effect is only of the magnitude of a few per cent. The flux
0, however, increases almost directly with the load. Therefore
the speed must drop, in order that the back emf. be of the proper
value, which is usually a few per cent, less than the terminal
voltage. Both effects tend to slow down the motor. The re-
sistance drop is ordinarily from 2 to 6 per cent, of the terminal
voltage V so its effect on the speed is only of this magnitude.
The speed is, however, inversely proportional to the flux <l> and a
given percentage change in <l> produces the same percentage
change in the speed.

Wheii the load is decreased, the flux <t> correspondingly decreases
and the armature must speed up in order to develop the required
back emf. If the load be removed altogether, 0 becomes ex-
tremely small, resulting in a very high speed. It is dangerous
to remove the load from series motors, as their armatures are
almost certain to reach speeds where centrifugal action will
wreck them.

Fig. 295 shows the characteristic curves of a series motor
plotted with current as abscissas. The torque curve concaves
upward for the reasons which have just been stated. The speed
is practically inversely as the current, that is, at large values
of current the speed is low and at small values of current the

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326

DIRECT CURRENTS

speed is high. The characteristics cannot be determined for
small values of current because the speed becomes dangerously
high.

The efficiency increases rapidly at first, reaches a maximum
at about half load and then decreases. This is due to the fact
that at light loads the friction and iron losses are large as com-
pared with the load. The effect of these becomes less as the load
increases. The field and armature loss varies as the square
of the current (PR), so that these losses increase rapidly with the
load. The maximum efficiency occurs when the friction and iron

Fig. 296.-

Current
-Typical series motor characteristics.

losses are practically equal to the copper losses. These curves
should be carefully compared with the corresponding character-
istic curves of the shunt motor, Fig. 293.

Series motors are used for work which demands large starting
torque, such as street cars, locomotives, cranes, etc. In ad-
dition to the large starting torque, there is another character-
istic of series motors which makes them especially desirable for
traction purposes. ' Assume that a shunt motor is used to
drive a street car. When the car ascends a grade, the shunt
motor maintains the speed of the car at approximately the
same value that it has when the car is running on level ground.
The motor therefore tends to take an excessive current. A
series motor, on the other hand, automatically slows down

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THE MOTOR

327

100

90 86

80 82

TO 28 I 1400

S 60 §24 « 1200

S ^ •-

• 60. I20 I 1000

fi s **

$ 40 .3I6 i 800

*^ 30 12

2b 8

10 4 200

0 0 0

i I I I I 1 I I I I 1 I I M

40 H. P. OUTPUT AT 12 AMP. IkPUT

VOLTS AT MOTOR TERMlNAl.8 BOO

DIAMETEB OF CAR WHEEL 83"

ARMATURE S TURN0, HELD SPOOLS n«.» TURNS

PINION t9, GEAR 67, RATIO 8. B»

Fig. 296. — Typical railway motor characteristics.

Fio. 297. — Railway motor with frame lowered for inspection.

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328

DIRECT CURRENTS

upon reaching such a grade, because of the increased current.
It therefore develops more torque at reduced speed. The
drop in speed allows the motor to develop a large torque with but
a moderate increase of power. Hence, a series motor could be
smaller than a shimt motor operating under the same
conditions.

When the characteristics of railway motors are plotted, the
curves refer to the output at the track and not at the motor shaft.
Fig. 296 gives such characteristics for a 500-volt, 40-hp., Gen-
eral Electric railway motor. It will be noted that tractive effort
is plotted rather than torque. The speed of the car in miles per
hour is given rather than the r.p.m. of the motor armature.
These curves differ from the curves of torque and r.p.m.
respectively by a constant quantity, determined by the gear ratio
and by the diameter of the driving wheels. The efficiency curve
is also the efficiency at the rails. These curves resemble closely
the characteristic curves of Fig. 295. Fig. 297 shows a typical
railway motor with half of the casing lowered.

218. The Compound Motor. — ^A shunt motor may have an
additional series winding in the same manner as a shunt gener-
ator. This winding may be
connected so that it aids the
shunt winding, in which case
the motor is said to be cumulat-
ive compound; or the series
winding may oppose the shunt
winding, in which case the motor
is said to be differential compound.
The characteristics of the
cumulative compound motor
are a combination of the shunt
and series characteristics. As
the load is applied the series turns
increase the flux, causing the
torque for any given current to be greater than it would be for
the simple shunt motor. On the other hand, this increase of
flux causes the speed to decrease more rapidly than it does in
the shunt motor. These characteristics are shown in Fig. 298.
The cumulative compound motor develops a high torque with

rpB

ij^

.—

■^

h—

:;.-

-!._

—

.Sfa

J:&l

"Ot

tni

^;!j

™

lie

s.*

Fig. 298. — Torque and speed char-
acteristics of shunt and compound
motors.

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THE MOTOR 329

sudden increase of load. It also has a definite no-load speed, so
does not run away when the load is removed.

Its field of application lies principally in driving machines
which are subject to sudden applicatipns of heavy load, such as
occur in rolling mills, shears, punches, etc. This type of motor
is used also where a large starting torque is desirable but where
a straight series motor cannot be conveniently used. Cranes
and elevators are representative of such loads. In elevators
the series tmns are usually short-circuited when the motor
reaches speed.

In the differential compound motor, the series field opposes
the shxmt field so that the flux is decreased as the load is applied.
This results in the speed remaining substantially constant or
even increasing with increase of load. This speed characteristic
is obtained with a corresponding decrease in the rate at which
the torque increases with load. Such motors are used where a
very constant speed is desired. Because of the substantially
constant speed of the shunt motor there is little occasion to use the
differential motor. In starting a differential compound motor
the series field should be short-circuited, as the large starting
current passing through the series field may be suflSciently large
to overbalance the shunt field ampere-tums and cause the motor
to start in the wrong direction. Typical torque and speed
curves of the differential compoxmd motor are also shown in
Fig. 298.

To reverse the direction of rotation in any motor, either the
armature alone or the field alone must be reversed. If both are
reversed the direction of rotation remains unchanged. There-
fore, in so far as the direction of rotation of the motor is concerned,
it is immaterial which line is positive.

219. Motor Starters. — It was shown in Par. 214 that if a
10-hp., 110-volt motor were connected directly across 110-

volt mains, the resulting current would be ^-^ or 2,200 amperes.

Such a current would not be permissible under commercial con-
ditions. Hence, resistance should be connected in series with
the motor armature when starting. This resistance may be grad-
ually cut out as the armature comes up to speed and develops a
back electromotive force.

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330 DIRECT CURRENTS

Fig. 299 shows the use of a simple resistance R for starting
a motor. It will be noted that this resistance is in the armature

circuit and that the field is connected
directly across the line and outside
the resistance. K the field were con-'
nected across the armature terminals,
putting the resistance R in series
with the whole motor, there would
be little or no voltage across the field
^'''•^!f;;;;^1^^^^r''^'''' »* starting. There would be Uttle

Rtarung purposes. ^^

torque developed and difficulty in
starting would be experienced.

Fig. 300 shows a 3-point starter. This does not differ fun-
damentally from the connections shown in Fig. 299. One line
connects directly to an armature and a field terminal tied to-
g^her. It makes no connection whatever with the starting box.
The other line goes to the line terminal of the starting box
which is connected directly to the starting arm. The starting
arm moves over contacts set in the slate front of the starting
box. These contacts connect with taps distributed along the
starting resistance. The armature terminal of the starting box,
which is the right-hand end of the starting resistance, is con-
nected to the other armature terminal of the motor. The field
connection in the starting box is connected from the first starting
contact, through the hold-up magnet, to the field terminal of the
box. This field terminal is connected directly to the other ter-
minal of the shunt field.

When the starting arm makes connection with the first contact,
the field is put directly across the line and at the same time all
the starting resistance is in series with the armature. As this
arm is moved the starting resistance is gradually cut out. When
the arm reaches the running position, the starting resistance is
all cut out and, to insure good contact, the line and armature
conductors frequehtty are connected directly by a laminated
copper brush, shown in Fig. 300. The field current now feeds
back through the starting resistance. This resistance is so low
compared with the resistance of the field itself that it has no
material effect upon the value of the field current. A spring
tends to pull the starting arm back to the starting position.

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When the arm reaches the running position, it is held against the
action of this spring by a soft-iron magnet (hold-up magnet),
connected in series with the shunt field. (A soft-iron armature is
often attached to the starting arm as shown in the figure.) If
for any reason the line is without voltage, the starting arm will

Fig. 300. — Three-point starting box.

spring back to the starting position. Otherwise, if the voltage
again came on the Une after a temporary shut-down, the station-
ary motor armature would be thrown directly across the Une and
a short-circuit would result.

The advantage of connecting the hold-up coil in series with the
field is that, should the field circuit become opened, the arm

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DIRECT CURRENTS

springs back to the starting position and so prevents the motor
running away.

The 3-point starting box cannot be used to advantage upon
variable speed motors having field control. Such motors fre-
quently have a speed variation of five to one. This results in
the field current having approximately this same range. The

■J ^^JUi^mt

Fig. 301. — Connections for a 4-point starting box.

hold-up magnet may be too strong, therefore, at the higher values
of field current and too weak at the lower values. To obviate
this difficulty a 4-point box is used. Fig. 301. It is similar to
the box shown in Fig. 300, except that the hold-up coil is of high
resistance and is connected directly across the line. The only
difference in the connection is that the ''line terminal" must be
connected to the side of the line which runs directly to the com-

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mon armature and field terminals. When the voltage leaves
the line, the hold-up coil becomes dead and allows the arm to
spring back to the starting position.

Sometimes the field resistance is contained within the starting
box. The box then has two arms, as shown in Fig. 302. The

Fig. 302. — Westinghouse starting and speed-adjusting rheostat.

shorter arm is pushed up by the longer arm and cuts out the
armature resistance in the ordinary manner. During the starting
period the field rheostat is short-circuited by the finger S, Fig.
302. When the starting resistance is all cut out, the shorter
arm is held by the magnet and the short circuit of the field re-
sistance is removed by this arm pushing S to the right. The
longer arm, which has no spring, inserts resistance into the field
circuit when moved backward. When the voltage goes off, the
shorter arm springs back carrying the longer one with it.

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In stopping a motor, the line switch should always be opened
rather than throwing back the starting arm. With shmit motors,
the line switch can be opened with no appreciable arc, since the
motor has a back electromotive force and the field can discharge
gradually through the armature. On the other hand, if the
starting arm is thrown back, the field circuit is broken at the
last contact button. Owing to the inductive nature of the field,
this results in a hot arc which bums the contact. To prevent

(&) Series starter, no-load release
Fig. 303. — Series motor starters.

the contact from being burned, a small finger breaks the arc,
Fig. 302.

The series motor starter needs no shunt field connection. There
are two principal types, one having a no-load voltage release,
shown in Fig. 303 (a), and one having a no-load release, shown in
Fig. 303 (6). In the former type, the hold-up coil is connected
directly across the line and releases the arm when the voltage
goes off the line. In the latter type, the hold-up coil consists
of a few turns in series with the motor. When the motor current
falls below the desired value, the starting arm is released. This

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last type is particularly adapted to series motors where there is
a possibility of the load dropping to such a low value that the
motor speed may become dangerous.

Controllers are used where the operation of the motor is con-
tinually under the direct control of an operator, as in street car,
crane and elevator motors. The controller must be more rugged
than the starting box, since the controller is used for constant
starting, stopping and reversing the motor while operating.
Such controllers usually have an external resistance which is
cut in and out by fingers in the controller. A shunt motor field
rheostat may also be incorporated in the controller. Controllers
are usually fitted with a ''reverse," so that the motor may be
run in either direction.

Fig. 304. — Cutler-Hammer automatic starter — dash-pot type.

Automatic starters are often used in practice. They have
many advantages over the hand-operated starter. They cut out
the starting resistance at a definite rate, so that the blowing of
fuses and the opening of circuit breakers, due to too rapid accelera-
tion, are avoided. In many installations where a motor is used
intermittently, it may be started and stopped by merely turning
a snap switch. Employees will be more likely to shut the motor
down when the power is not being used, because of the ease with
which starting and stopping are effected In the larger sizes of
motors, especially when extremely rapid operation is necessary as
in rolling mills, automatic starters alone can give satisfactory
results.

Fig. 304 shows an automatic starter of the sliding contact type,
with remote control. When the control switch is closed, the
solenoid S becomes energized through L2, i, the control switch.

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C, H, D, to Li. This closes the line switch and energizes the
solenoid through the auxiliary contact Fy to A, B, to Li. The
energizing of this solenoid pulls up the starting arm against the
action of a dash pot. The armature current flows from L2, the
switch, Aij through the series field and armature of the motor
back to Aif to N, through the starting resistance to H, B',
and Li. As soon as the arm starts to move, the line switch sole-
noid circuit becomes L2, L, the control switch, C, -ff, D, B'j Li.
This inserts the additional resistance DB' into this circuit, re-
ducing its current and therefore its power consumption. When

D !-■

R\ Ri

^8 LHu

H5 (h

6- Arm.

A.O.

1 2 (C)3

Fig. 305. — Electric Controller and Mfg. Co. automatic starter.

the starting resistance is all cut out, a brush on the arm makes
contact with B, making direct connection between Li and Ai-
At the same time the contact short-circuiting resistance A-B
is opened. This reduces the current in the solenoid to a value
sufficient to hold up the starting arm. The motor is stopped by-
opening the control switch.

By using 3- and 4-way switches, this type of controller may be
operated from widely separated points. Instead of a simple
snap switch, the motor may be controlled by a float switch, a
pressure switch, or any other automatically-operated switch.

Fig. 305 shows a simple and ingenious type of starter of the
contactor type. The contactors themselves operate as follows:

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Fig. 305 (a) and (5) represent a rectangular iron frame, FFy and
plunger, P. The plunger, P, is narrower at the bottom than at
the top and the narrow part of it fits loosely in an opening in the
bottom of the frame, FF, There are two air-gaps, DD, between
the plimger, P, and the bottom of the frame, PP, and one air-
gap, [/, between the plunger, P, and the top of the frame, FF.
A coil is placed around the plunger, P, as shown in the figure,
where the black circles represent the cross-sections of the wires
of the coil, CC. If a heavy current flows through the coil, mag-
netic lines will stream through the plunger, P, across the air-gap,
f/, back through the frame, FF, and through the narrow part of
the plunger P, and also across the air-gaps, DD Fig. 305 (a) . The
reason that some of the lines go through the air-gaps, DD, is that
the narrow part of the plunger, P, is saturated, or, in other words,
it cannot easily carry any more magnetic lines. These lines, there-
fore, are forced to pass through the air-gaps, DD, when a large
current flows through the coil. The magnetic lines in the air-gap,
f/, cause an upward pull on the plimger, but the weight of the
plunger and the downward pull of the magnetic lines in the air-
gaps, DDj hold the plunger down. In Fig. 305 (6) conditions are
the same as in Fig. 305 (a) except that less current flows through
the coil CC, with the result that there are not so many Unes exist-
ing through the plunger P, the air-gap U, and the frame FF.
Most of these lines now pass through the narrow part of the
plunger, but there are still a few in the air-gaps, DD, The down-
ward pull, due to the Unes passing through the gap, DD, is now
small and the pull in the gap, U, is suflScient to raise the plunger.

The operation of the switch is shown in Fig. 305 (c). When the
line switch is closed, the current flows from the positive main
through the coil Ci of contact 1, the resistances Ri, R2, and R^
in series and the motor armature to the negative main. A shunt
coil. She, on contactor CCz is also put across the line but it is not
sufficiently strong to raise the plunger of 3.

When the current falls to a suflGiciently low value, the plunger
CPl rises, as already described, closing the contact points Bl,
which short-circuits Rj. This causes an increase of current
which now passes through the coil C2. When the current drops
again, due to the motor coming up to speed, contactor CP2
operates, short-circuiting R2 and causing the current to feed

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through Cz. When the current drops again, Cs operates and short-
circuits all the resistances and coils so that the plungers of 1
and 2 fall back. 3 is held up by the shunt coil She.

220. Magnetic Blow-outs. — Controllers and circuit breakers
are often equipped with magnetic blow-outs. Their function
is to extinguish the arc, resulting from opening a circuit, so that
the arc does not persist and so bum the contacts. The principle
of blow-outs is as follows: The contacts between which the arc
is to be broken are placed between the poles of a magnet, as shown
in Fig. 306. When the contacts open, the current tends to per-

Arc

Fig. 306. — Magnetic blow-out.

sist in the form of an arc. This arc finds itself in a magnetic
field so that motor action immediately follows. The arc starts
to move across the field according to Fleming's left-hand rule.
In doing so it draws itself out to such an extent that it is broken.
221. Resistance Units. — Starting boxes are usually designed
for starting duty only. They can carry the starting current
of the motor safely for the short period of starting, but they can-
not carry such a current continuously. The box resistance units
are usually of the type shown in Fig. 307. In the smaller types
the wire is wound in the form of a helix. It may be self sup-
porting or it may be wound on asbestos or porcelain forms, as
shown in Fig. 307. In the larger types, cast-iron grids are used.

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These grids are bolted together. Current lugs are clamped on at
suitable points so that the desired ranges of resistance are readily-
obtainable.

Some types of starter are built in the form of controllers. The
resistance, usually of the grid type, is designed to carry the rated
current of the motor continuously so that it may be used to
secure speed control.

^Mm»

Fig. 307. — Starting box resistance units.

222. speed Control. — In the equation for motor speed, S =
KE/<l)j there are but two factors that can be changed to secure
speed control without making changes in the motor construction.
These factors are the back electromotive force E and the flux <t).

Armature Resistance Control, — In this method the speed control
is obtained by connecting a resistance directly in series with the
motor armature, keeping the field across the full line potential,
as shown in Fig. 308 (a). A wide range of speed can be obtained
by this method and at the same time the motor will develop
any desired torque over its working range, for the torque depends
only upon the flux and armature current

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The principal objections to this method of speed control
are that an excessive amount of power is lost in the armature
series resistance and the speed regulation is very poor. In
Fig. 308 (6) there is shown for comparison the speed-load curves
of a shunt motor with and without resistance in series with the
armature. The speed-load curve with series armature resistance
shows that half speed is obtained at rated load. It will be

(a) W)

Fig. 308. — Speed control and regulation — armature resistance method.

observed that the speed at no load rises to a value which is
practically equal to the speed of the motor when there is no
series armature resistance. The speed regulation with resistance
is about 60 per cent, and about 50 per cent, of the power supn
plied to the armature is lost; in the series resistance. Without
series resistance the speed regulation is the usual 3 or 4 per cent.

Example, — ^A 220-volt, 7-hp. motor has an armature resistance of 0.25
ohm. When running without load at 1,200 r.p.m. the armature takes 6 amp.
(a) What resistance should be connected in series with the armature to
reduce the speed of the motor to 600 r.p.m. at its rated load of 30 amp.?
(6) How much power is lost in the resistance? (c) What percentage of
the power delivered to the armature circuit is delivered at the armature
terminals? (d) What is the speed regulation of the armature? Neglect
armature reaction.

(a)

El (at no load) = 220 - (6 X 0.25) = 218.6 volts.

BOO
E2 (at 600 r.p.m.) -

Total {R + Ra) =

218.5 = 109.3 volts.
110.7

1,200

220 - 109.3

30 30

Subtracting the armature resistance,

R = 3.69 - 0.25 = 3.44 ohms.

= 3.69 ohms.

Ana.

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(6) Power lost in the series resistance

Pi = (30)2 X 3 44 ^ 3 Q9g wQ,tiH, Ana.

Pi = 220 X 30 = 6,600 watts.
Power delivered to armature

Pi = 6,600 - 3,096 = 3,504 watts.
Percentage power delivered to armature
^ 3,504
6,600

(d) Speed regulation

1,200 - 600

= 53.1 per cent. Ana,

1,200

= 50 per cent. Ans,

Multi-voltage System. — In this system several different volt-
ages are available at the armature terminals of the motor.
These voltages are often supplied by a balancer set, Fig. 309.

Balancer Set
Fid. 309. — Multi-voltage speed control.

The shunt field of the motor is connected permanently across
a fixed voltage and, with the 4-wire system shown, six voltages
are available for the armature. Intermediate speed adjustments
can be made with a limited field control. Owing to the necessity
of having a balancer set, or its equivalent, and due to the large
number of wires necessary, this system is little used in this
country.

Ward Leonard System, — In this system, shown in Fig. 310,
variable motor voltage is obtained by means of a separate
generator, 6, driven by a motor. Mi. By varying the field of the
generator, the desired voltage across the motor terminals, Af 2,
is obtained. The motor field is connected across the supply
mains in parallel with the fields of the other two machines.

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In Fig. 310, Ml is a motor driving generator G, (? in turn sup-
plies variable voltage to the armature of motor Jlf 2 whose speed
is to be varied. This system is very flexible and gives close
adjustment of speed. The chief disadvantages are the necessity
of having the two extra machines and the low over-all eflSciency
of the system, especially at light loads. This system has been

Mains

Fi^ld

Fig. 310. — Ward Leonard system of speed control.

used extensively for turning the turrets of battleships, but is
now superseded for this purpose.

Field Control. — In the foregoing methods of speed control,
the armature volts have been varied. A change of speed may also
be obtained by varying the flux, <^, by means of a field rheostat.
This method is very efiicient so far as power is concerned and for

->\ Field Flti*

Fig. 311. — Eiffect of a weak field upon brush position.

any particular speed adjustment the speed regulation from no
load to full load is excellent. The range of speed obtainable by
this method with the ordinary motor is limited by commutation
difficulties. Referring to Fig. 311, F is the field flux at low
speed and Fa is the corresponding armature flux. The resultant
flux is Fo. If it be attempted to double the speed of the motor
by weakening its field, the new field flux will be F\ The brushes
will now have to be moved farther backward so that the armature

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flux will be at the position shown at F a. The resultant field
is F\.

It is evident that the neutral plane has been moved backward
to a considerable extent and that the armature flux is about
equal to the field flux. In addition to severe sparking at the
commutator, the strong armature field may so weaken the main
field that the motor tends to run away. In order to eliminate
the demagnetizing action due to the moving of the brushes,
commutating-pole motors only
should be used where the
speed range is large. A range
of 5 to 1 in speed variation
is obtainable with properly
designed machines having com-
mutating poles.

The Stow Motor. — In this
type of motor, shown in Fig.
312, the field cores slide in
and out of the yoke and are
actuated by a hand wheel
through a rod and bevel gear
mechanism. By varying the
length of the air-gap, the flux,

and therefore the speed of the motor, may be varied. As the
reluctance to the armature flux is increased at the higher speeds,
with the increased air-gaps, there is little difficulty with commu-
tation. In other words, the ratio of field ampere-turns to
armature ampere-turns does not change.

The Lincoln Motor, — In the Lincoln motor, made by the Re-
liance Electric and Engineering Company, the flux entering
the armature is varied by moving the rotating armature in and
out of the field structure, as shown in Fig. 313. As the armature
is moved out of the field the length of armature conductor
cutting flux is reduced. Therefore the armature must rotate
faster in order to develop the requisite electromotive force.
This gives a finely graduated speed control over wide ranges,
ratios as high as 10 to 1 being obtained. These motors are
provided with commutating poles.

Fig. 312. — The Stow motor.

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Fig. 313. — ^Lincoln adjustable speed motor.

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223. Railway Motor Control. — ^In a 2-inotor trolley car, two
different si)eeds can be efficiently obtained. The motors are
first connected in series through a starting resistance 22 as shown
in Fig. 314 (a). This resistance is gradually cut out by the con-
troller as the car comes up to speed and then each motor receives
one-half the line voltage. This is the first running position. For
any given value of armature current each motor will run at half
its rated speed. As there is no external resistance in the circuit,
the motors are operating at an efficiency very nearly equal to
that obtainable with full-line voltage across the terminals of each.

Trolley

( }>) Running with
Motors in Parallel

ICall

Fig. 314. — Series-parallel control of series motors.

When it is desired to increase the speed of the car, the two
motors are thrown in parallel with each other and in series with
a portion of the resistance J?. This resistance is gradually cut
out and when the running position is reached, each motor
receives full-line voltage, as shown in Fig. 314 (6).

In a 4-motor car, the motors are usually divided into two groups,
each group consisting of two motors which are always in parallel
with each other. In starting, these two groups are connected
in series, each group taking the place of the single motor of a
2-motor car. This starting condition is shown in Fig. 314 (c).
When the full-speed running position is reached, both groups are
connected in parallel across the line. Each motor then receives
full-line voltage.

Multiple-unit Control. — In the heavier electric cars and locomo-
tives, the currents become so large that direct platform control is
out of the question from the standpoint of the size of controller,

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safety, and expense. Moreover, when cars are operated in
trains, it is necessary that the motors on all the cars shall be under
a single control and that they shall operate simultaneously.

In the multiple-unit system, all the heavy-current switching
is done by solenoid-operated contactors located beneath the car.
These contactors in turn are operated by an auxiUary circuit
called the train line, which runs the entire length of the train (Fig.
315) . The train line is made continuous through plug and socket

Trolley Wire

, S^ Coupler „ . , .

J " / Train Line 1

Ground —

"^-S^i

"=?=" Ground

Car 2 Oar 1

Fig. 315. — Principle of multiple-unit control.

connectors located in the car couplers. The wires of this train
line receive their power through the master controller operated
by the motorman. As this train line current is only of the mag-
nitude of 2.5 amperes, a small platform controller can be used.
Another distinct advantage of this system is that the rate of
cutting out the starting resistance during the acceleration periods
is outside the control of the motorman, being accompUshed by
automatically-operated contactors which close in sequence at the
proper times. This insures uniform acceleration and eUminates
the opening of the car circuit breakers and the shocks to the equip-
ment caused by too rapid acceleration when manual operation
is used.

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Fig. 315 shows the underlying principle of the system, no
attempt being made to give the many details which must neces-
sarily accompany such a system. Each car has its own trolley
or third-rail shoe for collecting the current A train line of small
wires runs the entire length of the train, the connections being
made by the use of couplers between cars. This line usually
consists of six wires. Solenoids, operating contactors, are con-
nected across the train lines. Some of the contactors are oper-
ated directly by the controller in the hands of the motorman and
others operate automatically after the controller has been turned
to the desired position. For example, in Fig. 315 are shown two
motors, one in each car. One line of the train line is shown run-
ning between cars and connected by the coupler. It is assumed
that the train is to be operated from car 1. If the switch Si
in the controller of car 1 be closed, train line 1-1 becomes alive.
This energizes relay (1) (1) in each car and both relays simultan-
eously close the motor circuits, the starting resistances 2Ji, R2
being in series with each motor respectively. As the motors
''pick up,'' the current drops and relays (2), (2), become auto-
niatically energized and some of the starting resistance /?i, Ri is
cut out in each car. The next set of relays become energized in a
similar manner, until all the starting resistance is cut out and the
motors are across the line.

The above is merely an abbreviated description of the sys-
tem. In the complete system there are six train lines, some of
which reverse, change from series to parallel, etc. The great
advantage of this system is that every motor on the train can be
operated from either controller on any one car, that all the motors
act simultaneously, the acceleration cannot exceed a certain value
irrespective of the motorman, and as there are driving wheels
on every car, high accelerations can be obtained. This system
is also used extensively on single cars.

224. Dynamic Braking. — It is often desirable to brake a motor
when it is being driven by its load, as in the case of descending
elevators, cranes, etc. This is often done by using a controller
which leaves the field connected across the Une and at the same
time puts a resistance load across the armature terminals. This
produces generator action and therefore retards the armature.
If series motors are used, their fields must be connected across

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DIRECT CURRENTS

the line in series with a resistance. Such braking is not effective
for completely stopping the motor armature, as the braking action
ceases when the armature is stationary.

Dynamic braking for a series motor is shown in Fig. 316. In
(a), which shows the holding or "ofif" position, the motor is
totally disconnected from the line. The solenoid of the mechan-
ical brake becomes de-energized, resulting in the brake being
set. (See Fig. 33, page 23.) In (6), the brake solenoid and the
series field are connected across the line in series with a resistance.

(&) Braking or Lowering
Fio. 316. — Dynamic braking.

The armature has a resistance connected across its terminals
through the brake solenoid and series field on one side. The
brake is released, the armature acts as a generator sending cur-
rent through the braking resistance and so is retarded.

Regenerative braking is based on this same principle, except
that the power is returned to the line rather than wasted in re-
sistance. Such a system is used on the electric locomotives of
the Chicago, Milwaukee and St. Paul Railroad.

226. Motor Testing — Prony Brake. — It is often necessary to
determine the efiiciency of a motor at certain definite loads and

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frequently over its entire range of operation. A knowledge of
the efficiency may be necessary, as in the case of an acceptance
test; further, the motor may be used as a power-measuring device
for determining the power taken by some machine, such as a
generator, pump, blower, etc. Knowing the motor input, which
can be measured with an ammeter and a voltmeter, and also

|BHfiWBMi.

>; stops

Fig. 317. — Typical prony brake.

knowing the motor efficiency, the output for any given input
can be computed. This output will be the power delivered to
the generator, the pump, etc.

The most common method of making direct measurements of
efficiency in motors up to about 50 hp. is to use a prony brake.
Such brakes are made in various forms. One typical form is
shown in Fig. 317. It consists of a wooden arm of the proper
length, a canvas brake band and a hand wheel for applying ten-
sion to the brake band. By means of this hand wheel the motor

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load can be controlled. An oil dash pot is advisable, to prevent
vibrations of the brake arm.

The balance measures the pull on the arm due to the rotation
of the drum, plus the dead weight of the arm. By multiplying
the net balance reading by the distance L, the torque of the motor
can be determined.

There are two simple methods for determining the dead weight
of the brake arm. The brake band is loosened and some sort
of knife edge, such as a pencil, is placed between the top of the
drum and the brake carriage. This acts as a substantially fric-
tionless fulcrum, so that the balance registers the dead weight of

the arm alone. Another and
easier way is to turn the drum
toward the balance by hand,
stop and read the balance.
In this case the friction of
the brake causes the balance
to read too high. If this
operation be repeated by ro-
tating the drum in the opposite
direction, the balance reading
will be too low, due to the
same friction. The average
of these two balance readings
will give very nearly the cor-
rect value for the dead weight
of the arm.
Brakes oi this type are cooled ordinarily by pouring water
into the hollow brake drum. This water prevents the drum from
becoming excessively hot. As the maximum temperature which
water can reach in the open air is 100° C, the drum temperature
cannot much exceed this. The heat developed in the drum is
utilized in converting the water into steam. As a considerable
number of heat units are required to convert a small amount of
water into steam, a moderate amount of water will keep the drum
comparatively cool.

To determine the equation for the horsepower developed by
such a brake, consider Fig. 318. Let F be the net force in pounds
acting at a perpendicular distance L feet from the center of the

Fig. 318.-

-Work developed by a prony
brake.

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THE MOTOR 351

drum. First assume that the drum is stationary and that the arm
is pulled around the drum by means of the force F. The distance
per revolution through which the force F acts is 2tL, The work
done in one revolution of this arm around the drum is the force
times the distance = F(2tL).

The work done is S revolutions = F{2TrL)S.

If /S is the revolutions per minute, the horsepower

27r(FL)S

33,000
but FL is the torque T, therefore

2irTS

Hp. =

lierefore

Hp. =

33,000
^'^ = 0.00019

33,000
Therefore Hp.= 0.00019 TS (115)

Obviously, the same amount of work is done on the brake sur-
face whether the drum is stationary and the arm rotates or the
arm is stationary and the drum rotates. Therefore, equation
(115) applies to brakes of the type shown in Figs. 317 and 318.
It will be noted that in this particular type of brake the horse-
power is independent ot the diameter of the drum.

Example, — In a brake test of a shunt motor, the ammeter and voltmeter
measuring the input read 34 amp., 220 volts. The speed of the motor
is found to be 910 r.p.m. and the balance on a 2-ft. brake arm reads
26.2 lb. The dead weight of the arm is found to be +2.4 lb. (a) What
is the output of the motor? (6) What is its efficiency at this particular
load?

(o) Net reading of balance = 26.2 - 2.4 « 23.8 lb.
The torque T = 23.8 X 2 = 47.6 Ib.-ft.
Hp.' output = 0.00019 X 47.6 X 910 = 8.23 hp. Ana.
(6) Output = 8.23 X 746 = 6,140 watts.
Input = 220 X 34 = 7,480 watts.

Efficiency 17 = -' 100 = 82.1 per cent. Ans,

In brakes of this type, the brake arm should be kept approxi-
mately level.

Another simple type of brake is the rope brake shown in Fig.
319. A rope is given a turn and a half around a drum and the
two free ends are each held by a spring balance. The larger bal-

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352

DIRECT CURRENTS

ance is on the end of the rope which is being pulled downward by
the rotation of the drum. Let Fi be the reading of the larger
balance and Fi that of the smaller balance. As Fi and F2 pull
in opposite directions with respect to the rotation of the drum,
the net pull at the drum periphery is Fi — Fj.
The torque in Ib.-ft. is

r= (Fi-F2)R
where R is the radius of the pulley in fed.

FiQ. 319. — Rope brake.

Fio. 320. — Jagabi Tachoscope.

Example. — In a rope brake of the type shown in Fig. 319, Fi = 32.4 lb.
and Ft = 8.2 lb. TTie drum is 10 in. in diameter. If the motor speed
is 1,400 r.p.m., what horsepower does the motor develop?

The torque

T = (32.4 - 8.2) ^ = 24.2 X 6/12 = 10.08 lb.

The horsepower

Hp. - 0.00019 X 10.08 X 1,400 = 2.68. Ana.

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THE MOTOR

353

226. Measurement of $peed. — The measurement of the speed
of machines is as a rule much simpler than the measurement of
torque. The most common method is to use a simple revolution
counter having a conical rubber tip which fits into the counter-
sink of the shaft. The Veeder type is a convenient form of
revolution counter. The revolutions are recorded directly on the
counter. As this counter cannot be set to zero, the actual speed
must be found by subtracting the counter reading before from
that after the measurement.

The Jagabi tachoscope, Fig. 320, is a combination of speed
counter and stop watch. The spindle may be inserted in the
counter-sink of the shaft without recording. A little pressiu-e.

•b

Volts

(6) Speed-voltage curve of
magneto.

Speed measurement with magneto and voltmeter.

however, causes the counter and stop watch to start simultan-
eously. They also stop simultaneously when the pressure on the
tachoscope is removed. Measurements made with this type of
instrument are free from personal error.

Tachometers indicate the instantaneous value of speed.
There are mechanical tachometers, where the indicator is
actuated by centrifugal action. This type should be carefully
checked at each occasion of use, as it is especially subject to
error after having been in service for some time.

A simple and convenient type of tachometer is the combination of
a direct-current magneto and a voltmeter, as shown in Fig. 321 (a).
In the magneto the flux is produced by permanent magnets and
so is constant. Therefore, the voltage induced in the magneto
armature is directly proportional to the speed. If this voltage

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354 DIRECT CURRENTS

be measured with a voltmeter, the voltmeter reading multiplied
by a constant gives the speed directly. The relation of speed
to volts may be plotted as shown in Fig. 321(6) and the speed
read directly from the plot. This plot is ordinarily a straight
line through the origin, which makes one point accurately de-
termined. It is convenient to attach the magneto to the shaft
of the machine whose speed is being measured, by a piece of
rubber tubing. It is usually necessary to thread a small stud
into the end of the shaft whose speed is to be measured, as
shown in Fig. 321 (a).

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CHAPTER XIII
LOSSES; EFFICIENCY; OPERATION

228. Dynamo Losses. — A certain portion of the energy de-
livered to any motor or generator is lost within the machine itself,
being converted into heat, and therefore wasted. This represents
not only energy lost, but has the further objection that it heats
the machine and so limits its output. If the energy loss in the
machine becomes excessive, the resulting temperature rise may
injure the insulation by carbonizing it.

As a motor and a generator are similar, they have the same
types of losses throughout. Therefore, the following applies to
either a motor or a generator.

COPPER LOSSES

Amidlure, — The armature windings have a certain resistance
and when current flows through them a certain amount of

Fig. 322. — Measurement of armature resistance.

power must be lost. In addition to the loss in the armature
copper, there is an electrical loss in the brushes and in the com-
mutator. Let this tptal power loss be Pa- Then,

Pa = la^'Ra - (116)

where /« is the armature current and fia i& the armature resist-
ance measured between the terminals of the machine and includ-
ing, therefore, the brushes and their contact resistance. This
contact resistance is not exactly constant, but little error is made
in assuming it to be so. (See Par. 196.) The resistance measure-
ment is often made by the connections shown in Fig. 322. The

355

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356 DIRECT CURRENTS

resistance R is inserted to limit the current flowing through
the stationary armature. (See Par. 118.) The measurement
should be made with the armature in three or four different posi-
tions in order to obtain an average value of resistance. As the
low-reading scale of the voltmeter is ordinarily used in making
this measurement, the instrument may be injured on opening the
circuit by the rise of voltage due to the self-inductance of the
armature. Therefore, the voltmeter should be disconnected
when the circuit is being opened or closed and when the armature
is being turned.

Shunt Field, — ^The field takes a current 1/ at the terminal vol-
tage V of the generator or motor. Therefore, the power lost in
the field is

Pf = Vis (117)

This includes the power lost in the field rheostat as this is
chargeable to the field circuit.

Series Field. — ^The series field loss is

P. = I.^R. (118)

where I, is the series field current, which may or may not be equal
to the armature current, depending on whether the machine is
long or short shunt.

Rt is the series field resistance. If a series field shunt or
diverter is used, R» is the equivalent parallel resistance of this
diverter and the series field and 7, is the current of the series
field plus that of the diverter.

The losses in the commutating pole circuit are determined in
the same way as are those of the series field.

The foregoing losses are all copper losses and can be either
measured directly or calculated with a high degree of precision
from instrument readings.

IRON LOSSES

Eddy Currents, — As the armature iron rotates in the same
magnetic field as the copper conductors, voltages are also induced
in this iron. As the iron is a good conductor of electricity and
the current paths are short and of large cross-section, large cur-
rents would be set up in the armature iron were it a solid mass as
shown in Fig. 323 (a). These currents represent an excessive

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LOSSES; EFFICIENCY; OPERATION 357

power loss which could not be tolerated in a commercial machine.
By laminating the armature iron in the manner indicated in Fig.
323(6), the paths of these currents are broken up and their magni-
tude is reduced to a very low value. Laminating does not en-
tirely eliminate these eddy current losses, but it does reduce them
to a small value. It will be noted that although the laminations

(«) (6)

Fio. 323. — Eddy currents in armature iron without and with laminations.

break up the eddy current paths, they do not interpose reluctance
in the magnetic circuit, since they are parallel to the direction of
the magnetic flux.

These eddy currents are proportional to both the speed and
the flux. As the loss varies as the square of the current (PR),
the eddy current loss varies as the square of both the speed and
the flux.

Example. — The eddy current loss in a certain machine is 600 watts when
the total flux is 2,000,000 lines per pole and the speed is 800 r.p.m. What
is the loss when the flux is increased to 2,500,000 lines and the speed increased
to 1,200 r.p.m.?

^• = ™x(IS)'x(S)"-.™"'---

Hysteresis, — ^It was shown in Chapter VIII that when iron is
carried through a cycle of magnetization (Par. 143) there results an
energy loss proportional to the area of the hysteresis loop. The
iron in an armature undergoes a similar cycUc change of magneti-
zation when the armature rotates. Consider the small section
of the armature iron at (a). Fig. 324, when it happens to be under
a north pole. This small section has a north and a south pole at
its ends. When the section reaches position (6) its poles have
become reversed, as shown. Obviously, nearly all the armature

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358 DIRECT CURRENTS

iron is continually going through similar cycles of magnetic
reversals. Therefore, there results a hysteresis loss in the arma-
ture iron as the armature rotates. This loss is directly propor-
tional to the speed and is proportional to the 1.6 power. of the
maximum flux density, by the Steinmetz formula, (^uation 72,
page 183) Laminating the iron does not affect the hysteresis loss.

Fio. 324.— -Reversal of magnetic flux Fia. 325. — Pole-face loss due to tufts
in armature iron. of flux from teeth.

Pole-face Loss, — The flux enters and leaves the armature in
tufts through the teeth as has already been shown (page 29, Fig.
40, Chap. II). As these tufts of flux pass across the pole face,
•they produce flux pulsations in the pole face. These pulsations
set up eddy currents in the pole face, as shown in Fig. 326. This
results in a power loss. A hysteresis loss also accompanies these
flux pulsations. These combined losses are some function of the
flux and of the speed. They are reduced, being in part due to
eddy currents, by laminating the pole faces. (See Fig. 221.) (^

FRICTION LOSSES

These losses consist of bearing friction, brush friction and
windage, and all are functions of the speed.

SUMMARY

The foregoing losses may be summarized as follows :
Copper losses:

Armature la^Ra
Shunt field 77/ •
Series field L^R,

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stray
power

LOSSES; EFFICIENCY; OPERATION 359

Iron losses (armature and pole face) :

Eddy current — function of flux and speed.

HyBteresis — function of flux and speed.

Friction losses (bearings, brushes, windage) —

function of speed.

The copper losses can be accurately measured or can be calcu- '

f lated. The iron and friction losses can neither be so accurately,'

calculated nor so readily measured as separate losses. Moreover,.

since they are all some function of the flux, or speed, or both)

these losses are combined and are called stray losses; the power

that they represent being called stray power.

As stray power is a function of the speed and the flux only, it

will be constant in a given machine provided the flux and the

speed be kept constant. Therefore, no matter what the load is,

the stray power does not change unless either the flux or the '

speed changes.

In distinction to the copper losses the stray power is all supplied

mechanically. For example, in a motor, a mechanical torque is

required to supply these losses, making the torque available at

the pulley less than that developed by the armature. In a

generator these losses are supplied by the prime mover and not i

by the generator itself. On the other hand, the electrical losses

are supplied by the generator itself.

229. Efficiency. — The efficiency of a machine is the ratio of

output to input. Thus:

Eff. = 2HtP^
input

This may also be written in either of the following ways :

- Eff. = — -2Ht2Ht (119)

output + losses

Eff. = iJ^ILi????? (120)

mput

Therefore, if the losses in a machine be known, the efficiency
may be found for any given input or output.

Example. — A shunt motor takes 40 amp. at 220 volts. The total motor
losses are 1,800 watts. What is the motor efficiency?
Using equation (120)

^„ (220 X 40) - 1,800 -^ ^ ■ .

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360

DIRECT CURRENTS

As electrical units rather than mechanical quantities are
ordinarily used in eflSciency determinations, equation (119) is
used for generators (output is electrical) and equation (120) for
motors (input is electrical).

230. Efficiencies of Motors and Generators. — The efficiency
of electrical apparatus is high as a rule. For instance, a 1-hp.
motor has an efficiency of about 65 per cent. ; a 5-hp. 75 per cent. ;
a 10-hp. 82 per cent., and a 20-hp. 88 or 89 per cent. A 500-kw.

machine may have an efficiency of 94
per cent.

The efficiency of a motor may be
determined from simultaneous meas-
urements of its input and its output
as was shown in Par. 225, where a
prony brake was used.

Theoretically, the efficiency of a
generator may be determined in a
similar manner by measurements of
its input and output. The output
is readily measured with an ammeter
and a voltmeter. The input, how-
FiQ.326. — Cradle dynamometer, ever, is Very difficult to measure.

The difficulty Ues in the measure-
ment of the torque transmitted to the- generator. Torsion
dynamometers have been devised but they are unsatis-
factory as a rule. The generator may be suspended in a
'* cradle,'' as shown in Fig. 326. The ends of the generator
shaft are supported in bearings, so that the frame is free to
turn. The torque is determined by measuring the torque
necessary to prevent the frame's turning. Such a cradle is
expensive, is not readily adaptable to all generators and neces-
sitates the generator shafts' protruding beyond both generator
bearings.

In any direct measurement of efficiency any percentage error
in the measurement of either output or input introduces the
same percentage error into the efficiency.

In the direct measurement of efficiency the power necessary
for the test must be equal to the rating of the machine. In
addition to supplying this power there must be means for ab-

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LOSSES; EFFICIENCY; OPERATION

361

sorbing it. This is not a serious matter with small machines,
but when large machines are tested, supplying and absorbing
the necessary power may be difficult, if not quite impossible.

Because of the foregoing reasons, it is often desirable and even
necessary to obtain the efficiency by determining the losses.

Exam'ple. — A 250-kw. 230-volt d.c. generator is delivering 800 amp.
at 230 volts. The field current is 20 amp. The armature resistance is
0.005 ohm and the series field resistance is 0.002 ohm. The stray power at
this load is 2,500 watts. The generator is connected long shunt. What is
the generator efficiency at this load?

Output = 230 X 800 = 184,000 watts.

Sh. field loss = 230 X 20 = 4,600 watts

Armature loss = 820« X 0.005 = 3,360 watts

Ser. field loss = 820^ X 0.002 = 1,340 watts

Stray power = 2,500 watts

E£F.=

Total loss = 11,800 watts
184,000 184,000

184,000 + 11,800 195,800

= 94 per cent. Ana.

231. Measurement of Stray Power. — It is necessary merely to
duplicate. the flux and the speed in a motor or a generator in
order to duplicate the stray
power loss. As the speed from
equation (111) is S = KE/it>y
it is only necessary to duplicate
the speed's and the electro-
motive force E in order to
obtain the proper value of 0.

To measure stray power, the
machine, whether it be a motor
or a generator, is run Ught
(without load) as a motor, as shown in Fig. 327. The field is
connected across the line in series with a rheostat..

The total power input to the machine is:

This power is distributed as follows: Some goes to supply
the field loss, some supplies the armature la^Ra loss and the re-
mainder is the stray power, S.P. , the output being zero. Therefore

Via + VI f = VI J + la^a + S.P.

Fig.

327. — Determination of
power in a dynamo.

stray

S.P. =(YIa)- I.

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362

DIRECT CURRENTS

The stray power is equal to the total input to the armature
minus the armature resistance loss.

Example, — A shunt generator when running light as a motor takes 12
amp. from 115- volt mains. The field current is 7 amp. and the arma-
ture resistance is 0.03 ohm. What is the stray power loss of the machine at
this particular value of flux and speed?

The armature current /« = 12 — 7 = 5 amp.

The stray power, S.P. = 116 X 5 - (5)* 0.03 =

675 - 0.75 = 574 watts. Arts.

It will be observed that the armature la^Ra is negligible in this
instance. -

Assume that the above generator is delivering 100 amperes at
110 volts at 1,000 r.p.m. The field current is 7 amperes. It
is desired to determine the value of its stray power under these
conditions.

If the full-load electromotive force E and speed S be dupli-
cated when the generator is running light, the stray power will
be the same in both cases. When the machine is running
light as a motor the stray power is readily measure d as follows :

When carrying the above load, the induced emf.

^ = 110 + (107 X 0.03) = 113.2 volts
S = 1,000 r.p.m.

E and S, the generator
is run as a motor, con-
nected as shown in Fig.
328. A rheostat R and
an ammeter are con-
nected directly in the
armature circuit and a
voltmeter is connected
directly across the arma-
ture terminals. The
rheostat R is first ad-
justed so that Vi =

113.2 volts, the small armature drop at this load being negligible.

The field rheostat is then adjusted to give a speed of 1,000 r.p.na.

The machine is now operating at the same value of speed and flux

as it did under load. Therefore, the stray power is the same in

the two cases and is equal to VJa — la^Ra^

To make these adjustments of

I Ami R z 1

-|fc_J-A/VWW

Fig.

328. — Connections for stray power
measurement.

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LOSSES; EFFICIENCY; OPERATION 363

As an example, assume that the current /« is 4.8 amperes and
Vi = 113.2 volts. (This neglects the small drop in the armature,
4.8 X 0.03.) The stray power

S.P. = 113.2 X 4.8 - (4.8)2 0.03 = 543 watts. Ans.

The efficiency of the generator can now be determined.
Output under load = 110 X 100 = 11,000 watts

la^Ra = (100 + 7)2 0.03 = 1344 Watts

VI f =■ 110 X 7 =r77QJwatts

S.P. = j_543, watts

Total loss == 1,657 watts

w 11,000 11,000 _g ^ ,

^^- = 11,000 + 1,657 = ^ipeo = ^^^ P"^ ^"^*- ^^^-

232. Stray-power Curves. — It is sometimes desired to deter-
mine the stray power of a machine over a considerable range,
in order to have sufficient data for obtaining the stray power
under various operating conditions. Stray power is a function
of two variables, flux and speed, and a single curve cannot
express the relationship under all conditions. To plot the rela-
tion, one quantity, either flux or speed, is held constant and the
other is varied. Because it is more convenient, the flux is
(isually held constant and the speed is varied, the connections
being shown in Fig. 328. The flux is held constant by means
of the field rheostat and the speed is varied by means of the rheo-
stat R in the armature circuit.

Since the induced voltage in a machine is JE? = K<t>S, the flux

* = i f (122)

That is, the fliix is equal to a constant multiplied by the ratio
of voltage to speed. The field current can be used without great
error in determining the flux. If this is done, the errors intro-
duced are that the flux under light and under full load may be
different for the same value of field cmrent, owing to armature
reaction; and the flux for a given value of field current may vary
diie to hysteresis. Therefore, if the field current instead of the
flux is used to determine stray power, the stray power may be
too large with the machine running light owing to armature
reaction. This is in part compensated by the fact that the flux

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364

DIRECT CURRENTS

''peaks" under load so that the loss for any value of total flux
is increased, even though the average flux be the same. (See
page 273, Fig. 244.)

In a stray-power run, the field current may be held at a
definite value and the speed varied over the probable working
range of the machine. The field current may then be adjusted
to another value and the run repeated. At least three values

600
660
600

—

— 1

1 —

450
»400
|860

-. ^

t^i

0« 260
2200

j)>

160

100

^y"

"y^

.-"

^-^

r^c

,'^^

^0^'^

100 2D0 300 400 600 600 700 800 900 1000 1100 1200 1800
R. P. M.
FiQ. 329. — Typical stray power curves.

of field current should be used, the maximum 'and the minimum
value under which the machine is likely to operate and an inter-
mediate value. Curves similar to those shown in Fig. 329 are
obtained in this manner.

Example. — The curves of Fig. 329 were obtained from a 10-kw. 230-volt
generator by the method just described, the generator being run as a motor
when these curves were obtained. The rated current of this machine is
43.5 amp. and its armature resistance is 0.14 ohm.

Determine its efficiency as a generator at half load and at rated load, the
voltage being the same in each case, the respective values of field current
being 1.5 and 1.8 amp. The speed is constant at 1,000 r.p.m.
At half load,

/ = 43.5/2 = 21.8 amp.
la = 21.8 + 1.6 = 23.3 amp.
la^Ra = (23.3)2 0.14 = 76 watts
VI f = 230 X 1.5 = 345 watts

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LOSSES; EFFICIENCY; OPERATION 365

Prom Fig. 329, on the 1,000 r.p.m. ordinate, one-third the distance from
curve I to curve II ( =* 1.5 amp.), the stray power is found to be 230 watts.
The efficiency at this load is:

230 X 21.8 5,000 ^^ ^ ^ ^

Eff. = = = 88.6 per cent. Ans.

230 X 21.8 + 76 + 345 + 230 5,650 ^

At rated load,

/ = 43.5

la = 43.5 + 1.8 == 45.3 amp.
la^Ra = (45.3)^0.14 = 287 watts
VI f = 230 X 1.8 = 414 watts.

In Fig. 329, on the 1,000 r.p.m. ordinate, one-third the distance from
curve II to curve III, corresponding to 1.8 amp., the stray power is found
to be 330 watts.

j,„ 230 X 43.5 10,000 „^ - ^ .

^^- = 230 X 43.5 +287+ 414 -h 330 = UfiSO = ^^'^^'' ^"^*- ^^^-

Assume that it is desired to determine the efficiency of this machine when
running as a motor at 900 r.p.m. and taking 45 amp. at 230 volts from the
line. Under these conditions the field current is found to be 1.6 amp.

/a = 45 - 1.6 = 43.4 amp.

la^Ra = (43.4)2 0.14 = 264 watts

VI f = 230 X 1.6 = 368 watts

On the 900 r.p.ni. ordinate. Fig. 329, two-thirds the distance from curve I
to curve II (= 1.6 amp.), the stray power is found to be 225 watts.
^„ 230 X 45- 264 - 368 - 225 9,490 ^, ^ ^ .

^^' 230-X45 = Io;356=^'-^P^^^^^*- ^^-

It is also possible to determine the stray power of a machine
by driving it without load by means of a smaller machine whose
efficiency is known. In using this method it is possible to sepa-
rate the friction and windage losses from the core loss by measur-
ing the power delivered to the machine when the field circuit is
closed and again when it is opened.

233. Opposdtion Test— Kapp Method.— The objection to the
foregoing stray-power method of measuring losses is that the
machine is not under load when the losses are being measured, so
their values may be in error. If two similar machines are avail-
able, their losses may be determined when both machines are
loaded, and yet the line supplies only the losses of the two ma-
chines. The connections for making such a test are Shown in
Fig. 330.

'"^

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366

DIRECT CURRENTS

The two similar machines are coupled together mechanically
and are then connected to the line, as shown. The motor should
have a starting box. Five ammeters are used, one in each field,
one in each armature circuit and one in the line supplying the
two armatures. The fields are connected directly to the line
so that their currents are not indicated by the ammeter Ai.

Fig. 330. — Kapp opposition method for determining losses.

The operation of the set is as follows: The motor supplies
mechanical power to the generator. This in turn supplies elec-
trical power to the motor. The power delivered by the generator
is less than that required by the motor, owing to the losses in the
two machines. Therefore, this deficit must be made up by the
line which supplies the current 7.

The total input to the two armatures is VI .
This power is distributed as follows:

Motor armature loss = Ii^Ri
Generator armature loss = h^Ri
Motor stray power
Generator stray power
where Ri and R2 are the motor and generator armature resist-
ances.

As the generator field is necessarily stronger than that of the
motor, because it requires the higher internal voltage, its stray
power will be greater than that of the motor, as stray power
increases with increase of flux. As a close approximation, the
total stray-power loss may be divided between the two machines
in proportion to their induced voltages.

Let El equal the motor induced volts and E2 the generator
induced volts.

Ei = V - IiRi
E2==V + hR2

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LOSSES; EFFICIENCY; OPERATION 367

Let Pi and P2 be the values of stray power in the two machines.
Then:

k-k <'^''

The total input to the two machines goes to supply their
armature and stray-power losses, because the output of the sys-
tem is zero and the field power is supplied separately. By sub-
tracting the armature losses from the input, the total Tstray power
(Pi + P2) remains.

That is:

Pi + P2 = V7 - h^Ri - WR^

The field losses are mea^red directly by the anuneter in each
field circuit.

The advantages of this method are that each machine is
operating imder load conditions; the regulation of each machine
may be determined; the line need supply only the losses.

The principal disadvantage is that it requires two similar
machines. The assumptions made in regard to the stray power
distribution may be slightly in error.

The machines are brought into operation by first starting the
motor with the starting box. The generator voltage is then
made equal to the motor terminal voltage and the generator
terminals are then connected directly across the motor terminals,
just as generators are connected in parallel. Care should be
taken that the correct polarity is observed. The generator field
is then strengthened and the motor field weakened until the
desired conditions of load and speed are obtained.

Example. — Two similar 120-voit, 7.5-hp. motors are connected in the
manner shown in Fig. 330. The armature resistance of each is 0.12 ohm.
The fields are so adjusted that the motor current /i is 57 amp., and the
generator current li is 45 amp. Under these conditions the line is
supplying a current / of 12 amp. at 120 volts. Find the stray power of
each machine under these conditions of load.

The power supplied by the line

P = 120 X 12 = 1,440 watts

I^^R, = 57 2^ 0.12 = 390 watts

li^Ri = 46^ X 0.12 = 243 watt^

Total = 633 watts.

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368 DIRECT CURRENTS

Total stray power = 1,440 - 633 = 807 watts.

El = 120 - (57 X 0.12) = 113.2 volts
E2 = 120 + (45] X 0.12) = 125.4 volts.

The motor stray power

1100

^■°° 113.2 + 125.4 »<^°^^'^^^-
The generator stray power

^'' ° 113.2+125.4 807 =424 watts.

Knowing the stray power, and the armature and field losses, the efficiency
is readily calculated.

234. Ratings and Heating. — Practically all power apparatus,
whether it be steam engines, gas engines or dynamos, has
definite power ratings. These ratings are determined by the
manufacturer and are supposed to give the power which the appa-
ratus can safely or elBiciently deliver. It is interesting to con-
sider what, in general, determines the rating of various power
devices.

Both a steam engine and a steam turbine are usually rated at
the load for which their efficiency is a maximum. These two
types of prime mover can carry a high overload without
difficulty. Ordinarily, they can carry at least 100 per cent, over-
load easily, but at reduced efficiency.

Owing to their excessive weights and costs, large gas engines
are usually rated as high as possible, which is near the point
at which they cease to operate. Their thermal efficiency is
ordinarily so much greater than that of the steam engine or tiu-bine
that the question of weight is more important than the question
of efficiency.

Electrical apparatus is usually rated at the load which it can
safely carry without overheating. (Commutation may at times
limit the output of direct-current machines.)

If the temperature of electrical apparatus becomes too high,
the cotton insulation upon the armature and the field conductors,
and the insulating varnishes, become carbonized and brittle.
This may result ultimately in grounds and shOTt-circuits within
the machine. The A. I. E. E. Standardization Rules specify
safe temperature limits as follows:

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LOSSES; EFFICIENCY; OPERATION 369

(A) Cotton, silk, paper, all impregnated; enameled wire 105° C.

Above imtreated 95° C.

(B) Mica, asbestos 125° C.

It is very important, therefore, to be able to test a machine
in order to determine whether it is operating within safe tempera-
ture limits. The difficulty in making such tests lies in the fact
that the highest temperatures are within the coils, at points
which are not easily accessible. The highest temperature within
the machine is called the "hot spot" temperature.

The temperature at the surface of the winding may be mea-
sured by placing a thermometer bulb against the surface and
covering it with a small pad of cotton. It has been found that
15° C. added to this reading will give an approximate value of
the hot spot temperature.

It has already been shown that the resistance of copper con-
ductors changes with the temperature. By utilizing this prin-
ciple, an idea of the average temperature within a winding may
be obtained. The increase of resistance per degree rise of tem-
perature may be obtained from the formula 1/(234.5 + 0/ where
t is the surrounding or ambient temperature. For example,
at an ambient or room temperature of 30° C, the increase of
resistance per degree rise is 1/264.5 = 0.00378.

Example, — With an ambient temperature of 30° C. the resistance of the
field of a shimt generator increases lProm 104 to 112 ohms. What is its
temi)erature rise? ^

The fractional change in resistance is r^rj — = 0.077

Temperature rise = 0.077/0.00378 = 20°.4 C. Ans.

Owing to the long time required to reach a constant tempera-
ture, motors and generators should be run from 6 to 18 hours,
in order that an accurate test of their temperature may be made.
As such a long time is usually prohibitive, the heating is often
accelerated by running overload for an hour or so and then drop-
ping back to rated load. By this procedure a very good idea of the
ultimate temperature may often be obtained in a run of 2 or 3 hours.

To get an idea as to how close a machine is to its ultimate tem-
perature, it is often desirable to plot a curve of temperature

1 See Par. 48, page 43.

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370 DIRECT CURRENTS

rise during the test. A typical curve of this type for a shunt
field is shown in Fig. 331. At the beginning of the test, there is
but a sUght differenise of temperature between the field coils
and the room. Therefore, but a small amount of heat is given
out by the coils and as a result the temperature rises rapidly.
As the difference between the coil temperature and the room

temperature increases, more and
conitant Tempcratare^^ morc hcat is givcn out by thc

^ " coils, and the temperature rises

y^ less rapidly. Therefore, the rate

/ of temperature increase becomes

less as the time increases. This

is illustrated by the curve of Fig.

Hour* 331, When the curve becomes

FiQ. 331 -Curve of temperature rise practically horizontal, the total
with time, for a dynamo. ^ •^ i • i •■■ •

heat developed m the coils is
equal to the heat dissipated by the coils and the coils have
reached a constant temperature. Similar curves would hold
for other parts of the machine.

Care must be taken in measuring the armature resistance when
determining temperature rise. The object of this meaaurement
is not to determine the resistance with the idea of calculating
the loss, but to determine the change of resistance in the armature
coppery due to change of temperature. Therefore, it is essential
that the resistance of the copper alone be measured and that
the current path through the copper be the same in every mea-
surement. To exclude all resistance except that of the copper,
the brush and contact resistances must not be included in the
measurement. Therefore, the voltmeter leads must be held on
the commutator segments inside the brushes, as shown in Fig.
332 (a). Moreover, these segments should be marked and in
every subsequent measurement they should be directly under the
same brushes. This insures the same conducting path for each
measurement.

When a multi-polar armature is so measured, the division of
current in the various paths is determined in part by the brush
contact resistance. Thus in Fig. 332 (6), the current from brush
a to brush h is 7i and that from brush a to brush c is Jj.
The total current entering the brush a is their sum, / amp.

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LOSSES; EFFICIENCY: OPERATION

371

The division of the current I between brushes h and c is in
part determined by the contact resistance at these two brushes.
As contact resistance is a variable quantity, the current division
in the armature may change considerably with different measure-
ments. To keep the current in definite paths, two brushes may
be insidated as shown in Fig. 332 (c). In this case the current
paths are not symmetrical, but the division of current is deter-

31 — vsaAaL

FiG. 332. — Measurement of armature resistance for temperature test.

mined not by brush contact resistance but by the copper
resistance itself.

In measuring the shunt field resistance, the voltmeter should
be connected directly across the winding so as to exclude the
drop in the rheostat.

This resistance method gives an average value of the tempera-
ture of the windings. To find the hot spot temperature, 10° C.
should be added according to the A. I. E. E. rules.

In addition to measuring the temperature of the windings,
the rise of temperature of bearings and of commutator should be
measured with a thermometer.

In modern machines of large size, thermo-couples are inserted
in the windings and are connected to millivoltmeters on the
switchboard, so that the operator can determine the **hot spot'*

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372

DIRECT CURRENTS

temperatures at any time. If the thermo-couples are located
between coil sides or between coil sides and core in a i2-layer
winding 5° C. are added; if the thermo-couples are placed be-
tween coil sides and core or between coil sides and wedge in a
single-layer winding 10° C. are added, and 1° C. for every 1000
volts above 6000 volts terminal pressure. The hottest spot is
the highest value by either method after corrections have been
applied.

236. Parallel Rmming of Shunt Generators. — In most xx>wer
plants it is necessary and desirable that the power be supplied by
several small units rather than by a single large unit.

(a) Several small unitfe are more reUable than a single large
unit, for if a unit is disabled the entire power supply is not cut

oflf. (6) The units may be con-
nected in service and taken out
of service to correspond with the
load on the station. This keeps
the units loaded up to their rated
capacity which increases the effi-
ciency of operation, (c) Units
may be repaired more readily if
there are -several in the station.
.(d) Additional units may be in-
stalled to correspond with the
growth of station load. (e)
The station load may exceed
the capacity of any single available unit.

Shunt generators, because of their drooping characteristic, are
particularly well suited for parallel operation. In Fig. 333 are
shown the characteristics of two shunt generators which will be
designated as No. 1 and No. 2 respectively. It wiU be noted
that generator No. 1 has the more drooping characteristic.
If the two generators are connected in parallel. Fig. 334, their
terminal voltages must be the same, neglecting any very small
voltage drop in the connecting leads. Therefore, for a common
terminal voltage, Vi, Fig. 333, generator No. 1 deUvers/i amperes
and generator No. 2 delivers h amperes. That is, the machine
with the more drooping characteristic carries the smaller load.
. Assume that some condition arises which temporarily causes

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Fig.

Amperes

333. — Characteristics of shunt
generators in parallel.

LOSSES; EFFICIENCY; OPERATION

373

generator No. 1 to take more than its share of the load. This
condition might arise from a temporary increase in the speed of
its prime mover, or it might be occasioned by change of load on
the system. Generator No. 1 would immediately tend to oper-
ate at some point a on its characteristic. This results in a
drop in its terminal voltage, which tends to make it take less
load. Therefore, any tendency of one machine to take more
than its share of the load results in a change of voltage which
opposes this tendency. Therefore, shunt generators in parallel

Bw-bMB

Double Pole
Trip C C ^^'^^^^ Breaker

BbeoBtat

Fia. 3^. — Connections for the parallel operation of shunt generators.

may be said to be in stable equilibrium. The reactions of the
system are such as to hold the generators in parallel. Moreover,
if any change of load on the system occurs, each machine must
carry some of the increase or decrease of load.

The connections for operating shunt generators in parallel are
shown in Fig. 334. Each generator should have its own ammeter.
A common voltmeter is sufficient for all the machines. The
individual machines can be connected to the voltmeter or poten-
tial bus through suitable plug connectors or selective switches.
Assume that No. 2 is out of service and that No. 1 is supplying
all the load. It is desired to put No. 2 in service. The prime
mover of No. 2 is started and No. 2 is brought up to speed.

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374

DIRECT CURRENTS

Its field is then adjusted so that its voltage is just equal to
that of the bus-bars, which condition may be determined by
the voltmeter. The breaker and switch are now closed and
No. 2 is connected to the system. Under these conditions,
however, it is not taking any load, as its induced voltage is just
equal to the bus-bar voltage and no current will flow between
points at the same potential. Its induced voltage must be
greater than that of the bus-bars in order that it may deUver
current. Therefore, the field of No. 2 is strengthened xmtil the
generator takes its share of the load. It may be necessary to
weaken the field of No. 1 simultaneously in order to maintain the
bus-bar voltage constant.

To take a machine out of service, its field is weakened and that
of the other machine is strengthened until the load of the first

-^^Bng

FiQ. 335. — Compound generators in parallel.

machine is zero. The breaker and then the switch are opened,
clearing the machine. Connecting in and removing a machine
from service in this manner prevent any shocks or disturbance
to the prime mover or to the system.

If the field of one generator be weakened too much, current
will be delivered to this generator, which will run as a motor
and tend to drive its prime mover.

It is evident that if shunt generators are to divide the load
properly at all points, their characteristics should be similar, that
is, each should have the same voltage drop from no load to full load,

236. Parallel Running of Compound Generators. — Fig. 335
shows two over-compounded generators connected to the bus-
bars, positive and negative terminals being properly connected
as regards polarity. Each generator is taking its proper share
of the load.

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LOSSES; EFFICIENCY; OPERATION

375

Assume that for some reason generator No. 1 takes a slightly
increased load. The current in its series winding must in-
crease, which strengthens its field and raises its electromotive
force thus causing it to take still

„ No.l

more load. On the other hand,
as the system load is assumed to
be fixed, generator No. 2 will at
the same time drop some of its
load, resulting in a weakening
of its series field and a conse-
quent further dropping of its
load. In a very short time
No. 1 will be driving No. 2 as
a motor, and ultimately the
breaker of at least one of
machines will open.

This condition is again illustrated by Fig. 336, which shows the
individual characteristics of the two machines. Assume that
the machines are operating at a voltage Fi, which corresponds

Amperes

the ^°' ^^^' — Characteristics of compound
generators in .parallel.

Fig. 337. — Typical connections for two compound generators operating in

parallel.

to the respective currents /i and Iz- Assume that No. 1 takes
a slightly increased load. Its voltage will then tend to rise to
some point a. This increased voltage means that the machine

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376

DIRECT CURRENTS

takes still more current and the efifect will continue until ulti-
mately the breaker opens.

These compound generators may be considered to be in un-
stable equilibrium. That is, any action tending to throw the
machines out of equilibrium is accentuated by the resulting
reactions.

The machines may be made stable by connecting the two series
fields in parallel, Fig. 337. This connection, which in Fig. 337
ties the two negative brushes together, is a conductor of low
resistance and is called the equalizer. Its operation is as follows:
Assmne that generator No. 1 starts to take more than its proper
share of the load. This increased current will pass not only
through the field of generator No. 1 but also, by means of the

+ BaB

-^Jqaa^i«^^

Series
Field

Fia. 338. — Compound generators requiring two equalizers.

equalizer, some of it will pass through the field of generator No. 2.
Therefore, both machines are afifected in a similar manner and
No. 1 is unable to take the entire load.

To maintain the proportionate division of load from no load
to full load, the following conditions must be satisfied:

(a) The percentage regulation of each armature must be the
same.

(b) The series field resistances must be inversely proportional
to the machine ratings.

It is not always possible to adjust compound generator charac-
teristics by means of series field diverters so that they divide the
load properly. Suppose , Fig. 337, that the series field of generator

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LOSSES; EFFICIENCY; OPERATION 377

No. 1 is shunted by a diverter. If the equalizer and bus-bar have *
negUgible resistance, this diverter shunts the series field of gene-
rator No. 2 as well as that of No. 1. Therefore, the diverter
merely drops the characteristic of the entire system but does
not afifect the division of load. The proper load adjustments
may be made by means of a very low resistance in series with
one of the series fields.

It should be noted that the desired division of load among
either shunt or compound generators at any one load may be
obtained by adjusting their field rheostats. However, it is
usually desirable that this division remain constant at all loads,
especially if an operator is not in continuous attendance. There-
fore, it is desirable that generators operating in parallel have
similar characteristics.

A compound generator with a single series field usually has a
3-pole switch, one blade of which connects the equalizer, as shown
in Fig. 337. If a 3-wire generator (see page 394) having two
series fields is to be connected, a 4-pole switch is necessary as
there are two equaUzers. (See Fig. 338.) The load ammeter
in a compound generator should always be connected between
the armature terminal and the bus-bars. If it is connected in
the series field circuit, the ammeter may not indicate the gen-
erator current, due to the fact that some of the generator current
may be passing through the equalizer.

Compound generators are put in service and taken out of
service in the same manner as shunt generators, that is, the load
is adjusted and shifted by means of the shunt field rheostat.

237. Circtiit Breakers.--<jrenerators, motors and electric cir-
cuits in general require protection from short-circuits and over-
loads. The sudden load imposed by a short-circuit may injure
the generator or its prime mover. Wires may overheat under
the short-circuit current, resulting in fire hazard. Two common
devices are used for opening short-circuits and overloads, the
fuse and the circuit breaker. The fuse has a much lower first
cost and occupies less space. On the other hand, it is worthless
after being blown (unless it is of the refillable type) and con-
siderable inconvenience often results from not having spare fuses
at hand. The circuit breaker has a higher first cost and requires
more space. On the other hand, it operates an indefinitely great

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378

DIRECT CURRENTS

Fia. 339. — ^Two pole, 2000-ampere circuit breaker (Condi t).

Fig. 340. — 6000-ampere, electrically-operated circuit breaker (Gondii).

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LOSSES; EFFICIENCY; OPERATION 379

number of times without injury and is readily re-set. The action
of a breaker is faster than that of a fuse. That is, it opens the
circuit more quickly.

Practically all breakers operate on the same principle. The
mechanism which presses the breaker contacts together is held
by a trigger. This trigger is actuated by a solenoid plunger, the
turns of the solenoid itself being in series with the circuit. When
the current becomes excessive, the plunger is raised and so
trips the trigger, allowing the breaker to spring open. Many
breakers also have shunt solenoids, which allow them to be tripped
from remote points.

The current in the breaker is usually carried by copper lamina-r
tions which bridge the copper blocks, as shown in Fig. 339. On
closing the switch, these laminations press on the blocks making
a wiping contact. The carbon blocks parallel these copper con-
tacts. They ordinarily carry a negligible portion of the current,
but when the breaker opens they break contact later than the
copper, and so interrupt the arc, which would otherwise burn the
copper. The carbon contacts are cheap and easily renewed.
Circuit breakers of large capacities are more or less complicated
mechanisms, as is illustrated by the 6,000^ampere breaker shown
in Fig. 340.

Circuit breakers should always be mounted at the top of the
switchboard. If they are placed at the bottom, the arc which
rises may cause personal injury or may damage the switchboard
equipment.

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CHAPTER XIV

TRANSMISSION AND DISTRIBUTION OF POWER

238. Power Distribution Systems. — Under modern conditions,
most central stations generate power on a large scale as alternat-
ing current and transmit this power as alternating current. The
reason for using alternating current in transmitting the power is
that the voltage may be efficiently raised and lowered by means
of transformers. Much less copper is required to transmit power
at high voltages. The Thmy system does transmit power as
direct current at high voltage (see Par. 204), but is not used in
this country.

FiQ. 341. — Typical power system.

Power is ordinarily utilized at comparatively low voltages
(110, 220, 600 volts), but it cannot be economically transmitted
to any considerable distance at these voltages. In fact, direct
current for commercial use can be economically transmitted and
distributed only in the most congested districts of large cities.
Its advantages under these conditions are the absence of induct-
ive and capacitive effects, which are present with alternating
current, and also the absence of eddy current losses in the cables.
Another advantage of direct current is that a storage battery
reserve can be readily utilized. Fig. 341 shows the general
method of power distribution. Power is generated at the power

380

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TRANSMISSION AND DISTRIBUTION OF POWER 381

station, is transmitted as alternating current at high voltage to
the sub-station (66,000 volts is shown; the transmission voltage
is seldom less than 6,600 volts). At the sub-station it is either
transformed to 2,300 volts alternating current by transformers
or to 600 volts or 230 volts direct current by motor-generator
sets or synchronous converters. (Fig. 341 shows the sub-station
supplying a trolley with 600 volts direct current; a 2,300-volt
alternating-current circuit supplies power for lighting, the voltage
being transformed near the consumer's premises to a 230-115-
volt 3-wire system; a 3-phase 2,300-volt alternating-current
power Une supplies a factory, the voltage being transformed to
550 volts, 3-phase, by transformers. These systems are discussed
more fully in Chap. XII, Vol. II.) The sub-station receives the
power in large amounts and distributes it to the various con-
sumers in smaller quantities. It bears the same relation to the
power system as the middleman or retailer does to an industrial
system.

239. Voltage and Weight of Conductor.— rA6 weight of con-
ductor varies inversely as the square of the voUage, when the power
transmitted, the distance and the loss are fixed.

Let it be required to transmit the power P at the voltage Vi
and current /i over wires having a resistance Ri.

The current

■-k

The power loss

Pi = Ii^Ri

Assume that the voltage is raised to V^, the power, the loss
and the distance remaining fixed.
The current

The power loss

P, = h*Ri = Pi

Therefore,

hmi = h*Rt

(hy (P/v,y 7i«
" \ij ~ iP/v,r F2«

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382 DIRECT CURRENTS

That is, the conductor resistance varies directly as the square
of the voltage. But the volume or the weight of a conductor
of given length varies inversely as the resistance.

Let the weight of copper in the two cases be W\ and 1^2, re-
spectively.

Therefore, the condudor weight varies inversely as the square of
the voUa^e, when the power, the loss and the distance are fixed.

If the voltage of a system is doubled, the weight of the copper
is quartered, other conditions being the same.

Exam-pie. — 50 kw. are delivered at a distance of 500 ft. at 110 volts
over a 400,000 CM. feeder, (o) What is the power loss? (&) Repeat for
220 volts.

(a) The current

J 50,000 ...

/i = —f^ = 454 amp.

If the Cable had 454,000 CM. (see Par. 69) the loss would be 454,000

(454\ ^
— 1 X 1,000

X 10-8 X 400,000 = 5,150 watts. Ans,

The loss is

(5)/2-^^ = 227amp.

(lis) ' ^'^^ ^ ^'^^^ ^***®' ^'^'

The loss in (6) is one-fourth that in (a). Therefore, a 100,000 CM. feeder,
having just one-fourth the weight of the feeder in (a), would transmit the
same power, the same distance, with the same loss.

240. Size of Conductors. — In transmitting or distributing
power by direct current, four factors must be considered in de-
termining the size of conductor.

(a) The wires must be able to carry the required current
without overheating.

This is particularly important with inside wiring where fire
risk exists. Tables of the permissible ciurrent-carrying capacity
of wires are given in the Appendix, page 410.

(b) The voltage drop to the load must be kept within reason-
able limits. This is particularly important when incandescent
lamps constitute the load.

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TRANSMISSION AND DISTRIBUTION OF POWER 383

(c) The wires must be of sufficient mechanical strength. This
is important when the wires are strung on poles. It is not advis-
able to use wires smaller than No. 8 A.W.G. for pole Unes.

(d) The economics of the problem must be considered. In-
creasing the size of conductor means higher investment costs
but less energy loss in transmission. That size of conductor
should be chosen which makes the cost of the energy loss plus the
interest on the investment a minimum. This* may be modified
in view of the considerations stated in (a), (6) and (c).

CONSTANT POTENTIAL DISTRIBUTION

241. Distribution Voltage. — About 110 volts has been found
to be the most convenient voltage for incandescent Ughting.
It is not so high as to be dangerous to persons. Incandescent
lamp filaments for voltages in excess of 110 volts become so
long and of so small a cross-section that they are fragile. An
even lower voltage than this would be desirable from the stand-
point of the filament, but a lower voltage would be accom-
panied by an increase in the required weight of copper. There-
fore, 110-115 volts has been standardized for Ughting and for
domestic use as being the most desirable when all factors are
taken into consideration. Six hundred volts is commonly used
for trolley distribution, because it is not so high as to give operat-
ing difficulties and it saves considerable copper as compared with
systems of lower voltage. At the present time, 1,200, 2,400, and
even 3,000 volts are used at the trolley in railway electrification,
these higher voltages being for trunk line electrification, not for
municipal traction.

242. Distributed Loads. — The load on a feeder or main may
be concentrated at one or two points, as is generally the case with
feeders, or may be distributed uniformly or non-uniform ly
along the conductors, as when lamp loads are located at various
points along mains. (See Fig. 342.)

The conductors may be of uniform cross-section throughout
their entire length, Fig. 342(a). This occurs where the mains
are short and the voltage drop is small.

Where the mains are of considerable length, the minimum
amount of copper for a given voltage drop is obtained when the
mains are uniformly tapered, Fig. 342(6).

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384 DIRECT CURRENTS

As it is impracticable to have a uniformly tapering conductor,
a conductor of constant cross-section is run for a part of the
distance, followed by another uniform conductor of lesser cross-
section, and so on, as shown in Fig. 342(c). A good rule to re-
member is that the current density in each section should be the
same. For example, the first section may consist of a 250,000
CM. conductor, carrying 200 amperes; assume the second sec-

150
tion carries 150 amperes; it should be a ^7)^*250, 000 = 190,000

CM. conductor. Ordinarily 4/0 wire would be used for this
second section.

(a) Uniform Copper Section

QOOOOQO^

(6) Tapered Conductor

243. Systems of Feeding. — ^In order to keep a number of
lamps at the same voltage without excessive copper, the return
loop or anti-parallel system shown in Fig. 343(a) is often used.
The two feeding wires are connected to opposite ends of the
load. This system allows all the lamps to operate at nearly
the same voltage and yet the voltage drop in the feeding wires
may be large.

The objection to the return loop system is the extra length of
wire required. This objection is often overcome by arranging
the loads in the manner shown in Fig. 343 (fe), called the open
spiral system. Where large groups of lamps are switched off
and on at the same time, as in theaters and auditoriums, it is
often possible to arrange the lamps in this way.

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TRANSMISSION AND DISTRIBUTION OF POWER 385

The open spiral may be closed at its ends, resulting in the
closed loop system of Fig. 343(c).

VMdiag
Point

0006660 g

{d) Return Loop or Anti-Parallel System

( 6 ) Open Spiral System (c) Closed Loop

Fig. 343. — Systems of feeding.

244. Series-Parallel System. — Doubling the voltage of a
system results in the weight of required copper being reduced to
one-fourth its initial value. If 110-volt lamps be arranged so
that two are always in series, as shown in Fig. 344, the system
may be operated at 220 volts. The copper section will then be
one-fourth that required for straight 110-volt distribution. The
obvious disadvantages of the series-parallel system are that

Fig. 344. — Series-parallel system.

lamps can only be switched in groups of two and if one lamp
burns out, the lamp to which it is connected ceases to operate.
. Also, both of the lamps in series must be of the same rating.

THE EDISON 3-WIRE SYSTEM

246. Advantages. — The objections to the series-parallel sjrstem
may be eliminated by running a third wire, called a neiUral,
between the two outer wires. This neutral maintains all the
lamps at approximately 110 volts. The advantage of a higher

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386

DIRECT CURRENTS

voltage in reducing the weight of copper is obtained by the use
of this system. If there were no neutral wire, the 220-volt
system would require one-fourth the copper of an equivalent
1 10-volt system. If it be assumed that the neutral of the Edison
system is of the same cross-section as the two outer wires, the
total copper for the Edison system is ^ or 373^ per cent, that
for a 110-volt system of the same kilowatt capacity. Therefore
the saving in copper is 62)^ per cent. In practice, the neutral

can be made smaller than the

iio| y.

N y

.10 ^mp.

1=0

) Amp.

FlQ.

— 10 Amp.

345. — Edison 3-wire
balanced loads.

system —

two outer wires so that the
saving in copper is even
greater than 623^ per cent.

The general plan of the
system is shown in Fig. 345.
Two wires A and B have 220
volts maintained between
them, A being the positive
and B the negative. A third wire N is maintained at a
difference of potential of 110 volts from each of the other two
wires. Therefore N must be negative with respect to A and
positive with respect to B. That is, current tends to flow from
A to Nj and from iV to B.

Fig. 345 shows the conditions which exist when the load on
each side of the system is the same. Each of the loads a and h

10 Amp ^

5 Ampr— *.

> Amp.

110 Amp.

110

Is Amp.

5 Amp-— >

6 sl 10 Amp.

S Amp.

< 10 Amp.

(o) (b)

Fig. 346. — Unbalanced 3-wire systems.

takes 10 amperes. The 10 amperes taken by load a
through to load h and then back through wire B to the source.
This is equivalent to a series-parallel system as both loads are
equal and are in series. Under these conditions the current in
the neutral wire is zero and the loads are said to be balanced.
Fig. 346 (a) shows the conditions existing when the load a on
the positive side of the system is 10 amperes, and the load h on

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TRANSMISSION AND DISTRIBUTION OF POWER 387

the negative side is 5 amperes. Under these conditions the extra
5 amp. taken by load a must flow back through the neutral
to the generator or source. Therefore there are 5 amperes in the
neutral returning to the generator. In Fig. 346 (fe) the load b is
now 10 amperes and load a is 5 amperes. Under these conditions
the extra 5 amp. must flow out to the load through the neutral.
It will be observed that the current in the neutral may flow in
either direction, depending upon which load is the greater.
Therefore, if an ammeter is used in a neutral it should be of the
zero-center type. Moreover, it will be observed that the neutral
carries the difference of the currents taken by the two loads. In
practice the loads are usually so disposed that they are nearly
balanced. Twenty-five per cent, unbalancing (that is, a neutral
current which is 25 per cent, that in the outer wires) is usually
allowed. Therefore, the current in the neutral is ordinarily
much smaller than that in the outers and a much smaller con-
ductor can be safely used. + j2a.
In practice the neutral
usually grounded.

Effect of Opening the Neu-
tral.— In practice, it is very
desirable to keep the neutral ~ *~^*

^ . 1 o • X 1 J Fio. 347. — Effect upon the balancing of

Of the S-WU'e system closed a 3-wire syetem of opening the neutral.

under all conditions. The

reason for this is illustrated by the following example.

Fig. 347 shows two lamp loads on a 3-wire system. The
load on the positive side consists of 6 lamps each taking 2 amp-
eres, making a total of 12 amperes. The load on the negative
side consists of 4 lamps each taking 2 amperes, making this load
8 amperes. The voltage across each load is 110 volts so the
resistance Ri of the positive load is

Ri = -T^- = 9-17 ohms.

The resistance of the negative load is
IJIO
8

If the neutral now be opened at the point S, the two loads R\
and R2 are in series and therefore each must take the same cur-
rent. The total resistance ft is JBi -}- ^2 = 22.92 ohms.

fta = -^ = 13.75 ohms.

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388 DIRECT CURRENTS

There is now 220 volts across these two loads in series, so that
the current

J 220 ^ ^

^ = 2"2:92 = ^'^ ^"'P-

The voltage Vi across load Ri

Vi = 9.60 X 9.17 = 88.0 volts.
The voltage V2 across load 222

Vi = 9.60 X 13.75 = 132 volts.

This assumes that the resistance of the lamp filaments does
not change. It will be observed, however, that the larger bank
of lamps is operating at a much reduced voltage, resulting in a
material decrease of candlepower, and that the smaller bank is
operating considerably above rated voltage, which would result
in the lamps burning out in a short time.

For the above reason the neutral of the 3-wire system is
usually grounded and one rarely sees circuit breakers in the
neutral wire of power plants.

246. Voltage Unbalancing. — The voltage on the two sides of a
3-wire system may become considerably unbalanced if the
loads on the two sides of the system become unequal, as shown
in Fig. 349.

IKOY.

0.1 Q eo amp. — ». I6,V.

^ 00 amp. lio V.

0.2 n I-O

io«r.

11 ov.

>90 amp. lip V» 104 v.

0.1 0 ^ r i

fOT

< CO amp. r

(a) (b) f

Fig. 348. — Voltage drop in a 3-wire system having balanced loads.

In Fig. 348 (a) a load of 60 amperes exists on each side of the
system. Each outer wire has a resistance of 0.1 ohm and the
neutral has a resistance of 0.2 ohm. The generator voltage is
220 volts across the two outer wires.

As the two loads are equal, there is no current in the neutral
wire. Therefore, the voltage drop per wire for the outers is

e = 60 X 0.1 =6.0 volts.

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TRANSMISSION AND DISTRIBUTION OF POWER 389

The voltage across each load is 104 volts. There is no voltage
drop along the neutral, as it carries no current. Fig. 348(6)
shows a plot of the voltage distribution.

110 V.

_. 100 a

0.2 n.t—80fl

0.1 n .e_20a

(«) (6)

Fio. 349. — Voltage unbalancing in a 3-wire system having unbalanced loads.

Assume that the loads are as shown in Fig. 349, 100 amperes
on one side of the system and 20 amperes on the other side. This
represents the same total amperes as in Fig. 348.
The drop in the positive wire

ei = 100 X 0.1 = 10 volts.
The drop in the neutral

6a = 80 X 0.2 = 16 volts.
Voltage across positive load

Vi = 110 - 26 = 84 volts. Ans.
The drop in the negative wire

62 = 20 X 0.1 = 2 volts.
Voltage across negative load

7, = 110 -r 2 + 16 = 124 volts. Ans.

There is now 40 volts difference between the voltages on the
two sides of the system.

Under these conditions, the voltage across the load on the
negative side is greater than the voltage on the negative side of
the system at the power station. This rise in voltage from power
station to load is obtained at the expense of the drop in the neu-
tral. Fig. 349(6) shows these conditions graphically.

When motor loads are to be connected to a 3-wire system
they are usually connected between the two outer wires rather
than between an outer wire and neutral so that they will not
produce any voltage unbalancing In fact some power com-
panies will not permit motor loads exceeding one horsepower to be
connected to neutral.

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390 DIRECT CURRENTS

METHODS OF OBTAINIKG A 8-WIRE SYSTEM

There are various methods of obtaining a 3- wire system which
are as follows :

247. Two-generator Method. — Two shunt generators may be
connected in series as shown in Fig. 350. The positive terminal
of one should be connected to the negative terminal of the other

110

'' 0 0 0 0

110

0 0 0 0

Fig. 350. — Two generators supplying a 3-wire system.

that is, the generators are in series between the outers. Both
generators may be driven by the same prime mover. When
connected in this manner, each machine supplies only the load
on its own side of the line. The obvious objection to this method
is that two separate machines are required.

-*— 86a
FiQ. 351. — Storage battery giving neutral in a 3-wire system.

248. Storage Battery. — A storage battery may be floated across
the line as shown in Fig. 351. The neutral wire is connected to
the middle point of the battery. When the load is unbalanced,
that half of the battery on the more heavily loaded side will
discharge and the other half will be charged. Fig. 351 shows an

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TRANSMISSION AND DISTRIBUTION OF POWER 391

unbalancing of 10 amperes. In this particular case the upper
half of the battery supplies 5 amperes, and the other 5 amperes in
the neutral go to charge the lower half of the battery. The objec-
tions to this method of obtaining a neutral are the high main-
tenance cost of a storage battery and the difficulty of maintaining
both halves of the battery at the same condition of charge.

249. Balancer Set. — A balancer set is a very common method
of obtaining the neutral. This set consists of a motor and a
generator mechanically coupled together. They are connected
in series across the outer wires and the neutral is brought to their
common terminal, as shown in Fig. 352.

52.2 a

QO.QOp

"'^- QOpO"*

-• S22a -* — 40a

Fig. 352. — Balancer set giving neutral in a 3-wire system.

The action of this set may best be illustrated by the hydraulic
analogy shown in Fig. 353. Water is supplied by the canal A.
This water falls over a weir into canal B and may be made to do
useful work in so doing. All this water is not needed between
the canal B and the tail race C at the point of utilization D.
Some of the water which is not needed at D passes to C through
the water wheel shown in the figure. This water wheel is belted
to a centrifugal pump operating between B and A. In virtue
of the water passing through the water wheel some of the water
in the canal B is pumped back to A by the pump, where it may
be utilized again. The water wheel corresponds to the motor
or machine B, Fig. 352, and the centrifugal pump to the generator
or machine A.

If in Fig. 353 more water is required between canals B and C
than can be supplied by the weir at A, the centrifugal pump may
act as a water wheel and the water wheel as a pump. Some of

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392

DIRECT CURRENTS

the extra water required in B will be supplied through the upper
machine operating as a water wheel and discharging into B. In
so doing the upper machine drives the lower machine as a pump.
The lower machine then pumps water from C back to B. This
condition corresponds to an excess of load on the negative side
of the system of Fig. 352.

If in Fig. 352 there is an excess of load on the positive side
of the system, as represented by 20 amperes in the neutral, 12.2
amperes of this 20 amperes flows through the motor and in dropping

Waterfall!

Fia. 353. — Water-wheel analogy of balancer set.

through 110 volts gives up its energy. The motor then causes
the generator to pump 7.8 amperes back to the positive side of the
line. This current distribution is determined in the following
manner:

Each of the machines A and B is assumed to have 80 percent,
efficiency. Let Ji be the generator current in machine A and h
be the motor current in machine B. The generator output will
be 0.8 X 0.8 = 0.64 times the motor input. Assmning that the
voltages are equal, actually they will be slightly unbalanced,

IIOJ2 X 0.64 = 110/1
/i + /2 = 20
Solving Zi = 7.8 amp.

I2 = 12.2 amp.

The machines will respond more readily to unbalanced loads
if their fields are crossed, that is, if the motor field is across the

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TRANSMISSION AND DISTRIBUTION OF POWER 393

generator side of the line and the generator field is across the
motor side of the line. In order that a generator may supply-
additional current, its terminal voltage must drop or its induced
voltage must rise. In order that a motor may take additional
load, either its terminal voltage must rise or its induced voltage
must drop. The (excess load on the positive side of the system
(Fig. 352) tends to reduce the field of machine A and to increase
that of machine B. These effects are the reverse of what is
desired. If the generator field is across the motor side of the Une,
the increased voltage is now across the generator field and will
raise the generator induced voltage. Therefore, its terminal
voltage need not drop so much to take care of unbalanced cur-

FiG. 354. — Connections of a 3-wire system using a balancer set.

rents. The same result may be obtained by compounding the
two machines. The series fields should be so connected that the
machine acting as a generator is cumulatively compounded, and
that acting as a motor is differentially compounded.

Fig. 354 shows standard connections for a balancer set with
series fields. The machines are started in series, with the neutral
switches open and the shunt fields in series across the line. When
the machines are up to speed the neutral switch S is closed. If
the voltages on the two sides of the system become widely differ-
ent, the currents in the two halves of the differential relay become
unbalanced. This relay then closes the tripping coil circuit
of the main generator breaker, resulting in the main generator
circuit opening, even though its load is not excessive.

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394

DIRECT CURRENTS

260. Three-wire Generator. — The three-wire generator or
Dobrowolsky method is a very efficient method of obtaining
a neutral. The details of the method can be understood better
after alternating currents and the synchronous converter have
been studied J The principle of the method is as follows: Alter-
nating current is generated within a direct-current armature as
has already been shown. If slip rings be employed, alternating
current can be obtained from the machine. A coil wound on an
iron core which therefore has high inductance and offers a
high impedance to this alternating current is connected across
the slip rings. The center of such a coil is at the cen-

(a) (6)

Fia. 355. — ^3- wire generator connections (Dobrowolsky method).

ter of gravity of the voltages generated within the armature.
Further, this inductance coil offers very little resistance to
the flow of direct current. Therefore, if the three-wire neutral
be connected to the center of this coil, the voltage to either brush
from the neutral will be the same. Moreover, any current flow-
ing back through the neutral can readily flow back into the
armature through this reactance. The connections of such a
generator are shown in Fig. 355 (a). Sometimes, to obtain better
balancing, two and even three reactances are employed. All
have their neutrals tied together, as shown in Fig. 355 (6). Oc-
casionally the reactances are placed within the armature. This
arrangement requires but one slip ring, but increases the weight
of the armature.

The Edison 3-wire system may be extended to 4, 5, 6, and 7-
wire systems. (See Fig. 309, page 341.) The complications and
number of wires prevent these multi-wire systems being exten-
sively used.

1 See Vol. II, Chap. XI, page 375.

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TRANSMISSION AND DISTRIBUTION OF POWER 395

261. Feeders and Mains. — In congested districts the mains
form an undergromid network. This network is supplied at
various points, called centers, by feeders connected to the direct-
current bus-bars at the power station. It requires a careful
study of the various loads, amount of copper, etc., in order to
determine the most advantageous feeding points or centers.
Two or more sub-stations may simultaneously feed the same
centers. In order that the voltages at these centers may be
determined and so maintained at the desired values, pilot or
pressure wires nm back to the station voltmeter. By means of
a dial switch the operator is able to read the voltages at the va-
rious centers. Fig. 356 shows the cross-section of a concentric

Oater CouiJuctor-
Paper laiglttJcm-
fiabb«r^Ini u U(«d
Lead SticaLh

Fia. 356. — Cross-section of a 220-volt, 1,000,000 CM. concentric cable.

1,000,000 CM. cable. The outer and inner conductors are the
outer wires of the Edison 3-wire system. The neutral is usually
a separate wire of much smaller cross-section, or there may be
one large neutral common to several feeders and mains. The
three pilot or pressure wires are connected, one to each outer
wire and one to the neutral at the feeding point. If the operator
finds that the voltage is too low at the feeding point, he throws a
feeder to a bus-bar of higher voltage. A large voltage drop can
exist in such feeders, as no loads are taken off at intermediate
points.

In practice the following are the percentage drops usually
allowed: In feeders, 5 to 10 per cent.; in the distribution mains,
3 per cent.

The services are usually taken directly from the distribution
mains and occasionally, extra large loads may be taken directly
from junction boxes. Junction boxes are circular iron castings

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396 DIRECT CURRENTS

containing a set of insulated bus-bars, to which either the dis-
tribution mains or the feeders are connected. Distribution
mains are connected, through fuses, to suitable terminals al-
ready installed in the junction boxes. A junction box thus
provides a convenient method of connecting the single feeding
wires to the several distribution wires. The mains are always
fused, but only disconnecting links are used for the feeders, it
being deemed advisable to allow the feeders to burn themselves
clear of any short-circuits.

262, Electric Railway Distribution. — Electric railway genera-
tors are generally compounded, the series field being on the nega-
tive side. The negative terminal is usually connected directly
to ground or to the rail through a switch. The positive terminal
feeds the trolley through an ammeter, a switch, and a circuit
breaker.

On short lines, with light traffic, the trolley alone may suffice
to carry the current to the car, as shown in Fig. 357 (a). Except
in small installations, the trolley is of insufficient cross-section to
supply the required power and at the same time to keep the vol-
tage drop within the necessary limits. As the size of the trolley
wire is limited by the trolley wheel, it cannot be conveniently
increased. The same effect as increasing the size of the trolley
may be obtained by running a feeder in parallel with the trolley
and connecting the feeder to the trolley at short intervals, as
shown in Fig. 357 (6). This is called the ladder system of feeding.
The trolley and feeder together may be considered as forming a
single conductor.

Where the density of traffic requires several feeders, the best
results are obtained by connecting the feeders in the manner
shown in Fig. 357 (c). Each feeder is protected by a circuit
breaker.

The objections to the preceding methods of feeding are that
trouble, due to a ground, for example, at any point on the trolley,
involves the entire system. In cities where traffic is particu-
larly dense, it is not permissible to take chances of having the
entire system shut down due to a ground at one point only.
Therefore, the trolley is sectionalized as shown in Fig. 357 (d) . In
this method the trolley is divided into insulated sections, each of
which is supplied by a separate feeder. Trouble in one section

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TRANSMISSION AND DISTRIBUTION OF POWER 397

is not readily communicated to the other sections. This in-
creased reliabiUty is obtained at the expense of a less efficient
tise of the copper, as the feeders are unable to assist one another.
In the preceding systems this mutual help is obtained.

Trolley

(a) Simple Trolley
Feeder.

I I I I I I I

1,, 1,1 I I I I I I I I

Trolley

(b) Ladder System - Single Feeder

Feeders

-t\

::i

Trolley

(d) Multiple Feeders - Section^lized Trolley
FiQ, 357. — Methods of feeding a trolley system.

263. Electrolysis. — Most trolley systems use the track as the
return conductor for the current taken by the car. The return
currents not only pass through the tracks themselves, but seek
the paths of least resistance by which they may return to the nega-
tive terminal of the station generator. Such currents in spread-

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398 DIRECT CURRENTS

ing through the earth follow such low resistance conductors as
water pipes, gas pipes, cable sheaths, etc., as shown in Fig. 358.
The fact that the current enters and flows along these conductors
in itself does no harm. However, it is obvious that such ciurrents
must ultimately leave these pipes as at (a), Fig. 358. In so doing
they tend to carry the metal of the pipe into electrolytic solution,
which ultimately results in the pipe being eaten away. To de-
crease the effects of electrolysis several expedients have been
devised. The two most successful methods are the following:
(a) Provide as good a return path through the track as is pra<;ti-
cable. This is done by good bonding and by using insulated

Tfollay ^^

Fig. 358, — Electrolysis by earth currents.

negative feeders, that is, heavy copper feeders that are run back
to the negative bus from various points along the track. Fig.
358 shows how poor rail bonds may cause the current to leave the
track and enter the pipe. In some cities the total permissible
drop in the ground return circuit must not exceed from 10 to
15 volts. (6) Discourage the current's entering the pipes by
inserting occasional insulating joints in the pipes.

In testing for electrolysis the usual method is to measure the
voltage existing between the track and the water pipes (as at a
hydrant). The magnitude of this voltage indicates roughly the
magnitude of the current which must be flowing from one to the
other. The polarity shows which way the current is flowing.
For example, if the track is positive to the pipe, current must be
flowing from the track to the pipe.

The electrolysis situation is still in an unsettled state both, as
regards its mitigation and as to the ultimate responsibility for
the damage resulting.

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TRANSMISSION AND DISTRIBUTION OF POWER 399

STORAGE BATTERY SYSTEMS

264. Central Station Batteries. — Fig. 359 shows a typical
load curve of a central station. Between 11.00 p.m. and 5.00
A.M. the load is comparatively small, consisting of street lights
and a few all-night conmiercial loads.

This portion of the load curve is called a " valley. " The load
increases rapidly from 5.00 to 7.00 a.m. due to commercial power
loads, lights and perhaps to the beginning of street car service.
The morning peak occurs about 8.00 a.m. The load drops ofiF
gradually until noon.

ai tcry

•

3ii

(Charge

bhar{

e J

ISNt. 6A.M ISNn. 6 P.M. IS Nt

Fig. 359. — Battery smoothing out central station load curve.

The valley between 12.00 and 1.00 is due to the shutting down
of the commercial loads because of the luncheon hour. The
evening peak, which is usually the largest, occurs between 5.00
and 6.00 p.m. This peak may hold up for an hour, after which
it drops to the evening load, which consists mostly of lighting.
This load gradually diminishes to the all-night value.

Obviously the power company must have sufficient station
and distributing capacity to carry the peak. Even although this
apparatus is in use only one hour a day, the investment charges
are in eflFect 24 hours a day.

The ratio of the average load to the maximum load of a station
is called the load factor.

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400 DIRECT CURRENTS

Example. — A station delivers 192,000 kw.-hr. in a day and its peak load
is 20,000 kw. What is the daily load factor?

The average load = tlf^ ^ g^ooo kw.

8 000
The load factor « oQliOO ™ ^ ^' ^®°*' ^^'

Obviously a high load factor is very desirable. In fact
power companies welcome any loads that will fill in the
valleys of the curve and are usually prepared to offer attrac-
tive rates for such loads in order to improve their load factors
and thus to utilize apparatus at times when it would other-
wise be idle.

The load curve of a station may be smoothed out by the use
of a storage battery. The battery may be charged at night and
early morning and so fill in the valley of the load curve and then
be discharged on the peak of the load curve, as shown in Fig.
359. This equalizes the load on the station and increases its
load factor.

As a rule, batteries are not installed for the purpose of smooth-
ing out the load curve. A storage battery operating under the
best conditions is good for only a limited number of complete
charges and discharges. Therefore, the battery maintenance
is usually found to more than offset the economies effected by
taking some of the load off the peak. Such batteries may be
very useful in office buildings and other isolated plants, because
it is often possible to shut down the entire Ughting plant and run
on the batteries at night, thus eliminating considerable labor
charge.

Batteries are commonly installed as reserve in large central
station systems. They are placed, therefore, near the center of
the load. In case of a shut-down in the generating system or in
the transmission system, the battery can help maintain service.
For this reason pasted plate batteries are more often used,
because of their high overload capacity. (See page 103, Par. 94.)

Storage batteries are also useful in taking care of unexpected
loads. For example, a thunder storm may result in a sudden
demand which could not be foreseen and so cannot be met im-
mediately by the generating station, as it takes time to get up
steam and put a generator on the lipe. A battery may be put

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TRANSMISSION AND DISTRIBUTION OF POWER 401

on the line immediately and so carry the sudden load increase
until boilers and turbines can be brought into service. If the
battery is already floating across the line it takes the load in-
crease automatically.

266. Resistance Control. — In order to control the load taken
by a generator connected to the bus-bars, it is necessary to change
its induced voltage by adjusting the field current. It is not
possible to adjust the voltage of a storage battery in this manner.
One method of controlling the battery ^

load is to have the battery voltage sey- -|«

eral volts higher than the bus-bar voltage -;L_

and to insert resistance in series with -^-

the battery, as shown in Fig. 360. By -=-

adjusting this resistance, the load de- ~-

liyered by the battery may be con- 1=1

trolled. The disadvantage of this . -r-

method is the loss of power in the re- '

J ., ,, , • .1 FiQ. 360 — Resistance con-

sistance and the voltage drop m the troi of battery discharge,
resistance, which depends upon the load.

Even with constant load the resistance must be adjusted
occasionally to compensate for the drop of battery voltage
during discharge.

Example. — It is desired to discharge a storage battery, consisting of 115
cells each having an electromotive force of 2.1 volts and an internal resist-
ance of 0.001 ohm, into 220 volt bus-bars so that the battery delivers 100
amperes. To what value must the series resistance be adjusted?
The total battery electromotive force

E = 115 X2.1 = 242 volts.
The bus-bar voltage

V = 220 volts.
The battery resistance

r = 115 X 0.001 = 0.115 ohm.
Let R = the added external resistance

242-220
"" 0.115 +i2
lOOi^ = 22 - 11.5 = 10.5
R = 0.105 ohm. Arts.

266. Counter Electromotive Force Cells. — If an electric cur-
rent be sent through two plain lead plates immersed in dilute
sulphuric acid, a simple storage battery is formed which immedi-

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402 DIRECT CURRENTS

ately develops a counter electromotive force of about 2.0 volts.
(See page 97, Par. 92.) Neglecting the small IR drop in such a
cell, the counter electromotive force is practically independent of
the current. This principle is utilized in controlling the current
delivered by a battery.

Plain lead plates are immersed in dilute acid and are connected
in series with the battery. If it is desired to decrease the dis-
charge rate of the battery more of these cells are cut in. To
do this an end cell switch, similar to that shown in Fig. 361, is
used. The advantage of this method over the resistance control
is that the opposing or control electromotive force is independent
of the load.

267. End Cell Control. — A battery usually consists of a suffi-
cient number of cells to give an electromotive force exceeding that

p. of the bus-bars by an ample

n.a ""^-fl^ ^.Bat margin. To charge such a bat-

ciu T — PT^^ tery a booster may be used.

-^=^\a»tu»y oont«ct (See page 111, Par. 102.) The

"=- electromotive force of the bat-

-=- tery, and hence its load, may

-S- be controlled by cutting in or

"=" out the cells at the end of the

T -»" battery.

Fio. 361. — End csell control of storage It is essential tO do this with-

battery. ^^^ Opening the circuit. For

this purpose a switch similar to that shown in Fig. 361 is used.
The main contact is connected to the auxiliary contact by a
resistance R. When sliding from one battery contact to the next
the auxiliary contact maintains the circuit connections through
the resistance R. Were there zero resistance between the main
contact and its auxiliary contact the individual cells would be
dead short-circuited during the transition period. The* resis-
tance R is usually so chosen as to allow the normal battery current
to flow during the transition period. The end cell switches
become rather massive in large battery installations and are often
operated by a motor-driven worm. This also permits remote
control.

The end cells, not being in continuous service, are discharged
to a lesser degree than the others. Therefore they require in-
dividual attention on charging. / ^^^i^

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TRANSMISSION AND DISTRIBUTION OF POWER 403

268. Floating Battery. — A battery is occasionally used to
equalize sudden fluctuations of load such as commonly occur in
railway systems. The battery voltage should be such that with
an average load on the station it is just equal to the bus-bar
voltage. The battery is then delivering no current and is merely
"floating."

When a sudden load comes on the station, the bus-bar voltage
drops. The battery then discharges and assists the generators.
On the other ha^id, if the load drops to a low value, the bus-bar
voltage rises and the battery charges.

As a rule the bus-bar voltage does not change enough to cause
the battery to respond sufficiently to the load changes. In
fact with over-compounded railway generators the reverse

nM *'^

Motor* Driven
Booster

Fig. 362. — Regulating storage battery discharge with a booster set.

action might well occur. There are several methods of causing
the battery to charge and discharge at the proper time. One
typical method is shown in Fig. 362. The battery is connected
to the bus in series with a motor-driven booster. Two carbon
rheostats Ri and ^2 are connected in series across the battery. The
booster field is connected from their common point to the middle
of the battery. If Ri = R2, the booster field is connected across
two points of equal potential, the field current is zero and no
voltage is induced in the armature. An increase of load, how-
ever, causes solenoid P to pull down on the lever. This com-
presses Ri and releases the pressure on R2- The resistances Ri
and ^2 now differ considerably so that the booster field is no
longer across points of equal potential. A current now flows
through the booster field causing the booster to generate an elec-
tromotive force of such polarity as to assist the battery to dis-
charge. /S is a spring.

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404 DIRECT CURRENTS

In order to reduce the current flowing through Ri and Rt and
the battery, the change of booster excitation is often accomplished
through an intermediary exciter, whose field is connected in the
same manner as the booster of Fig. 362.

Battery at End of Line. — ^Very poor voltage regulation may
occur at the end of a trolley Une, due to insufficient copper.
Rather than to install more copper, it may be more economical
to install a storage battery at the end of the line. This battery
not only steadies the trolley voltage but tends to reduce violent
fluctuation of the power station load as well.

As the voltage at the end of the line requiring a battery under-
goes fluctuations of considerable magnitude, the battery is
usually self regulating both as regards charge and discharge.
With little load on the line, the voltage at the battery should be
high enough to charge it. On the other hand, when a car is near
the battery, the Une voltage should drop to such a value as to
allow the battery to discharge and assist the power station.

•)- Bni-bar

X Bfcttetj

Track
Fig. 363. — Battery floating at end of trolley line.

Example, — The bus-bar voltage (Fig. 363) at the station is mwntained '
constant at 600 volts. A 4/0 trolley having a resistanoe of 0.26 ohm per
mile extends out 4 miles from the station. The resistance of ground and track
is 0.05 ohm per mile. At the end of the line a storage battery consisting of
240 cells is ''floated.'' Each cell has an average electromotive force of
2.0 volts and a resistance of 0.002 ohm. At what rate will the battery charge
when there is no load on the line? When the load at the battery is 150
amp., how much current does the battery supply and how much does the
station supply?

The total resistance of the trolley and track

R = 4(0.26 + 0.05) = 1.24 ohms.

The battery resistance

Ri = 240 X 0.002 = 0.48 ohm.

The total battery electromotive force

^ = 240 X 2.0 = 480 volts.

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TRANSMISSION AND DISTRIBUTION OF POWER 405

When there is no load between the station and the battery the current
to the battery

. 600 - 480 120 ^^

^' ' 1.24 + 0.48^72 = ^^ ^"'P- ^^-

To find the division of the 150-amp. load at the battery, first find the
current at which the battery will just "float,"

,, 600 — 480 120 ^^
^^° 1.24 -r24 = ^-^>"^P'

That is, with a load of 96.8 amp. at the battery the line drop from the
power station to the battery will be 120 volts, making 480 volts at the battery.
Under these conditions the battery will neither charge nor discharge but
will "float."

The remaining 53.2 amp. will be divided inversely as the trolley and
battery resistance.

Let II be the line current and Is the battery current.

Ib ^ 1.24
II 0.48
Ib +Il ^ 53.2
Solving Ib = 38.4 amp.

Il — 14.8 amp.

The station is already supplying 96.8 amp.

The total station current is then 96.8 + 14.8 — 111.6 amp. and the bat-
tery current is 38.4 amp. Ans,

This may be checked by calculating the voltage at the battery.
600 - (111.6 X 1.24) = 461.6 volts
480 - (38.4 X 0.48) = 461.6 volts. (Check)

269. Series Distribution. — In the parallel system of distri-
bution the loads are all independent of one another. That is, a
load applied at any one point does not affect any of the other
loads, provided the voltage does not change. In the series
system the loads are all in series with one another so that the
same cmrent passes through each. Therefore if the circuit of any
one load be opened the current to all the other loads will be inter-
rupted. As this is not permissible in practice, a load must be
short-circuited when it is desired to remove it from service.

Power is usually supplied to a constant current system by one
of two methods; the series generator, of which the Brush arc
and Thomson-Houston machines are examples, and the constant
current tranrformer operating in conjunction with the mercury
arc rectifier. (See Chap. VII, Vol. II.) Both of these methods
tend to maintain constant current under all conditions of load.

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406

DIRECT CURRENTS

Therefore, if the circuit be opened and a very high resistance
thus introduced j a constant current is maintained across a high re-
sistance and a very high voltage results. For this reason the lamps
used on a constant current system are protected by having a thin
disc of paper between the lamp terminals (film cut-out). If the
lamp burns out, the high voltage across this paper punctures it
and so prevents the circuit being opened.

The advantage of the series system is the small amoimt of
copper required. This is due to the fact that the copper carries
only the current of any single load. As the loads are in series the

rj station

Fia. 364. — Open loop series circuit.

11 Statioa
Fig. 365. — Parallel loop series circuit.

resulting voltage is high. Therefore, this system is applicable
only to outside work, such as street lighting, because it would ordi-
narily be dangerous to have such high voltages in buildings.

There are two general methods of connecting such series
loads. In the open loop system, shown in Fig. 364, the circuit
is connected to the loads without reference to the separation of
the two conductors. This system is economical of copper.

In the parallel loop system the outgoing and return con-
ductors are always kept near each other, as shown in Fig. 365.
This system requires more copper than the other but facilitates
testing for faults and reduces inductive disturbances.

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APPENDIX A
Relations of Units

Length

1 inch « 2.54 cm.
1 foot = 30.48 cm.
1 mile = 1.609 kilometers

Area
1 circular mil == 0.7864 sq. mil.
1 circular mil = 0.000507 sq. mm.
1 sq. inch = 6.452 sq. cm.
1 sq. meter = 10.76 sq. ft.

Volume
1 cubic inch = 16.39 cu. cm.
1 liter = 1,000 cu. cm.

» 0.2642 gallon
1 gallon = 231 cu. in.

W&ight
1 gram =981 dynes

1 ounce (av.) = 28.35 grams
1 kilogram = 2.205 lb.
1 ton = 2,000 lb.

1 long ton = 2,240 lb.
1 metric ton = 1000 kg.
« 2205 lb.

W<yrk
1 joule (watt-second) = 10,000,000 ergs

1 gram deg. Cent, (gram calorie) = 4.183 joules
1 pound deg. Fahr. (B.t.u.) = 252.1 gram deg. Cent, (gram calorie)

= 777.5ft.-lb.
J kilogram-meter = 9.81 joules

= 7.233 ft.-lb.
1 foot-poimd = 1.356 joules

1 horse-power-second = 178.3 gram deg. Cent, (gram calorie)

= 0.7074 lb. deg. F. (B.t.u.)

= 550 ft.-lb.

Pressure
1 atmosphere = 14.70 lb. on sq. in.

= 29.92 in. of mercury at 32** F.
= 760.0 mm. of mercury at 32** F.
= 33.94 ft. of water at 60** F.
1 lb. on sq. in. == 702.9 kg. on sq. meter
407

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408

DIRECT CURRENTS
APPENDIX B

Sp«

ciflc Gravities

Aluminum

2.67

Mercury 13.60

CJopper

8.96

Nickel 7.83

Gold

19.26

Platinum 20.30

Iron, bar

7.48

Silver 10.55

Iron, wrought

7.79

Tin 7.29

Steel
Lead

7.86
11.46

Zinc 6.86

1 cu. ft, of water weighs 62.6 Ih.

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APPENDIX

APPENDIX C

Table of Turns per Sq. In. ; Solid Layer Winding*

(The Acme Wire Co.)

409

Si»e.
A.W.G.

Single-cotton
covered

Enamel
and cotton

Single-silk
covered

Enamel
and silk

Enamel

12
13
14

15
16
17
18
19

20
21
22
23
24

25
26
27
28
29

30
31
32
33
34

35
36
37
38

87.5
109
136
169
210

260
321
396

488
598

772

947
1,155
1,410
1,720

2,080
2,500
3,020
3,630
4,270

5,100
5,920
6,950
8,120
9,430

10,850
12,350

84.5
105
130
161
199

248
304
374
456
556

722

890
1,075
1,303
1,575

1,910
2,310
2,770
3,300
3,910

4,630
5,330
6,300
7,300
8,410

9,610
10,850

865
1,075
1,330
1,650
2,045

2,520
3,090
' 3,810
4,690
5,650

6,950

8,410

10,000

12,080

14,500

17,300
20,400
23,700
27,800

807
1,010
1,230
1,510
1,860

2,290
2,830
3,460
4,220
5,100

6,200

7,300

8,900

10,650

12,600

14,900
17,300
20,400
23,700

92.5
117
147
184
231

292
366
458
572
715

907
1,150
1,425
1,780
2,220

2,800
3,540
4,440
5,570
6,950

8,730
10,650
13,500
16,900
21,000

26,000
31,900
40,000
49,300

♦Standard Handbook, Sec. 5, Par. 9&

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410

DIRECT CURRENTS

APPENDIX D

Table of Current-carrying Capacity in Amperes of Wires and Cables
Under Various Conditions

Ma.tionA.1 P^lAf*friftA.1 f!nHA

Lead-oovered cables

Si>e.
A.W.a. or

Single conductor

Three-c onductor

cir. mils

Rubber

iD8.

Slow burning
ins.

Rubber. 30

deg. Cent.

rise

bnc. 40deg.
Cent, rise

paper ins.,

45 deic. Cent.

rise

4
3

1
0

000

0000

250,000

300,000

400,000

500,000

750,000

1,000,000

1,500,000

2,000,000

15
20
25
35
50

100

125

150
175
225
235
275

325
400
525
650
850

1,050

20
25
30
50
70

90
100
125
150
200

225
275
325
350
400

500

600

800

1,000

1,360

1,670

20
30
50

121

145

169

192
245
285
320
370

460
550
750
900
1,200

1,400

22
34
56

87
110
134
160
187

210
270
316
360
415

515

605

830

1,030

1,450

1,590

110

132

150
190
225
255
300

370

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QUESTIONS ON CHAPTER I

1. What metal is the most useful for magnetic purposes? Why? What
other substances show magnetic properties?

2. Distinguish between a natural magnet and an artificial magnet.
Under what conditions should soft iron be used for magnetic purposes?
Hardened steel?

3. What is a magnetic field? What are lines of induction and do such
lines actually exist? Distinguish between a north-seeking pole and a north
pole. In what way does the magnetic circuit differ from the magnetic
field?

4. What is the effect of breaking a bar magnet near the neutral zone?
Explain how the newly created poles come into existence.

6. What is Weber's molecular theory of magnetism? How does it ex-
plain the phenomenon that occurs when a bar magnet is broken?

6. What is meant by consequent poles? Distinguish between consequent
poles and poles obtained by breaking a bar magnet.

7. When a freely suspended south pole is brought into the presence of a
north pole, what effect is noted? What effect is noted if the freely suspended
south pole is brought into the presence of a south pole? What is the general
law governing attraction and repulsion between like poles, and between un-
like poles?

8. What is a unit pole? How is pole strength determined? What is
the law governing the force of attraction and the force of repulsion between
magnetic poles?

9. Distinguish between lines of force and lines of induction. Are both
closed lines? In what way are lines of force and lines of induction similar?
At what part of the magnetic circuit is the magnetic force quite distinct
from the lines of induction?

10. What is unit field intensity? How are the lines of force related
to field intensity? What relation exists in the magnetic field between field
intensity and "lines of force per square centimeter?"

11. How many lines of force emanate from a unit pole? From a pole
of strength m units? If B is the flux density within a steel rod of 1 sq. cm.
cross-section, what is the pole strength at the end of the rod?

12* Cyf what does a compass needle consist? How is it used in practice
to determine the correct polarity of motors and of generators? How is it
I>ossible to obtain accurate indications from a compass when it is used upon
steel ships? How may a compass needle be used to map out an electrical
field in the vicinity of a magnet?

18. How may the flux distribution in a certain region be determined by
iron filings? Explain the relation of pole attraction and repulsion to the
distribution of the magnetic lines existing near the poles.

411

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412 DIRECT CURRENTS

14. What is magnetic induction? What is the relation between the
inducing and the induced pole? How does magnetic induction explain the
attraction of soft iron to magnetic poles? How may a compass become re-
versed? What is the use of a "keeper" in connection with a horseshoe
magnet?

16. What general law governs the path taken by the lines of induction?
How does this law explain the attraction of iron to the poles of a magnet?

16. What is the objection to the use of the bar magnet in practical work?
What advantages have the ring and the horseshoe magnet over the bar
magnet?

17. What is the principle underlying the compound or laminated magnet?
Where are laminated magnets used in practical work?

18. How may sensitive instruments be shielded from stray magnetic
fields that may exist in their vicinity?

19. How may a steel bar be magnetized by means of a bar magnet? By
means of two bar magnets? In practice, how may magnets be magnetized
by the use of electromagnets and also by means of electric current?

20. State why the compass needle does not point to the true north and the
true south in most places on the earth's surface. What information is
necessary in order to determine the true north from the indication of the
compass needle? What is the dip of the needle?

PROBLEMS ON CHAPTER I

■ . 1. Sketch the field around two bar mag-

^ £J nets arranged as in Fig. lA,

2. A uniform field is produced between
two parallel polar surfaces of a magnet.
A bar magnet is inserted in this field parallel
to the lines of induction and with its north
pole pointing to the north pole of the magnet.
Sketch the resulting field.

3. In problem 2 show the ultimate efifect
upon the magnetic flux distribution of in-

S I creasing the strength of the field due to the

Fig. lA. large magnet. What occurs to the bar

magnet?
4. In problem 2 sketch the flux distribution when the bar magnet is
perpendicular to the lines of induction.

6. Two poles of strength m = 800 and m' = 1,000 are 2 in. apart in air.
What force (in lb.) is acting between them?

6. The two poles of a horseshoe magnet are 4 cm. apart. If each has a
strength of 1,000 units, what force (lb.) is tending to pull them together?

7. A north pole of a bar magnet has a strength of 2,000 units. When
it is 4 in. from one end of a long soft-iron bar, it induces a pole of 300
units on this end of the bar. With what force is the magnet acting upon the
bar? Neglect the effect of the other poles. What pole is induced on the
bar, and in what direction does the force act? Make a sketch.

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QUESTIONS AND PROBLEMS 413

8. A pole strength of 150 units acts with a force of 1.2 grams upon another
pole 4 cm. away. What is the strength of the second pole?

9. A uniform magnetic field of 50,000 lines per sq. in. exists between
two parallel polar surfaces. What force (grams) is acting upon an N-pole
of 500 units placed in this field? Toward which pole will the N-pole tend
to be drawn?

10. A magnetic field of 160,000 lines and having the shape of a truncated
cone, exists between two parallel, plane surfaces having areas of 25 sq. cm.
and 60 sq. cm. respectively. What force is exerted upon a unit N-pole if
placed near the first surface? Near the second? Explain.

11. A N-pole has a strength of 100 units. How many lines of force
emanate from this pole?

12. What is the flux density at a distance of 2 in. from this pole?
What force would exist upon a unit S-pole placed at this distance from the
pole?

13. A pole of 500 units exists at the end of a steel rod of circular cross-
section and having a diameter of 0.8 cm. What is the approximate flux
density in the rod?

14. A long steel rod has a square cross-section of 0.5 in. per side. The
flux density in the crossnsection taken at the center of the rod is 15,000 lines
per sq. in. What is the strength of the poles at the end of the rod.

QUESTIONS ON CHAPTER H

1. What is the nature and general shape of the magnetic field about a
conductor carrying an electric current? What relation exists between the
direction of the current and the direction of the field produced about the
conductor?

2. How may the above relations be shown experimentally? What simple
rules enable one to remember the relation which exists between the cur-
rent direction and the direction of the magnetic field?

3. The current in a conductor flows from left to right. In what direction
will the north end of a compass needle point if held over the wire? If held
beneath the wire?

4. If two parallel conductors carry current in the same direction, do these
wires tend to separate or come together? Give two reasons for the answer.
Repeat for two conductors carrying current in opposite directions.

6. A single loop of wire lying in the plane of the paper carries a current
in a clockwise direction. What effect will be noticed if a compass is placed
within this loop? Has this loop any properties in common with those of
a bar magnet?

6. Show how several loops similar to the one mentioned in (5) may be
combined to form a long solenoid.

7. Give three methods whereby the poles at the ends of a solenoid may
be determined, provided the direction of the current through the solenoid
turns be known.

9* What are commercial uses of the solenoid? Name seven such uses.

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414

DIRECT CURRENTS

9. Explain by the fundamental laws of magnetism why the plunger is
drawn into a solenoid when current flows in the solenoid winding.

10. Plot the relation between the pull on the plunger and the position
of the plunger in the solenoid.

11. What effect does ''iron-cladding " have upon the pulling characteristic
of the plunger? State one practical application of the simple solenoid;
of the iron-clad solenoid. What effect does the stop have upon the solenoid
characteristic? State a practical use of this type of solenoid.

12. Show the principle whereby a U-shaped solenoid attracts an armature.
Explain the principle of operation of the telegraph relay; the ordinary elec-
tric door-bell.

18. Sketch a lifting magnet, showing its general construction. Where are
such magnets used commercially, and in what way are they more economical
than the older methods of handling material? Does the magnet itself do
appreciable work when it is being used to handle iron and steel?

14. What is the disadvantage of the early types of magnetic circuits of
dynamos, as represented by the Edison bi-polar type? How has the design
of the magnetic circuits of the more modem generators overcome some of
the disadvantages of the earlier ones. What should be the approximate
ratio of the cross-section of the field cores of a multi-polar generator to the
cross-section of the yoke? What general rule should be followed in the
placing of the exciting ampere-turns upon a magnetic circuit? Does
magnetic leakage between the poles of a generator represent a direct loss of
power?

PROBLEMS ON CHAPTER H

16. A portion of a direct-current feeder is shown in Fig. 15A. When a
compass is held above the feeder the needle deflects as shown. In what direc^
tion does the current in the feeder flow, in or out of the duct?

Fig. 15A.

Fio. 16A.

16. Fig. 16A shows two positive feeders of a trolley system running upon
a pole line and carrying current in the same direction. If the troUey wire
drops upon the track, causing an enormous current to flow in the feeders for
an instant, in what direction will these conductors tend to move and what
b the direction of the force acting upon the insulators?

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QUESTIONS AND PROBLEMS

415

17. In Fig. 17 A is shown the principle upon which one type of electric
hammer operates. Two coils C and C are connected in series and in the
positions shown. P, a soft-iron plunger running in guides, actuates the
hammering device. A coil D, encircling the plunger P, is excited continu-
ously with direct current. If the terminals a and h of the coil D are of the

a b

Fio. 17A.

Fio. 18A.

polarity shown, indicate the polarity of the ends of the plunger P. If terminal
^ is -f and terminal B is — , in what direction will the plunger P tend to
move? If the polarity of terminals A and B is reversed, in what direction
does the plunger tend to move?

■■}

=T'

1. .-r. TrJ

t ■'

''\

^^^ ,

V "

K^^ ,

^nn- magnetic
^jiDflaD«ie Steel

Fig. 20 a.

18. Fig. ISA shows two coils on a simple horseshoe magnet. Connect
these coils so that they aid one another. Sketch the magnetic field between
the poles.

19. Assuming that one of the field coils of Fig. 38, page 27 is reversed,
that is, the two coils **buck" each other, sketch the general appearance of
the magnetic field. Will the total flux be increased or diminished by this
method of connection?

20. Fig. 20A shows in cross-section a lifting magnet about to pick up
a heavy iron sphere known as a "skull cracker" (used in breaking up scrap

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416

DIRECT CURRENTS

iron). Sketch the magnetic lines and mark the poles existing under these
conditions. The horizontal section of the magnet is circular. Assume that
the current enters the coil in the right-hand section of the exciting coil.

21. Connect the coils 06, cdf ef^ gh, in the multi-polar machine shown in
Fig. 21 A, so that the proper sequence of poles is obtained. Make the left-
hand pole a north as shown. Sketch the paths of the magnetic lines.

Fig. 21A.

Fio. 22A,

22. Fig. 22A shows the pole face of a generator and an armature tooth.
Sketch the paths of the magnetic lines in passing from the pole face into the
tooth and from the tooth to the rest of the armature iron. What is
"fringing?"

QUESTIONS ON CHAPTER HI

1. What is the mechanical analogue of resistance? What is the unit of
resistance? How is it defined?

2. Distinguish between insulating materials and conductors. What is a
"megohm?" A "microhm?"

3. May two conductors, each of the same material and of equal volume,
have different resistances? Explain.

4. How does the resistance of a homogeneous material vary with its
length and with its cross-section? What is specific resistance or resistivity?

5. If the volume of a substance is fixed, how does its resistance vary with
its length? With its cross-section? If the volume is fixed and the length
doubled, how is the resistance affected?

6. What is conductance and how does it vary with the length and cross-
section? Distinguish between conductance and conductivity . What is the
general meaning of "per cent, conductivity?"

7. What is the relation of the total resistance of a circuit to the resist-
ances of its individual parts when these latter are connected in series?

8. What is the relation of the total conductance of a circuit to the conduct-
ances of its individual parts when these latter are connected in parallel?
From this relation show how resistances connected in parallel may be com-
bined into an equivalent resistance.

9. What is the meaning of the term "mil?" What is a square mil? A
circular mil? What relation does one bear to the other? Where is the cir-
cular mil usually chosen as the unit of crow-section? What are its advan-

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QUESTIONS AND PROBLEMS 417

tages over such units as the square mil and the square inch? What relation
does the number of circular mils in a circular cross-section bear to its
diameter?

10. What is a cir. -mil-foot? What is its approximate resistance for
copper? How may the resistance of a copper wire be determined if its
length in feet and its cross^section in cir. mils be known?

11. How is the resistance of most of the unalloyed metals affected by
temperature? What is the "temperature coefficient of resistance?" How
is it used?

12. At what temperature would the resistance of copper be zero if the
resistance decreased at the same rate that it decreases within ordinary ranges
of temperature? How may this principle be used to solve problems in-
volving resistance and temperature?

13. What relation do the cross-sections of the wires in the A.W.G. bear
to one another? How does this relation enable one to determine readily
the resistance and weight of any given size of wire? What is the resistance
of 1,000 ft. of No. 10 wire? What is the weight of 1,000 ft. of No. 2 wire?

14. What are the best conductors among the metals? Which is most
commonly used and why? Compared with copper what are the advantages
and the disadvantages of aluminum as a conductor? When are iron and
steel used as conductors? Explain.

PROBLEMS ON CHAPTER m

23. Two conductors, A and By of the same material, have the same length,
but the cross-section of A is twice that of B, If the resistance of A is 30
ohms, what is that of B?

24. Two conductors, C and D, of the same material, have the same length,
but the diameter of C is twice that of D. If the resistance of C is 30 ohms,
what is that of D?

26. If the resistance of copper is 1.724 microhms per cm. cube at 20** C,
what is the resistance of an inch cube at the same temperature?

26. A rectangular copper plate has a length of 18 in., a width of 6 in. and
a thickness of 0.5 in. If the resistance of copper is 1.724 microhms per cm.
cube, what is the resistance of the plate between the 6-in. edges? Between
the 18-in. edges?

27. A phosphor-bronze strip J^ in, X 1 in. and 4 ft. long has a resistance
of 0.000597 ohm. What is its resistivity per cm. cube? Per in. cube?

28. No. 16 copper wire has a diameter of 51 mils and a resistance of 4.02
ohms per 1,000 ft. at 20** C. What is the resistance of 5 miles of 000 copper
wire (diameter is 410 mils)?

29. A cylindrical conductor A has twice the diameter and twice the length
of a cylindrical conductor B. If the resistance of B is 5 ohms, what is the
resistance oi A?

30. What is the resistance of a copper bus-bar 40 ft. long, made up of 4
bars of copper each A in, X H ui,7 The resistance of copper is 1.724
microhms per cm. cube.

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418 DIRECT CURRENTS

81. If copper weighs 0.32 lb. per cu. in. and costs $0.20 per lb., what is
the cost of the bus-bar in problem 30?

82. (a) If aluminum bars H in. thick and of the same conductance were
substituted for the copper in problem 31, what would be the ratio of radi-
ating surfaces? Spacers are used between the bars. Neglect the ends as
radiating surfaces.

(6) What should be the cost of aluminum per lb. in order that the
aluminum bus-bars shall cost the same as the copper? Specific gravity of
copper s 8.89; of aluminum — 2.70.

88. A 000 copper conductor 800 ft. long and having a diameter of 410
mils is drawn down so that its diameter is 268 mils. If the resistance of the
000, 800-ft. conductor was 0.05 ohm, what is the resistance of the entire
length when its diameter has been reduced to 258 mils?

84. Determine the conductance of a copper rod, 1 in. diameter and 8 ft.
long. (Conductivity of copper = 580,000 mhos per cm. cube.

86. The resistance of a 4-ft. length of No. 8 wire is measured and found to
be 0.00241 ohm at 20** C. What is its per cent, conductivity?

86. A copper bar ^ in. X 1 in. and 3.5 ft. long rolled from electroljrtic
copper is found to have a resistance of 0.0000755 ohm at 20"* C. What is
its per cent, conductivity?

87. The resistance of 500 ft. of No. 18 wire is measured at a temperature
of 25** C. and found to be 3.35 ohms. What is its per cent, conductivity?

88. Three resistances of 4.2 ohms each, two resistances of 6.3 ohms each,
•and a resistance of 8.6 ohms are all connected in series. What is the total
resistance of the combination?

80. Two resistances of 8 and 4 ohms are connected in parallel. What is
the total resistance of the combination?

40. Three conductances of 6, 8, and 10 mhos respectively are connected in
parallel. What is the resulting total conductance? What is the total
resistance?

41. If the three conductances of problem 40 are connected in series, what
is the resulting conductance? Resistance?

42. If the individual resistances of problem 38 were all connected in
parallel, what would be the resulting resistance?

820

60 Q 120 Q Ww^m*vnA^^^>v

FiQ. 43il. Pio. 44A.

43. A resistance of 41 ohms is connected in series with a group of two
resistances of 60 and 80 ohms respectively, connected in parallel (Fig.
43A). What is the resulting resistance?

44. A group of two resistances, of 120 and 140 ohms in parallel, is con-
nected in series with another group of 82, 96 and 110 ohms in parallel
(Fig. 44A). What is the total resistance resulting from this combination?

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QUESTIONS AND PROBLEMS 419

46. A 0000 trolley wire (hard drawn) having a resistance of 0.0514 ohm
per 1^000 ft. extends 6 miles from the power station. It is paralleled for 3
miles by a 250,000 CM. cable, having a resistance of 0.0431 ohm per 1,000 ft.,
the feeder and the trolley being connected every half mile by taps
(Fig. 45A). What is the total resistance of the overhead circuit from the
power house to the end of the trolley line?

8BO.00O CM. Feeder

Power I?' -M." -'.I I I l^.>».T,.n.r

House.

Fig. 45il.

46. The resistance of each rail of the trolley system of problem 45 is
0.080 ohm per 1,000 ft. including bonding. An insulated negative feeder
consisting of a 0000 stranded cable of resistance 0.0509 ohm per 1,000 ft.
runs from the power house and is bonded to both tracks 2)^ miles out.
What is the total resistance of the return circuit, neglecting any con-
ductance of the earth itself?

47. How many cir. mils in a rod of 1 in. diameter? 0.75 in.? 0.5 in.?
0.25 in.? Min.?

48. What is the diameter of a wire having a cross-section of 168,000 CM.?
66,400 CM.? 62,500 CM.? 8,100 CM.? 400 CM.?

49. Assuming that the resistivity (per cir.-mil-ft.) of copper is 10 ohms,
determine the resistance of 2 miles of copper wire having a cross-section of
10,000 CM.?

60. Determine the resistance of a telegraph loop between two stations 30
miles apart, if the wire is copper having a diameter of J^ in.

61. Hard drawn copper wire, such as is used for trolley wire, has a resis-
tivity 2.7 per cent, greater than that of annealed copper. Determine the
resistance of 5 miles of 000 trolley wire, if the resistivity (cir.-mil-ft.) of
annealed copper is 10.37 ohms.

62. What is the resistance of 2 miles of 700,000 CM. stranded copper
cable?

68. What is the resistance at 0° C of a reel of 0000 annealed copper wire,
the wire weighing 400 lb., if the resistance per 1,000 ft. of 0000 copper is
0.050 ohm at 25° C?

64. What is the resistance of the wire of problem 53 at 50** C?

66. The resistance of a copper telegraph circuit was found to be 40 ohms
when the external temperature was 0° C What would be its resistance at a
maximum summer temperature of 40" C?

66. The resistance of a shunt field coil of a generator is 44 ohms at 22® C
What is its resistance at 0° C ? At 76° C ?

67. The resistance of an armature winding of a shunt motor is found to be
0.042 ohm at 25° C What is its hot resistance when it attains a tem-
perature of 70° C?

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420 DIRECT CURRENTS

68. A direct-current feeder has a resistance of 0.007 ohm at 20° C. What
is its change in conductance between the lowest winter temperature of*
—20° F. and the maximum summer temperature of 100° F.? What is the
percentage change?

Without consulting the Wire Table solve the following problems :
60. Estimate the resistance of 1,000 ft. of No. 13 bare copper wire; of
No. 16.

60. Estimate the resistance and weight of 1,000 ft. of No. 18 bare copper
wire; of No. 24.

61. Estimate the resistance and weight of 2,000 ft. of No. 8 bare copper
wire. Of 800 ft. of No. 1. Of 500 ft. of 0000.

62. Estimate the weight, resistance and cir. mils of 600 ft. of 0 bare copper
wire. Of 600 ft. of 00.

QUESTIONS ON CHAPTER IV

1. What is the unit of electric current and how is it related to the unit of
electric quantity? What is the nature of potential difference and of electro-
motive force? What are the mechanical analogies of electromotive force
and why?

2. What is the nature of voltage drop in a line? Can it be compared to
pressure drop in a pipe? Is it possible to supply power over a line and have
the voltage at the load equal to the voltage at the sending end of the line?
Explain. Is there a voltage loss in the return wire to the generator as well
as in the outgoing wire? Can potential exist without a current flowing?
Illustrate.

8. What is meant by "difference of potential"? Is it possible to have
two or more emf.'s and yet have no difference of potential between certain
points?

4. How should a voltmeter ordinarily be connected in a circuit? Is an
ammeter connected in the same way as a voltmeter? Why should an am-
meter never be connected across a line?

5. What fundamental relation does Ohm's Law express? In what three
forms is the law expressed? Under what conditions is it most convenient
to use each of these?

6. How are series-connected resistances combined to equal an equivalent
resistance? How are parallel resistances combined? What relation does
the division of current in a two-branch parallel circuit bear to the resistance
of each branch? What relation exists among the currents when the circuit
has three branches?

7. What is the unit of electrical power? How may it be expressed in
terms of volts, amperes and ohms, taken two at a time? Differentiate care-
fully between power and energy. What is the unit of electrical energy and
what relation does it bear to the unit of power ? What is the unit of mechan-
ical horsepower? What relation does it bear to the units of electrical power?

8. Discuss the various forms in which energy is stored or in which energy
may appear. Describe the energy cycle involved in a steam-driven electrical

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QUESTIONS AND PROBLEMS 421

power plant. In what form does the energy appear ultimately? Approxi-
mately what is the over-all efficiency of a modem power system?

9. How is a B.t.u. defined? A gram-calorie? What is the relation be-
tween a gram-calorie anda wattnsecond?

10. What simple relation exists between the voltages at the sending
and receiving ends of a power feeder and the efficiency of transmission?

11. Under what conditions is the voltage drop in each foot of wire inde-
pendent of the total current? How is this principle utilized in solving
electrical problems? Can this method be applied to obtaining the power
loss? Explain.

PROBLEMS ON CHAPTER IV

63. A storage cell has a constant potential difference of 2.1 volts at its ter-
minals. What current flows when 0.4 ohm isconnected across its terminals ?

64. A carbon filament incandescent lamp has a cold resistance of 330 ohms
and a hot resistance of 240 ohms. What current does it take when it is
first connected to 115-volt mains? At what current does it operate?

66. A 110-volt, 25-watt tungsten lamp has a cold resistance of 40 ohms
and a hot resistance of 480 ohms. What current does it take when it is
first switched to 110-volt mains and what current does it take when
it has attained normal operating conditions?

66. A 220-volt generator has a field resistance of 160 ohms, including the
rheostat. What current flows in the field?

67. A 550-volt generator has a field resistance of 350 ohms and the
rheostat has a resistance of 45 ohms. What current does the field take?
To what value should the resistance of the rheostat be adjusted in order to
reduce the field current to 1.2 amp.?

68. A carbon rheostat has a resistance of 0.24 ohm and carries a current
of 40 amp. What is the voltage across the rheostat?

69. What voltage must a generator develop to supply 25 amp. to an
electric oven in which the heating coils have a resistance of 8.5 ohms and
the connecting wires a total resistance of 0.25 ohm.

70. A telegraph relay is wound for 150 ohms and operates at 40 milli-
amperes. What should be the voltage of the circuit battery if it is to operate
the relay over a line having a resistance of 30 ohms?

71. A series lighting system consists of 118 lamps, each having a resist-
ance of 7.2 ohms and requiring 6.6 amp. If the line resistance is 100
ohms, what is the voltage of the generator supplying this system?

72. An incandescent lamp takes 0.25 amp. at 110 volts. What is its hot
resistance?

73. When a copper bus-bar carries 1,580 amp., the voltage drop across a
6 ft.-length is found to be 1.26 millivolts. What is the resistance per ft. of
the bus-bar?

74. The voltage drop across the series field of a compound generator de-
livering 250 amp. is 0.7 volt. What is the resistance of the series field?

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422 DIRECT CURRENTS

76. A direct-current multiple arc lamp takes 6.0 amp. at 110 volts. If
the drop across the arc is 70 volts, what is the resistance of the '^ballast?"

76. Fig. 76ii shows a lamp bank, having a total resistance of 7.5 ohms,
being supplied from a 115-volt generator over connecting wires having a re-
sistance of 0.15 ohm per wire. What current does the lamp bank receive?

0.16J1.

QOQQQ^''^

0.15-n-

FiG. 76A.

77. An electromagnet has four spools of 1 ohm each, all connected in
series. Two wires, each having a resistance of 0.05 ohm, connect the magnet
to ll&-volt mains. What current does the magnet take? If one coil
becomes ''grounded" so that half of its resistance is short-circuited, what
will the magnet current be?

78. Determine the equivalent resistance of a circuit having four resist-
ances of 16, 20, 30 and 40 ohms in parallel. If the current in the 16-ohm
resistance is 2 amp., determine the current in each of the other resistances.

70. The series field winding of a generator has a resistance of 0.004 ohm

and is shunted by a diverter having a resistance of 0.012 ohm. What is

the voltage drop across the series field when the generator delivers 400 amp.?

80. In problem 79 how will the 400 amp. divide between the diverter

and the series field?

' 81. A 000 hard-drawn trolley wire is 6 miles long and is paralleled by a

0000 annealed copper feeder. What is their combined resistance? What

is the voltage drop in the feeder when a total

current of 100 amp. is flowing in the two?

82. Four selective relays connected in
^^o parallel are supplied by a common wire shown
in Fig. S2A. If their resistances are 20, 25,
31 and 37 ohms respectively, what is the
voltage across their terminals when 2 amp.
Fia. 82 A. ^^^ supplied by the common wire?

: - 83. In problem 82 how will the 2 amp.
divide among the four relays?

84. Two ammeters, one having a 50-amp. scale and the other a 100-amp.
scale, are connected in parallel so as to measure a current greater
than 100 amp. If the 50-amp. instrument has a resistance of 0.002 ohm
and the 100-amp. instrument a resistance of 0.0012 ohm, what will each read
when 130 amp. flows in the circuit?

86. To feed a trolley wire at a given point, two feeders, one 350,000 CM.
and the other 250,000 CM., parallel the 0000 hard-drawn trolley wire.
When the current demand upon the system is 600 amp., how does it divide
among the feeders and trolley?

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QUESTIONS AND PROBLEMS

423

86. Fig. 86A shows a 120-volt generator supplying lamp loads over
mains, the mains having a resistance of 0.3 ohm each. The loads are as
follows: 6 gem lights, 305 ohms each; 10 tungsten lamps, 290 ohms each;
and 4 tungstens, 150 ohms each. What current does the generator deliver?

0.3 .n.

FiQ. 86 A.

87. Fig. S7A shows a 115- volt generator supplying lamp loads. Indi-
cate the currents at each part of the system, and the voltage at the various
lamp terminals. What is the voltage at a57

0.2 -TV

0.2-

Lamp Load 10 -TL Lamp Load IS-Tl.

0.2 jO. ' 0.2 -fl.

Fig. S7A.

0.1 -H-

O.l-TL

88. A resistance of 50 ohms is connected in series with two parallel re-
sistances of 75 and 100 ohms. These are in turn connected in series with a
group of three parallel resistances of 120, 150 and 180 ohms. What is the
total current of the system when it is connected across 100-volt mains?
How much current does each resistance take and what is the voltage across
each resistance?

120 V.

12A

Gen. Field
80A

- 89. Fig. 89A shows a drop wire, used for regulating the field current of
a generator from zero to its maximum value. The total resistance of the
drop wire ab is 12 ohms and that of the field is 30 ohms. If the line voltage
is 120 volts, what current does the generator field take when the contact x
is H the distance from a to 6? }4 the distance? ^i the distance?

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424 DIRECT CURRENTS

90. A gas-filled lamp takes 6.8 amp. from 110-volt mains. What is
its rating in watts?

01. A generator delivers 1,670 amp. at 600 volts. What is its kilo-
watt rating?

92. The resistance of the series field of a compound generator is 0.002
ohm. What power is lost in this field when the generator delivers 1,200 amp.?

98. A 2,000-amp. shunt has a resistance of 0.000025 ohm. What power
is lost in this shunt when carrying its rated current?

94. A tungsten lamp having a hot resistance of 202 ohms is connected
across 110 volt mains. What is its watt rating?

96. Four street car heating units, each having a resistance of 55 ohms,
are connected in series. If the trolley voltage is 600 volts, what power do
the heaters take?

96. A shunt motor takes 75 amp. at 220 volts and delivers 20 hp.
What is its efficiency?

97. A generator delivers 250 amp. at 230 volts. If it has an efficiency
of 92 per cent., what horsepower engine is required to drive it?

98. An electroplating bath takes 80 amp. at 20 volts for 2.4 hours, and
then 50 amp. at 60 volts for 1.5 hours. How much energy is consumed?
At 4 cents per kw.-hr., what is the cost of the required energy under the above
conditions?

99. How many joules are supplied to a 25-watt lamp burning 4 hours?
K the supply voltage is 110, how many ampere-hours are delivered to the
lamp?

100. Energy costs 8 cents per kw.-hr. Determine the cost of heating
2 quarts of water at a room temperature of 25° C. to the boiling point
(100° C). Assume that the efficiency of the heater is 80 per cent. (Water
weighs 8.35 lb. per gal.)

101. A water-barrel rheostat contains 30 gal. of water. How long must
45 amp. at 220 volts flow before the temperature of the water is raised
to 200° F. from a room temperature of 70° F.? (1 gal. water weighs 8.35
lb.) Assume no losses.

102. A factory, 1 mile from a power station, takes a maximum current
of 120 amp. over a 250,000 CM. feeder. If the station voltage is main-
tained constant at 600 volts, what is the greatest change of voltage that
occurs at the factory from no load to the maximum load? What is the
efficiency of transmission at the maximum load and at half this load?

*^/////yi^ • 260.000 CM.

Powe^ I I I

SUtio]^

-S.gMk-

Fig. 103A.

103. An electric railway is fed by a 5-mile trolley line of 0000 hard-drawn
copper. A 250, 000 CM. feeder parallels this trolley for 3 miles, being tapped

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QUESTIONS AND PROBLEMS 425

in every half mile (Fig. lOSA). The resistance of the ground return may be
considered as 0.02 ohm per mile. If the station voltage is 600, what is the
voltage at the car when it is 3K miles from the station and is taking
110 amp. What is the voltage at the end of the line at this time? What
is the efficiency of transmission?

104. Fig. 104A shows two loads, one of 500 amp. }i mile from the
power station, and another of 200 amp. 1,000 ft. farther along. A 1,000,000
and a 500,000 CM. cable are in parallel to the first load; a 750,000 CM.
runs from the first to the second load. The voltage at the 200-ampere load
is 220 volts. What is the station voltage and the efficiency of transmission?

1,000,000 CM.

500.000 CM.

-)iMi^

760,000 CM.

Power I BWlAmp. «00 Amp.

Station I --j f
I 500.000 CM.

1,000,000 CM.

Fig. 104^.

Solve the following problems by the methods outlined in Pars. 68 and 69.
Do not consult the wire tables.

106. A 1,000-ft. length of 200,000 C M. cable supplies a certain power load
What is the total voltage drop in the cable if the load is such that the cable
operates at the normal densityf What is the power loss under these
conditions?

106. If, in problem 105, the cable operates at a density corresponding to
1,500 cir. mils per amp., what is the total voltage drop? What is the power
loss under these conditions?

^ 107. A 200-amp. load is to be supplied from the 600-volt bus-bars of a
power station at a distance of 0.5 mile. The voltage drop cannot exceed
10 per cent, of the station voltage. What size feeder is necessary, and what
is the efficiency of transmission?

108. A 40-hp. motor is to be supplied with power at a distance of 500 ft.
from 230- volt bus-bars. The voltage drop cannot exceed 15 volts. The
motor has an efficiency of 90 per cent. What size wire is necessary and what
is the efficiency of transmission?

QUESTIONS ON CHAPTER V

1. What is the effect upon the terminal voltage of a battery of applying
a load to its terminals? Explain. Why does the electromotive force of a
cell differ from the terminal voltage? Under what conditions are they
the same?

2. Is it possible to make a direct measurement of the internal voltage of
a cell when it is delivering current? How may this internal voltage be
calculated if the battery resistance be known?

8. To what is the internal resistance of a battery due?
Is this resistance a constant quantity?

4. If the electromotive force and the resistance of a battery be known,
how may the current delivered to an external resistance be calculaf<

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426 DIRECT CURRENTS

the battery becomes short-circuited what current does it deliver? What
becomes of the energy that the cell develops under these conditions?

6. Under what conditions may a battery be made to receive electrical
energy? What relation does the direction of current flow bear to its direc-
tion when the battery delivers energy? If a generator has a voltage equal
to that of the battery, what effects are noted when the generator is connected
to the battery, terminals of like polarity being connected together? What
effect is noted when the generator voltage is raised above this value ? What
is meant by the battery " floating? "

6. Before current can be sent into a battery, what voltage must first be
applied? Explain why the voltage in excess of that of the battery alone is
effective in causing the flow of current. What is a very common illustration
of a battery receiving energy?

7. K several cells are connected in series, what is the resultant electro-
motive force of the combination? What is the resultant resistance of the
combination? How may the current be found if the external resistance be
known?

8. Under what conditions do batteries operate most satisfactorily in
parallel? What is the electromotive force of the combination under these
conditions? What is the relation between the external current and the
current in the individual cells? What is the relation between the total
battery resistance and the resistances of the individual cells? K the resist-
ances of the individual cells are not equal, how may the resistance of the
entire battery be found? What relation does the current delivered by each
cell bear to the resistance of the cell? What relation exists among the ter-
minal voltages of individual cells connected in parallel?

9. What is a series-parallel grouping of cells? What is the voltage of the
entire battery? How may the resistance of the battery be found if the
resistance of the individual cells be known? How may the current in an
external circuit be found if the external resistance, the electromotive forces
and resistances of the individual cells and their arrangement be known?

10. In general, how should cells be grouped to obtain the best economy?
How should cells be arranged to obtain the maximum power output?

11. What two fundamental principles are stated in Kirchhoff's laws? If
several currents meet at a junction, how should their direction of flow be
taken into account?

12. How should a rise in potential be represented? A drop in potential?
When passing from a — to a + terminal of a battery, what should be the sign
of the potential change and why? When passing from -f- to — ? When
passing through a resistance in the direction of the current does a rise or a
drop in potential occur? What then should b^ the proper sign to use?
When passing along the resistance in opposition to the current what sign
should be used? Why?

13. If the assumed direction of a current in a network is in error, how is
this fact indicated in the result?

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QUESTIONS AND PROBLEMS 427

PROBLEMS ON CHAPTER V

109. A Le Clanch6 cell has an open-circuit voltage of 1.4 volts. When the
cell delivers 5 amp. its terminal voltage drops to 1.3 volts. What is the
internal resistance of the cell?

110. A starting battery, consisting of three storage cells connected in
series, has an open-circuit voltage of 6.4 volts. When delivering 90 amp.,
its potential drops to 5.0 volts. What is the internal resistance of the battery
and of each cell?

111. A gravity cell has an open-circuit voltage of 0.9 volt and an internal
resistance of 0.3 ohm. When its terminal voltage is 0.83 volt, what current
is it delivering?

112. When the storage battery of problem 110 is supplying only the light-
ing load of 14 amp., what is its terminal voltage?

113. An automobile starting battery, when being charged, has an electro-
motive force of 6.6 volts and an internal resistance of 0.03 ohm. What
voltage must the charging generator supply to the battery in order to charge
it at the 30-amp. rate?

114. A storage battery consists of 55 cells each having an electromotive
force of 2.1 volts and a resistance of 0.002 ohm. What current will it take
if connected across 120-volt bus-bars? What power

is being delivered to the battery? How much id s,-iov. Sg-av.
stored and how much is lost as heat? ri-o.»-rL ri^o.hJ\.

116. In Fig. 115A are shown two cells connected in ~||+ + |
series and in series with a 3.7-ohm resistance. De-
termine the current /, the power pi and pi developed

■V{-*u — iJj-*!

in each cell, the power Pi and P2 delivered by each ' — ^/WWVVV — '
cell, the power lost in each cell, the voltage Vi and 1^2 ^^ ^"^-^

across each cell and the voltage V across the re- -j^^ lis A.

sistance. (If a cell is absorbing energy, the power
developed is negative.)

116. Two Le Clanch6 cells, each having an electromotive force of 1.35
volts and an internal resistance of 0.15 ohm, are connected in series and
supply current to a 10-ohm resistance wire. What current flows in the wire
and what is the terminal voltage of the battery?

117. A station battery, consisting of 55 storage cells, all in series, each of
which has an electromotive force of 2.2 volts and a resistance of 0.0005
ohm, supplies current to a load having a resistance of 5 ohms. What cur-
rent does the battery deliver?

118. If the battery of problem 117 were accidentally short-circuited, what
current would flow?

119. Each of five dry cells has an electromotive force of 1.4 volts; three
have an internal resistance of 0. 1 ohm and two have an internal resistance of
0.12 ohm. If these cells are all connected in series and to a circuit having a

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428 DIRECT CURRENTS

resistance of 12 ohms, what current flows? What is the voltage across each
cell?

120. Each of two starting batteries has an electromotive force of 6.5 volts;
one has an internal resistance of 0.008 ohm and the other a resistance of
0.010 ohm. What is the equivalent resistance of the battery consisting of the
two connected in parallel? What is the terminal voltage of the combined
battery when it delivers 160 amp.? How does this current divide be-
tween the two individual batteries?

121. What is the resistance of a battery obtained by connecting all the
cells of problem 119 in parallel? When a resistance of 0.8 ohm is connected
across the terminals of this battery, what current flows? What is the ter-
minal voltage of the battery and how much current does each cell
deliver?

122. A battery consists of four storage cells all connected in parallel.
The internal resistances of these cells are 0.006, 0.004, 0.003 and 0.0026 ohm
respectively. K the electromotive force of each is 2.2 volts, what current
does the battery deliver when its terminal voltage is 1.9 volts?

123. Twenty-four dry cells are arranged in rows of six in series and the four
rows in parallel. The electromotive force of each cell is 1.4 volts and the
resistance of each is 0. 1 ohm. What is the total battery voltage and what is
its total resistance? If an external resistance of 2.6 ohms is connected
across its terminals, what current flows?

124. Arrange the cells of problem 123 so that the maximum amount of
power may be supplied to a load resistance of 0.6 ohm. Under these condi-
tions how much power is absorbed by the resistance and how much is lost
in the battery?

126. A certain load is such that the potential difference at its terminals
must not be less than 6 volts. Twelve storage cells, each having an electro-
motive force of 2.1 volts and a resistance of 0.02 ohm, are available. How
should these be connected so that the maximum efficiency is obtained?
When the load requires 10 amp., what is the battery terminal voltage?
What is the load resistance? What is the battery efficiency?

126. Arrange the cells in problem 125 so that the maximum amount of
current will be delivered to the load resistance. What is the efficiency of the
battery under these conditions?

127. A telegraph battery consists of 12 gravity cells, each having an
electromotive force of 0.9 volt and a resistance of 0.2 ohm. How should
these cells be connected so as to operate most satisfactorily a 20-ohm
relay over a 50-ohm circuit? What is the battery efficiency under these
conditions?

128. Two batteries A and B (Fig. 128A), having electromotive forces of
4 and 3 volts and resistances of 1.2 and 1.0 ohm respectively, are connected
in parallel, positive terminal to positive terminal. What current flows
through a 2-ohm resistance connected across the battery terminals? What
is the battery terminal voltage and how much current does each battery
deliver?

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QUESTIONS AND PROBLEMS

429

120. Two batteries, having electromotive forces of 6 and 6 volts and re-
sistances of 1.0 and 0.5 ohms respectively, are connected in parallel, positive
terminal to positive terminal (Fig. 129 A). These two supply current
through a 1.0-ohm resistance to charge a 2-volt battery of a resistance of 0.3

4V.

3V. ^

1.2r»-

1.0 rL >

? 1

Fio. 128A.

2-rv

6V.

[

5 1.0-0-

■■

0.6 -TL.

+ f«v.

IHM

0.3 .n-

Fig.

129A.

ohm ; the 2-volt battery is so connected that the current flows in at its posi-
tive terminal. What is the charging current of the 2-volt battery? What
is its terminal voltage?

130. Across the terminals of a 12-volt, 0.2-ohm battery, a resistance wire
of 10 ohms is connected. The negative terminal of a 6-volt battery, whose
resistance is 0.15-ohm, is connected to the

negative terminal of the 12-volt battery.
The positive terminal of the 6-volt battery
is connected to the resistance wire at a

point ?i its length from the negative ter- I *i^ + 8"^.

minals (see Fig. IZOA), Determine the
currents and the terminal voltages of each
battery.

131. (a) Determine the currents and the

terminal voltages of each battery in problem 130 if the 6-volt battery is
reversed.

(6) At what point must the contact a. Fig. 130A, be placed upon the
resistance wire so that no current flows in the 6-volt battery circuit?

132. Two sub-stations A and B feed into the same distributing center.
The voltage at the bus-bars of station A is maintained constant at 600 volts

t ^

8.<

• -^ 1

18 V.

0.8 -n- ]

7.J

s ^

IT-

Fig.

130 A.

-aooo ft.-

400,000 CM.

350,000 CM. }r^

. — T — i^

500 AmpB.

±i±

350,000 CM.

Fig. 132A.

and that at station B is maintained at 580 volts. Station A feeds a distance
of 2,000 ft. through 400,000 CM. cable and station B a distance of 1,000 ft.
through 350,000 CM. cable (see Fig. 132A). When the load at the distrib-
uting center is 500 amp., how much does each station supply? How

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430 DIRECT CURRENTS

much power does each station supply, and how much is received at the
distributing center?

188. Fig. 133A shows a distribution system. The voltage at the sub-
station A is maintained constant at 240 volts. A radial feeder extends from
A to each of the distributing centers B,C and D, The feeder to B is 2,300 ft.
J ft. >{v,iioo ft. »i ^°^8 aJid 2,000,000 CM. equivalent; that to C is
>--^y"-"-"*^ 1,800 ft. long and 2,600,000 CM. equivalent; that to
D is 2,000 ft. long and 2,000,000 CM. equivalent (per
wire in every case). A tie line 1,100 ft. long and of
500»000 CM. connects B and C and another similar
line connects C and £>. At £ is a load of 1000
amperes; at C a load of 500 amperes; and at D
Fi 133 A * ^°*^ ®^ ®^ amperes. Find the voltage at each

of the distributing centers B, C, and D,

QUESTIONS ON CHAPTER VI

1. What occurs if two copper strips be immersed in a dilute sulphuric
acid solution and a voltmeter connected between them? If the two copper
strips be replaced by two zinc strips? By two lead strips? Under what
conditions may a voltage between the strips be obtained?

Would a voltage exist if the sulphuric acid were replaced by some other
type of solution? Name three other such solutions?

2. What is meant by one metal being electrochemically positive to another?
If metal A is electrochemically positive to metal B, what will be the direc-
tion of the current flow between them within the cell? What will be the
direction of the current flow between them through the external circuit?

What is an electrode? What is the cathode? The anode?

3. In what form is the energy stored within the cell? What changes take
place in the electrodes when the cell delivers current? Distinguish between
a primary cell and a secondary cell.

4. What are the four requirements for a satisfactory primary cell?

6. What is the nature of the internal resistance of a cell ? In what manner
may this resistance be reduced? In what way does increasing the size of
the elements of a cell increase its current capacity? Its electromotive force?

6. What voltage does a voltmeter indicate when it is connected to the
terminals of a cell which is open-circuited? If the circuit is suddenly
closed, to what is the initial voltage drop due? To what is the excess
drop over this initial drop due? Explain the part* that hydrogen plays in
polarization. Describe two general methods of reducing polarization.

7. Describe the construction of the Daniell cell. What electrodes and
what electrolytes are used? For what type of work is it designed? What
is the electromotive force of this cell?

8. In what way does the gravity cell differ from the Daniell cell? Which
electrode requires replacing? What changes occur in the other electrode?

^ What is the cell electromotive force and for what type of work is the gravity
cell designed?

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QUESTIONS AND PROBLEMS 431

9. Describe the Edison-Lalaxide cell. What electrolyte and what elec-
trodes are used? In what way does its electrolyte differ from the cells
already described? What is the chief advantage of this type of cell?
What is its electromotive force and what is its terminal voltage when de-
livering a current?

10. What materials are used for the positive and for the negative elec-
trodes in the Le Clanch^ cell? What is the electrolyte? What is its elec-
tromotive force? When planning to use the cell commercially, what voltage
per cell should be allowed? What materials are introduced in the cell to
reduce polarization? How is the cell renewed? For what type of work is
this cell best suited?

11. What is the function of a Weston cell in distinction to the uses made
of other types of cells? In practice what two common electrical quantities
are most easily reproduced and maintained? What must be the character-
istics of a standard cell? How is the Weston cell constructed and how is its
permanency insured? In what way does the saturated cell differ from the
normal cell? Why Cannot the voltage of the Weston cell be measured with
an ordinary voltmeter?

12. In what way does a dry cell resemble a common type of wet cell?
Is a dry cell really "dry?" Of what is the positive electrode composed?
The negative? What is the electrolyte and how is it placed in the cell?
What materials are placed between the carbon and the zinc and what are
their functions?

13. What is the electromotive force of a dry cell when new? After it has
stood idle for some time? What is the magnitude of the internal resistance
when new and is it subject to change? How does the polarization effect
compare with the internal resistance effect? How much current should a
good cell deliver upon short circuit? What is the terminal voltage when a
cell delivers current?

14. To what cause is the cell's becoming exhausted principally due? Can
this cell be temporarily revived by any means? Name some of the commer-
cial applications of dry cells.

16. In what way is a storage cell renewed when it becomes discharged?
What condition concerning the materials of the cell is necessary for proper
functioning of the cell? What two general types of storage cells are in
commercial use?

16. Describe a very elementary experiment which illustrates the under-
lying principle of the lead cell. State the change that occurs in each of the
lead strips; what voltage is observed to exist at different times in the
experiment. What gases are evolved and from which plate does each
emanate?

17. Even although both of its plates are of lead, show that the existence
of an emf . in a lead storage cell does not in any way violate the principle
governing the emf. of electric cells in general. When the cell is approach-
ing discharge what change in the materials would account for the approach
of the voltage to zero? In what way is the 2.6 volts per cell utilized in the
process of charging?

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432 DIRECT CURRENTS

18. What change in the electrolyte during the charge and the discharge of
a cell is shown by the chemical equation? Why is a cell composed of
plain lead plates not useful in practice? Give two reasons. Describe
briefly the Plants process and describe two plates that are formed by this
process.

10. Describe the Faure or pasted process for making battery plates.
What are the advantages and the disadvantages of pasted plates over the
Plants plates? What commercial conditions demand a pasted plate and
why? How does the life of a pasted plate compare with that of a Plants
plate?

20. Describe briefly the construction of the "Iron-clad" Exide cell and
its principal use in practice.

21. What are the two general classes into which storage batteries may be
divided? What types of plate are best suited for regulating duty and for
emergency duty in stationary batteries? Why?

22. What two types of containing tanks are used for stationary batteries?
Under what conditions is each used and why? In what manner should the
joints and seams in lead-lined tanks be made non-leakable? How are the
plates suspended in the lead tank? What factors must be considered in
designing and installing a lead-lined wooden tank?

23. Whatthreetypesof separators are in general use? Name the advan-
tages and the disadvantages of each type. For what type of battery is each
kind commonly used? What one precaution must be taken in handling
wood separators? Why?

24. What should be the specific gravity of a fully-charged battery having
Plants plates? Pasted plates?. What precaution should be taken in dilut-
ing sulphuric acid for storage battery use? What simple device is used
for determining specific gravity? How is this device adapted for use with
vehicle and portable batteries?

26. What change takes place in the electrolyte during the charging period?
What is the effect of gassing on the specific gravity?. What change takes
place in the specific gravity after the charging has ceased? Explain. How
does the specific gravity of the electrolyte change during discharge? What
practical use is made of these changes of specific gravity?

26. When a battery is received, what special attention should be given to
the wood separators? In what manner should the jars be installed? How
should the plates be placed in position? Why is an initial charge necessary
and what should be its duration?

27. What happens to the active material in a cell if it is allowed to stand
idle over long periods? In what way may injury to the battery from this
cause be avoided? If it is desired to withdraw a battery from service for an
indefinite period, what procedure should be followed?

28. What are the requirements of a vehicle battery that make its design
different from that of a stationary battery? What changes are made in the
plates? Separators? Specific gravity of the electrolyte? How is a bat-
tery made up? In what way does a vehicle battery differ from a stationary

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QUESTIONS AND PROBLEMS 433

battery in the manner of shipment? What special attention should be
paid to the electrolyte?

29. In what manner is the rating of a storage battery expressed? What
is meant by the 8-hour rate? Can as many ampere-hours be extracted
from a cell at the 3-hour rate as at the 8-hour rate? To what is this differ-
ence due? If a cell is apparently exhausted after discharging at the 3-hour
rate, would it be possible later to extract any further current from it?
What can be said of the overload capacity of a storage battery?

30. What two general methods of charging are commonly employed?
In each method and with pasted plates what value of current should be em-
ployed when the charging commences? When does it become necessary to
reduce this current? What are the objections to pronounced gassing in a
cell? How does the charging rate with Plants plates differ from that with
pasted plates?

31. Name a very common example of constant-current method of charg-
ing. What care should be taken in the connecting up of the battery? De-
scribe a simple test by which the determination of the correct terminal
polarity may be ascertained.

32. What is the one great advantage of the constant-potential method of
charging? About what voltage per cell is necessary in this method?

33. When a battery is just floating on a bus-bar and it is desired to charge
it, in what manner may' the necessary excess potential for charging be ob-
tained? Does the generator employed supply the entire energy necessary
for charging?

34. What change occurs in the electromotive force of a cell during the
charging period? What corresponding changes occur in the terminal
voltage? To what is the discrepancy between the cell electromotive force
and the terminal voltage due? Can it be said that the voltage characteristic
of a storage battery is such that its use upon lighting circuits is practicable?

35. What is lost by a lead storage battery during its period of service?
With what should this loss be replaced except in rare instances? What cir-
cumstances justify the addition of acid to a cell? What care should be
taken in the selection of water for use with storage batteries?

36. In what manner can the freezing of the electrolyte in a storage battery
be absolutely prevented? How does a rise of temperature affect the rating
of a storage battery?

37. Compare roughly the kilowatts per pound of plate for a given cell at
different discharge rates. Repeat for kilowatts per pound of cell. Compare
the above factors for three different types of cell, stating the type of service
for which each type is best adapted.

38. Of what is the positive plate, the negative plate and the electrolyte
composed, in an Edison cell? In the chemical reaction that takes place both
on charge and on discharge, what part does the electrolyte play? How
does its specific gravity change during charge and discharge?

39. Describe briefly the. mechanical construction of the Edison cell,
stating the method of holding the plates and connecting them with the
binding posts. What kind of a tank is used for this cell? What is

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434 DIRECT CURRENT&

the advantage of this type of construction? For what puri>ose is the valve
necessary and what care does the valve require? How is the battery
mounted?

40. In what way does the normal rating of an Edison cell differ from that
of a lead cell? What is the voltage per cell? Is it possible to tell accurately
the condition of charge by readings of either voltage or of specific gravity?
How can complete charge be assured?

41. What should be used to replace evaporation of the electrolyte? Is
any greater care required in the selection of water for the Edison battery
than for the lead battery? Explain.

42. State the advantages of the Edison battery over other types of storage
batteries. What are some of the commercial applications of the battery and
what factors limit the applications of the battery? Compare the weights
per kw. with similar weights for the lead cell.

43. In what terms is the efficiency of a storage battery expressed? Is
the ampere-hour efficiency a true indicator of efficiency?

44. State the reason why the ratio of the kilowatt-hours of discharge at the
3-hour rate to those of charge at the 8-hour rate does not give the true
efficiency. Give some of the factors which determine the efficiency of a
battery.

45. What is the order of magnitude of the kilowatt-hour efficiency of
a lead storage battery? The ampere-hour? Why do the two differ? In
what manner does the cycle of operation of a storage battery affect the
efficiency?

46. What is the approximate kilowatt-hour and the ampere-hour efficiency
of an Edison battery?

47. State some of the factors which govern the selection of a storage
battery for any particular purpose.

48. State a simple method of producing copper plating upon a carbon
brush such as is used with generators. Which electrode is connected to the
positive terminal of the supply and which is connected to the negative ter-
minal? When copper is used in connection with a copper sulphate solution,
is there any marked change in the electrolyte? Explain.

49. Can copper be plated from a solution in which neither terminal is
copper? What voltages in the plating bath must the supply voltage over-
come? How are these voltages reduced to a minimum? Is electroplating
considered a high voltage or a low voltage process? In what way are plating
baths connected, when possible?

60. Show how the gravity cell is an electroplating bath which supplies
its own electroplating current.

51. Describe briefly the process of electrotyping.

PROBLEMS ON CHAPTER VI
134. A Daniell cell has an electromotive force of 1.07 volts and an internal
resistance of 0.2 ohm. (a) What is the maximum current which it can
deliver? The size of the cell is increased in such a manner that the plate
area is doubled. (6) What is the new electromotive force? (c) What is
the approximate maximum current that the cell can now deliver V

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QUESTIONS AND PROBLEMS

435

135. Two gravity cells have electrodes of the same materials and
solutions of the same kind, concentration, etc., but one cell has each linear
dimension twice that of the other, making its volume eight times greater.
The two cells are connected with terminals of like polarity together. If the
electromotive force of the smaller cell is 1.0 volt and its resistance 0.4 ohm,
how much current flows between the two cells? Give reasons for the answer.
If the plate area of the larger cell is eight times that of the smaller, what will
be its short-circuit current, approximately?

136. A Le Clanch^ cell has an electromotive force of 1.43 volts on open
circuit. A load of 2 amp. is suddenly applied and the terminal voltage
drops to 1.25 volts almost instantly. After a lapse of some time it drops to
1.06 volts. What is the actual internal resistance of the cell and what is the
** electromotive force of polarization"? What is the total apparent cell
resistance?

137. A telegraph relay has a resistance of 150 ohms and the loop resist-
ance of the sending circuit is 1,600 ohms. The relay requires 50 milli-
amperes for satisfactory operation. How many gravity cells, each having
an electromotive force of 1.09 volts and an internal resistance of 0.4 ohm, are
required at the sending end? How should they be connected?

138. A railway signal circuit consists of a 4-ohm relay, a track and connect-
ing resistance of 3 ohms. The relay requires 75 milliamperes to attract its
armature satisfactorily. Two gravity cells, each, having an electromotive
force of 1.05 volts and a resistance of 0.3 ohm, are used to operate this signal
relay. What extra resistance in series with the battery is required?

139. A certain signal motor requires 4 amp. at 10 volts at its terminals
for satisfactory operation. The leads from the battery to the motor have a
total resistance of 1.4 ohms. How many Edison-Lalande cells would be
necessary to operate this system? (See Par. 87.)

140. Three Le Clanch6 cells, connected in series, are used to operate a door
opener which has a resistance of 1 ohm. The resistance of the connect-
ing wires is about 0.5 ohm. What is the approximate current taken by
the door opener? (See Par. 88.)

0.91511

Std.cell

10 Ootls

Galv.

■e-

-E=1.01S3

R-mQ

FiQ. 141A.

141. A Weston cell having an electromotive force of 1.0183 volts is con-
nected to a potentiometer wire as shown in Fig. 141 A, in order to calibrate
the wire AC, Between A and B is a resistance of 0.915 ohm and 10 coils

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436 DIRECT CURRENTS

having a resistance of 5 ohnis each. The cell has a resistance of 180 ohms
and the galvanometer a resistance of 100 ohms. When the current in ^C is
0.021 amp., how much current passes through the galvanometer?

For what value of current in the wire AC will the current through the cell
and galvanometer be zero? Under these conditions what will be the voltage
across each of the 5-ohm coils?

142. A voltmeter having a resistance of 1,000 ohms is used in an attempt
to measure the electromotive force of the Weston cell in problem 141.
What will the voltmeter read? Is this a practicable method of using the
Weston cell as a standard?

143. The ignition system on an automobile requires 6 volts for satis-
factory operation. How many dry cells should be recommended for this
purpose?

144. A dry cell shows an open-circuit emf, of 1.2 volts and a short-circuit
current of 4 amp. What is its internal resistance? What does this
test show as regards the condition of this cell?

146. A certain flashlight has a 2 candle-power lamp whose efficiency is
1.2 watts per candle. If this lamp is operated by a single dry cell, in first
class condition, approximately what current does it take and what is its
voltage?

146. A storage cell is being charged at the normal rate as indicated by an
ammeter. A voltmeter across its terminals indicates 2.2 volts. At what
part of the charging period is the cell operating at this time? (See Fig.
104.)

147. A storage cell has an 8-hour rate of 40 amp. This rate is main-
tained constant for the 8 hours of charge. During this period the voltage
rises according to the curve shown in Fig. 104, page 113. How many
ampere-hours are delivered to the cell? How many watt-hours? (Note:
Mark several equally spaced points on the voltage curve and take their
average.)

148. If the cell of problem 147 discharges at the 8-hour rate and its volt-
age follows the 8-hour discharge curve of Fig. 104, page 113, how many
watt-hours are discharged?

149. It is desired to dilute a quart of concentrated sulphuric acid, sp. gr.
. = 1.84, to make acid having a specific gravity of 1.240. How much water
is needed and what is the total volume of acid when the solution is mixed?
State the procedure that should be followed in mixing the liquids.

160. A gallon of water weighs 8.3 lb. How much will 5 gallons of battery
acid (sp. gr. = 1.210) weigh?

161. What is the percentage by weight of acid (sp. gr. = 1.84) in the acid
solution of problem 150?

162. The hydrometer in a pilot cell of a stationary battery indicates a
specific gravity of 1.190. How many more hours should the battery be
left charging (Fig. 98, page 106)?

163. A hydrometer test of the electrolyte in a vehicle cell shows the specific
gravity to be 1. 185. If this cell is one of an electric vehicle battery engaged
in propellirig a vehicle, how near complete discharge is the battery?

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QUESTIONS AND PROBLEMS 437

164. A battery having a normal rating of 40 amp. at the 8-hour rate is
just received new. How many ampere-hours charge should be ^iven it
before it is ready for active service.

166. The average charging voltage per cell in problem 154 is 2.B volts.
There are 40 cells in series and a 0.5-ohm resistance in series with the battery.
At 5 cents per kw.-hr., what is the energy cost of getting the battery ready
for service?

166. If the specific gravity of a vehicle battery is found to be 1.200) what
would be a fair estimate of its condition of charge?

167. A battery is charged at the 80-amp. rate for 6 hours. How many
ampere-hours has it absorbed? If the ampere-hour efficiency is 95 per cent.,
for how many hours can it discharge 60 amp.?

168. If the battery of problem 157 is of the pasted plate type, what current
will it discharge at the 3-hour rate? How many ampere-hours does it
discharge at this rate? (See Par. 101.)

169. What current and how many ampere-hours will the battery of prob-
lem. 157 deliver at the 1-hour rate?

160. A battery has a rating of 320 ampere-hours. At what value of
current should the charging be started if the plates are of the pasted type?
Of the Plants type? (See Pars. 101 and 102.).

161. It is desired to charge a starting battery from llO^volt d.c. mains.
The battery consists of three cells each having a terminal voltage of 2.5 volts
when being charged at the normal rate of 12 amp. How much resist-
ance must be inserted in series with this battery? What percentage of the
power supplied is delivered to the battery?

162. If two batteries each similar to that of problem 161 are being charged
in series at the same rate, what series resistance is necessary? What per-
centage of the power supplied is delivered to the batteries?

163. A storage battery of 115 cells is floating on 230-volt bus-bars. It is
desired that the battery begin to discharge when the bus- bar voltage is
exactly 230 volts. On charge it is necessary to have 2.4 volts per cell.
What capacity of booster is required if the normal charging current is 60
amp.? How much power is delivered to the battery by the booster? How
much is supplied directly by the bus-bars?

164. If the booster generator of problem 163 has an efficiency of 78 per
cent, and the shunt motor which drives it has an efficiency of 80 per cent.,
what power does the booster set take from the bus-bars?

166. A storage battery of 50 cells has a total internal resistance of 0.6
ohm and is charged from 115- volt d.c. mains. At the beginning of charge
its electromotive force is 1.8 volts per cell, (a) What current does it take?
After charging 4 hours the electromotive force rises to 2.0 volts per cell,
(6) What current does it take at this time? (c) What must be the electro-
motive force per cell when the battery ceases taking current? What method
of charging is used and is this method a desirable one?

166. The specific gravity in a vehicle battery is found to be 1.240. Is
there any possibility of its freezing in the climate of the United States?
Give reasons.

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438 DIRECT CURRENTS

167. It IB desired to install a 110-cell stationary battery having a capacity
of 1.22 kw. per cell at the 4-hour rate, (a) What will be the approxi-
mate weight of the plates of this battery? (6) Of the total battery? (See
Par. 106.)

168. What will be the weight of a 24-cell vehicle battery composed of
** iron-clad" cells, this battery to have a total output of 1.28 kw. at the
8-hour rate?

169. Approximately how many Edison cells would be required for a 24-
volt lighting project?

170. It is desired to install a generator to charge a 60-cell Edison battery.
The normal charging rate is 20 amp. What size generator is necessary
(kilowatts, amperes, volts)? (See Fig. 108, page 117.)

171. What will be the weight in pounds, of a 50-cell Edison battery
designed to deliver 15 kw.-hr. at the 8-hour rate? (See Par. 108.)

172. What will be the weight in pounds, per kilowatt of this battery?

173. A lead cell is charged at a 40-amp. rate for 10 hours with an average
potential difference across its terminals of 2.3 volts. It discharges 45
amp. for 8H hours at an average terminal voltage of 1.95 volts. What
is its ampere-hour efficiency? What is its watt-hour eflSciency?

174. A storage battery in its discharged condition is charged at the 100-
amp. rate at an average voltage of 250 volts for 9 hours. It delivers 105
amp. at an average terminal voltage of 220 volts for 8 hours before it is
again in the discharged condition. What is its kilowatt-hour efficiency?

176. An Edison battery of 12 cells is charged for a period of 6 hours at the
25-amp. rate and the average terminal voltage per cell is 1.65 volts. The
battery discharges 5 hours at the 28-amp. rate with an average terminal
voltage of 1.2 volts. What is its ampere-hour and what is its watt-hour
efficiency?

176. One ampere-hour will deposit 0.843 gram of copper upon the cathode
in an electroplating bath. If the voltage across a plating bath is 12 volts
and the current is 12 amp. and the current is allowed to flow for 6 hours,
how many kilograms of copper are deposited and how many kilowatt-hours
are utilized in the process?

QUESTIONS ON CHAPTER VH

1. If a coil carrying a current be placed in a magnetic field, what efTect
is noticed? Give two explanations of this effect Of what importance is
this principle?

2. How is the principle of the moving coil adapted to measuring small
currents in the D'Arsonval galvanometer? How is the coil suspended?
How is the current led in and out of the coil? Why is a soft-iron core
placed between the poles?

3. What two common methods are used to read the galvanometer de-
flection? What is meant by the ** damping" of a galvanometer? How
may this damping be accomplished?

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QUESTIONS AND PROBLEMS 439

4. How may a galvanometer be protected from excessive currents?
Sketch the connections of two types of shunt. What are the advantages of
the Ayrton shunt?

6. What was the underlying principle of the early types of electrical
instruments? What two factors caused these instruments to be inaccurate?

6. Show that the movement of a Weston d.c. instrument is an evolution
of the D'Arsonval galvanometer. How is the moving coil pivoted? How is
the current led to the coil? What means are used to oppose the motion of
the coil? Is the coil damped? Explain. What is meant by a "radial
field " and what effect does it have-on the calibration of the instrument s cale?
WTiy are the top and the bottom springs coiled in opposite directions? Is it
possible to utiUze the movement of a Weston instrument as a galvanometer?

7. Of what order of magnitude is the current that will give full-scale
deflection in a Weston instrument? Is it possible to use the instrument for
measuring current in excess of this value? Explain.

8. Describe briefly the construction of a shunt. Why are four posts or
terminals necessary? Show that when a Weston instrument is used in
connection with a shunt, it is acting as a voltmeter.

9. What law does the current follow in dividing between the shunt and
the instrument? Why should the resistance of the ahunt and the resistance
of the instrument remain constant? What errors may be caused by the
heating of the shunt or of the instrument?

10. In what way may an ammeter be made to have several scales? In
general, when is an internal shunt used? An external shunt?

11. Does the movement of a voltmeter differ materially from that of an
ammeter? In what important respect does the voltmeter differ from the
ammeter? How is the current in the coil of a voltmeter limited when the
voltmeter is connected across the line?

12. Is it possible for a voltmeter to have more than one scale? Explain.
What is meant by a multiplier or extension coil?

13. In what manner may the heating effect of an electric current be util-
ized to measure the value of the current? State some of the advantages
and the disadvantages of hot wire instruments.

14. Show the connections that are used in measuring resistance with a
voltmeter and an ammeter. What precaution should be taken in connecting
the voltmeter? • What special type of voltmeter contact should be used in
measuring very low resistances?

16. Show the connections that can be used in measuring resistance by a
voltmeter alone. What is the order of magnitude of resistances that can
be measured by this method? What special type of voltmeter is often de-
sirable for this work and why? To what type of resistances is this method
especially applicable?

16. Sketch an arrangement of four resistances, a battery and a galvanom-
eter, whereby one of the resistances may be measured. How is the condi-
tion of "balance" in the bridge detected? Prove the law of proportionality
that exists when this condition of balance has been reached.

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440 DIRECT CURRENTS

17. What two types of bridge are in general use? Compare them
from the standpoint of ease of manipulation; plug-contact resistance;
convenience.

18. Give briefly the procedure which should be followed in obtaining a
balance with a plug bridge.

19. In what way does the slide wire bridge resemble the Wheatstone
bridge? Compare it with the Wheatstone bridge from the standpoint of
simplicity and accuracy.

20. Give the connections whereby the shde wire bridge may be put to
practical use in locating an earth fault in a cable. What is the name of this
method? Explain why the galvanometer and battery do not occupy the
same positions in the slide wire bridge of Fig. 133 as they do in Fig. 132.

21. Sketch the connections used in the Varley loop. Upon what arm is
the balance obtained? What additional factor must be known before the
position of the fault can be determined? Was it necessary to know this
factor in the Murray loop test? Which is the simpler method? What
possible sources of error exist?

22. Why is it desirable in practice to know the insulation resistance of
cables? Why is the voltmeter method not always practicable? What is
the general principle of the method described in Par. 124?

23. What method is used to obtain readable deflections of the galva-
nometer under all conditions of circuit resistance? Why is it desirable to
keep the 0.1 megohm in circuit continually and does it introduce any appre-
ciable error?

24. What other factor besides the resistance of the insulation affects the
value of the current flowing in the circuit? What time of electrification
has been adopted as standard in commercial measurements of insulation
resistance?. What precautions should be observed in the installation of
cable testing apparatus?

25. Upon what standard do potentiometer measurements primarily rest?
Against what is the standard cell balanced? What care as regards polarity
must be observed if a balance is to be obtained ? Why is a " nul ' ' method the
only one which will give satisfactory results when a standard cell is used?

26. Show how a wire may be calibrated and marked in volts, after the
standard cell balance has been obtained. Is it possible to measure other
electromotive forces with this standardized wire? What method is em-
ployed in such measurements?

27. Does the Leeds & Northrup potentiometer whose connections are
shown in Fig. 138 differ materially from the simple device sketched in Fig.
137? What minor changes are necessary? Where are the one-tenth volt
divisions located and how are they utilized when obtaining a balance? How
are the smaller decimal divisions obtained? What resistances are used in
each of these units? What is the working current of this potentiometer?

28. What provision is made for the variations in the voltages among
standard cells? What protection is aifforded the galvanometer during the
preliminary adjustments?

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QUESTIONS AND PROBLEMS 441

29. What is the maximum voltage measurable with this potentiometer
alone? By what means can voltages in excess of this be accurately mea-
sured? Is the device used for increasing the voltage range of the potenti-
ometer in any way complicated ? What is meant by a " drop wire " and how
may it be \ised to vary the voltage when the supply is at constant volt ate?

30. Is this potentiometer, as a voltage-measuring device, adapted to
measuring currents? What is the p.^ncple underlying the measurement of
current? What is a standard resistance? Why does it have four posts?
In what units of resistance are standard resistances generally manufactured?
Why is it desirable that their temperature remain normal and what means
are adopted to accomplish this?

31. What instruments are generally used in measuring the power in a
direct-current circuit? Do these instruments take any power themselves?
What should be the relative positions of the voltmeter and the ammeter
when the power delivered to a high resistance is being measured? When
that delivered to a low resistance is being measured?

32. Describe a wattmeter. In what way do the fixed and moving coils
differ in construction? In their manner of connection to the circuit? Why
are the instrument deflections a function of the power? What care is
necessary when using this type of instrument with direct currents?

33. What does a watthour meter measure? Upon what familiar electrical
device is it based? From what source are its field coils supplied? Its
armature? To what is the torquie acting upon the armature proportional?

34. Why is a retarding device necessary and what must be the law of
retardation? Upon what principle does this device operate?

36. At what values of meter load does friction produce the greatest error?
Explain. How is this friction error practically eliminated?

36. What methods are used to reduce friction in a watthour meter?
What are some of the causes of a meter running slow? How is the recording
dial of a meter actuated?

37. Why is it usually very important that a watthour meter register
accurately? What load and measuring devices are necessary in testing a
meter?

38. What is the fundamental relation between the revolutions of the disc
and the energy registered by the meter? What measurements are made in
checking the meter?

39. What two adjustments are made to change the meter speed? What
is the effect of moving the magnets nearer the center of the disc? Nearer
the periphery? At what loads is this adjustment made?

40. What adjustment is made to correct the meter registration at light
loads? Why is this adjustment made at light rather than at heavy
loads?

41. In what general respect does a three-wire meter differ from a two-wire
meter?

42. Describe in a general way the construction of a meter which makes the
meter practically astatic and therefore enables it to be used near bus-bars

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442

DIRECT CURRENTS

2000A

^ iA/WW\^^VVV\A'VVVVAVVWVVV%

carrying heavy currents. What two elements in a meter are most likely
to be affected by stray fields? How are these elements safeguarded from
these effects?

PROBLEMS ON CHAPTER VH

177. A galvanometer has a resistance of 351 ohms. What should be the
resistance of a shunt for use with this galvanometer if it be desired that
Ko the total current of the line pass through the galvanometer? If it
be desired that Koo of the line current pass through the galvanometer?

178. The resistance of a certain galvanometer is 495 ohms. Design a
shunt which will allow Ho» Koo, and Hooo the line current to pass through
the galvanometer.

179. An Ayrton shunt (Fig. 179A) has a resistance from A to B of 10,000

ohms. It is used to shunt a gal-
vanometer having a resistance of
2,000 ohms. When the shunt is
set at the 0.001 point (the re-
sistance AC = 10 ohms), de-
termine the current through the
galvanometer when 1 milliampere
flows in the line.

180. If the line contact be
moved to the 0.01 point at D (Fig.
179o), the resistance AD being
100 ohms, determine the current
through the galvanometer when
the line current is 1 milliampere.

181. Repeat problem ISO when the line contact is moved toB, Compare
this result with those of problems 179 and 180. If the shunt were removed
what current would now pass through the galvanometer? How much does
the shunt reduce the ultimate sensitivity of the galvanometer?

182. A 50-scale millivoltmeter has a resistance of 2 ohms. It is desired
that it measure a current of 75 amp. at full-scale deflection. What should
be the resistance of the shunt under these conditions? How much current
flows through the instrument and can it be neglected as compared with the
current in the shunt?

183. Find the resistances of shunts necessary for measuring currents of
150 amp. and 500 amp., fuUnscale deflection, with the instrument of problem
182.

184. An instrument has a resistance of 25 ohms. It is used to measure a
current of 60 amp. The shunt has a resistance of 0.00075 ohm. How
much current passes through the instrument? Through the shunt? What
is the rating of the instrument in millivolts?

185. It is desired to measure a current of 50 amp. An internal shunt,
5-scale ammeter alone is available. This instrument has a resistance of 0.01
ohm. What should be the resistance of a shunt to be used with this
instrument?

Line
1 HiUiamperv-

FlO. 179A.

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QUESTIONS AND PROBLEMS 443

186. A 15-scale voltmeter has a resistance of 160 ohms. What should be
the resistance of a multiplier to increase the range of this instrument to 150
volts?

187. A 160- 15-scale voltmeter has a total resistance of 17,500 ohms.
What is the resistance between its 15-volt scale binding posts? What
multiplier resistance will give this instrument a range of 600 volts?

188. It is desired to measure the potential difference between a trolley
and ground. A 40,000-ohm and a 10,000-ohm resistance are connected in
series between the trolley and the ground. Across the 10,000-ohm resistance
a 50-8cale voltmeter having a resistance of 5,100 ohms is connected. When
this instrument reads 45 volts, what is the trolley voltage?

189. What multiplier resistance would have been necessary in problem
188 to have obtained the same multiplying power for the voltmeter?

190. A 100-watt lamp when connected across d.c. mains is observed to take
0.9 amp. at 115 volts. What is its hot resistance?

191. When the armature of a 220-volt, 10-hp. motor is stationary, a current
of 40 amp. gives a voltage drop across its terminals of 8 volts. What is
the resistance of the armature?

192. The current in problem 191 was taken from d.c. mains whose poten-
tial difference was known to be 115 volts. What resistance was connected
in series with the armature?

193. The resistance of a sample of copper bus-bars is measured by the
method shown in Fig. 126. When the ammeter reads 140 amp., the milli-
voltmeter reads 3.5 millivolts. The bus-bar is 0.5 in. by 2 in. in cross-sec-
tion and the distance between voltmeter contacts is 3 ft. (a) What is the
resistance of the sample? (b) What is its resistance per cm. cube? (c)
What is its per cent, conductivity? (See Par. 39, Chap. III.) Res. of
standard copper = 1.724 microhm-centimeters.

194. It is desired to obtain the resistance of a 70-lb. rail. A current of
350 amp. is sent through the rail and a millivoltmeter is connected between
two contact points on the rail spaced 6 ft. apart. The millivoltmeter reads
5.1 millivolts. What is the re- nsvoiti
sistance per ft. of the rail?

195. A 300-scale voltmeter hav-
ing a resistance of 35,000 ohms
is connected across d.c. mains
and indicates 225 volts. It is
then connected in series with an
unknown resistance across these
same mains. It now indicates 48
volts. What is the value of the
unknown resistance?

196. A special 150-scale 100,000-
ohm voltmeter, when connected across d.c. mains, reads 115 volts. The
iron frame of a generator is connected to one wire of these mains and the
copper of the field coil is connected to the other wire through the voltmeter,
as shown in Fig. 196A. Under these conditions the voltmeter reads 8 volts.

Fig. 196A.

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444

DIRECT CURRENTS

What is the resistance in megohms of the insulation of the field circuit to
the frame of the machine?

197. In a Wheatstone bridge measurement the unknown resistance is
connected at X between one end of the arm M and the arm P. (See Fig.
128, page 141.) When a balance is obtained Af = 10 ohms; N = 1,000
ohms; P — 1,426 ohms. What is the value of the unknown resistance?

198. A resistance whose value is known to be between 10 and 20 ohms is
connected to a Wheatstone bridge at X as shown in Fig. 128, page 141.
What are the best values of M and N to use? If the unknown resistance is
16.72 ohms, what will P read when a balance is obtained?

199. An unknown resistance is measured by means of a 100-cm. slide
wire bridge. A known resistance of 100 ohms is inserted at the lOO^m.
end of the bridge. (See Fig. 132, page 146.) A balance is obtained when
the slider reads 32.4 cm. What is the value of the unknown resistance?

200. If a 10-ohm resistance be used as the known resistance in problem
199, what will be the reading on the slide wire when a balance is obtained?

201. A cable 1,200 ft. long, wound on a reel, is known to have a fault in its
insulation. It is immersed in a tank of water and the Murray loop test is
used to locate the fault. The slide wire bridge, 100 cm. long, reads 18.4
cm. when the balance is obtained. What is the distance from one end of the
cable to the fault?

202. An installed two-conductor cable of 4/0 copper is 3,200 ft. long.
Due to a bum-out both conductors are short-circuited and grounded at the
same point. To locate the fault a single 00 conductor of another cable which
parallels the faulty one is looped to one conductor of the faulty cable at the

"^L-qr

00 Cables,

x-^'-vI

"^ Fault
St- 0000 Condi

Fig. 202^.

far end, as shown in Fig. 202^4. The perfect conductor is connected to the
low reading end of the slide wire and one of the faulty conductors to the 100-
cm. end. A balance is obtained at 89.4 cm. How far out on the faulty
conductor is the bum-out located?

203. In a Varley loop test for a fault in a 1/0 conductor, 3,600 ft. long,
this conductor is looped back through a perfect 00 conductor. The ratio
arms are M = 10 and N = 1,000 ohms. (See Fig. 134, page 149.) A bal-
ance is obtained when P = 58.5 ohms. The bridge measurement of the
entire loop shows its resistance to be 0.70 ohm. How far out is the fault?

204. One conductor in a cable containing two No. 14 wires a and 6,
each 8,000 ft. long, is known to be grounded. The two are looped at the

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QUESTIONS AND PROBLEMS

445

far end and the Varley loop test made. P is connected in series with con-
ductor o. The two arms M and N (Fig. 134, page 149) are each set at 100
ohms. A balance cannot be obtained with P, as the galvanometer is found
to deflect the same way with P = 0 and P = » . P is then shifted over in
series with 6, the other conductor, and a balance obtained when P = 12.6
ohms. In which conductor is the fault and how far is it from the home end
of the cable?

206. In an insulation test of a cable the connections are made as in Fig.
135, page 150. When the cable is short-circuited and the A3ni»n shunt is
set at 0.0001, the galvanometer deflects 12.8 cm. The short circuit is then
removed, putting the cable in circuit, and the galvanometer deflects 19.8
cm. with the shunt set at 1.0 after the cable has been charged for 1 minute.
What is the insulation resistance of the cable?

206. The cable in problem 205 is 1,100 ft. long. What is its insulation
resistance in megohms per mile?

207. It is desired to measure the terminal voltage of a storage battery by
means of a standard cell. The ratio and rheostat arms of a Wheatstone
bridge (Fig. 207A) are connected across the terminals of the storage battery
and a standard cell having an electromotive force of 1.0176 volts is connected
across a 1 , 000-ohm coil . The galvanometer in the standard cell circuit stands
at zero when 1,050 additional ohms are unplugged in P. What is the ter-
minal voltage of the storage battery?

■ Storage

~ Batter7

20-0.

a
FiQ. 207^.

2266-0.

D.C. Snpplj

J-

Fia. 209^.

208. The storage battery of problem 207 is of such comparatively large
capacity that its electromotive force and terminal voltage are sensibly
the same when delivering the small current required by the resistance of
2,050 ohms. To measure the electromotive force of another cell which is
not capable of delivering any appreciable current, its negative terminal is
connected to the point a and its positive terminal to the point h through a
key and galvanometer (Fig.. 207 A). P is then adjusted until this galva-
nometer reads zero. P is then read and found to be 890 ohms. What is
the electromotive force of this cell?

209. It is desired to calibrate a voltmeter at the 115-volt point. No
potentiometer is available. The voltmeter is connected in parallel with the
arms of a bridge box (Fig. 209 A) and 115 volts is impressed upon this
circuit. A standard cell, which is known to have an electromotive force of
1.0180 volts, is connected across the two ratio arms in series with a key and

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446

DIRECT CURRENTS

galvanometer, the proper polarity being observed and 20 ohms are unplugged
in these two arms. The galvanometer reads zero with the key depressed when
2,266 ohms are unplugged in P. What correction should be applied to the
voltmeter at this point?

210. The power to a 25>watt tungsten lamp is being measured with a volt-
meter and an ammeter. The voltmeter, which has a resistance of 12,000
ohms, is connected directly across the lamp terminals. When the ammeter
reads 0,23 amp. the voltmeter reads 117 volts. What is the true power
taken by the lamp? What per cent, error is introduced if the instrument
power be neglected?

211. In measuring the power taken by a low resistance rheostat, an am-
meter having a resistance of 0.0008 ohm and a voltmeter having a resistance of
120 ohms are used. When the ammeter reads 70 amp. the voltmeter,
which is connected directly across the resistance, reads 2. 1 volts. What is
the true value of the resistance? What per cent, error is introduced by
the voltmeter current? What error is introduced by connecting the volt-
meter outside the ammeter?

212. In a test of a direct- current watthour meter the average voltmeter
reading is 118 volts and the average ammeter reading is 21.4 amp. Thirty
revolutions are counted and the time is found to be 42.6 seconds. If the
meter constant is 1.0, what is the per cent, accuracy of the meter at this load?
What adjustment should be made to bring it nearer the correct registration?

213. The meter load (problem 212) is dropped to 1.0 amp. but the
voltage is still 118 volts. It takes 62.6 seconds for the disc to make two
revolutions. What is the per cent, accuracy of the meter at this point?
What adjustment should be made in order to bring it nearer the correct
registration?

232 V.

T T T

-ffi

'^"

h-a

A/yv

Fig. 214^.

214. In order to make a laboratory test of a 2,000-amp., 220- volt,
astatic watthour meter, of the type shown in Fig. 148, page 167, its current
coils are supplied with current from a 4-volt storage battery and its armature
circuit, which has a resistance of 2,200 ohms, is connected across the 232-
volt mains, as shown in Fig. 214A. A calibrated voltmeter is connected in

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QUESTIONS AND PROBLEMS 447

parallel with the armature circuit and an external shunt ammeter is con-
nected in series with the current terminals of the meter. The meter con-
stant is 150. The corrected voltmeter reading is 232 volts and the ammeter
reads 1,980 amp. The meter makes 40 revolutions in 45.8 seconds. What
is the per cent, accuracy of the meter at this load? How much power is
required for this test? How much power would be required if the meter
current were supplied at 232 volts?

QUESTIONS ON CHAPTER VHI

1. In what way does the magnetic circuit resemble the electric circuit?
In what two ways do they differ from each other? Why cannot magnetic
flux be readily confined to definite paths? In a general way how does the
precision obtainable in magnetic calculations compare with that obtainable
in electrical calculations?

2. Are ampere-turns dependent on current alone? On turns alone?
What is the numerical relation between magnetomotive force and ampere-
turns? Which is the larger unit, the gilbert or the ampere-turn? To what
quantity in the electric circuit does magnetomotive force correspond?

What is reluctance and to what does it correspond in the electric circuit?
What is the basic unit of reluctance? How is permeance related to reluc-
tance and to conductance?

What is meant by permeability?

To what quantity in the electric circuit does magnetic flux correspond?

3. How is the reluctance of a magnetic path related to its length? To
its cross-section? To its permeability? How are reluctances in series
combined? In parallel?

4. Why is it usually necessary to represent the relation between magnet-
izing force, and flux, in iron and steel, by curves? What general shape does
the lower part of such curves have? The upper part? What is meant by
saturation? How may a permeability curve be obtained from a B-H curve?
How does the variation of permeability compare with such variation of
electric resistance as is due to heating?

5. State the simple law governing the relation between flux, mmf. and
reluctance. To what law in the electric circuit does this law correspond ?

6. Why is a method of trial and error sometimes necessary in solving
magnetic problems?

7. Upon what three factors is the magnetomotive force acting upon a
circuit dependent? How may the 0.4ir be eliminated from computations
in centimeter units? In inch units? How are magnetization curves plotted
in order to reduce computations to the simplest basis?

8. If the magnetomotive force acting upon a sample of iron be increased
from zero to some definite value and then decreased again to zero and the
relation between magnetomotive force and flux be plotted, does the curve
for increasing values of magnetomotive force differ from that for decreasing
values? If the excitation be decreased to zero, does the magnetic flux return
to zero? How may the magnetic flux be brought back to zero? What is a
cycle? A hysteresis loop? Remanence? Coercive force? What does
hysteresis represent in terms of energy? Digitized by vjOOglC

448 DIRECT CURRENTS

9. How is the hysteresis loss related to the loop area? How may the
loss be calculated under practical conditions? How is the loss related to
the maximum flux density? What is the Steinmetz Law?

10. How is the geometrical position of the lines of induction related to the
current in a circuit? Does this relation suggest the term ** linkages? "
How may these linkages be calculated? What relation does inductance
bear to the total linkages?

11. Is it possible to produce an electromotive force in a circuit which is
insulated from everything else? How, in a general way, is this electro-
motive force produced?

If an induced current is allowed to flow in a coil, what reaction will exist
between this current and the inducing agent? If the inducing agent as,
for example, a bar magnet, be withdrawn from a coil, will the induced electro-
motive force have the same direction as when the bar magnet was inserted
in the coil? What reaction will be produced between the induced current
and the inducing agent?

12. Upon what two factors does induced electromotive force depend?
Is it possible to determine the value of this electromotive force in volts if
these factors are known?

What is Lenz's Law?

13. If the flux linking a coil be made to change by altering the value of
the current in the coil itself, show that an electromotive force is induced.
What is the relation of this electromotive force to the direction of the current
flowing in the coil? How does this relation affect the rapidity with which
the current builds up to its Ohm's Law value?

14. What is the "time constant" of a circuit and by what two quantities
may it be expressed? In a general way, what does it indicate as regards
the circuit? Does the time lag of current in a circuit have any practical
importance?

16. If an inductive circuit carrying a current be short-circuited, why does
not the current die out immediately? To what is this tendency of the cur-
rent to persist due?

What is the nature of inductance as regards its effect upon circuit changes?
To what mechanical property does it correspond?

How does the effect of inductance manifest itself when the current of a
circuit is interrupted? How can it be shown that this condition is not
produced by the current alone? To what is this arc due? Under what con-
ditions in practice may it become a menace? How may this menace be
partially or wholly removed?

What personal dangers may result from opening inductive circuits?

16. Upon what three factors does the electromotive force of self induction
depend?

Does the establishment of a magnetic flux require an expenditure of
energy? Is energy expended in maintaining this flux after it is once estab-
lished? What becomes of the power required by electromagnet field coils?
Give examples of electromagnetic energy manifesting itself.

Is it possible to calculate thi.- energy? Upon what two factors does it

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QUESTIONS AND PROBLEMS

449

depend? How may the energy of generator fields be very materially re-
duced, before opening the circuit?

17. How does the gas-lighting spark coil utilize the electromotive force of
self induction in its operation? How is it connected in the circuit? Show
that the spark coil can be considered as a reservoir in which magnetic energy
is stored and later liberated. Explain why the coil produces a hot spark.

18. Is it possible for a magnetic flux produced by one coil to induce an
electromotive force in another coil from which the first is insulated? Does
this in any way correspond to the production of electromotive force by the
insertion of a bar magnet in the second coil? What is the relation of the
direction of the induced voltage in the secondary when the primary circuit
is closed to its direction when the primary circuit is opened? Upon what
three factors does this electromotive force depend?

19. Is it possible for all the flux produced by one coil to hnk another?
What is the definition of the " coefficient of coupling " of two circuits?

How is mutual inductance defined? How may it be utilized to determine
the induced voltage?

How may the mutual inductance of two circuits be materially increased?

20. Explain how the action of the induction coil depends upon mutual
induction? How is the primary current interrupted and why is it necessary
that this current be interrupted?

21. Upon what two factors does the pull between magnetized surfaces
depend? How does this pull vary with the flux density?

PROBLEMS ON CHAPTER Vm

215. A certain electromagnet has two exciting coils, each of which has
2,200 turns, (o) When these two coils are connected in series, what is the
total niunber of ampere-turns acting on the magnet if 3 amp. are supplied
from the line? (6) If the coils are connected in parallel and the total current
supplied is 3 amp., what is the number of ampere-turns?

216. If one of the coils in problem 215 has a resistance of 80 ohms and the
other a resistance of 60 ohms, what is the line voltage in (a)? What is the
line voltage in (6) and what are the ampere-
turns per coil and the total ampere-turns?

217. A certain exciting coil has 1,400 turns
and has a resistance of 160 ohms. What are
the ampere-turns when this coil is connected
across 120-volt mains? Another coil in every
way similar to this one is placed on the same
magnetic circuit and connected in series with
it across the same 120-volt mains. What am-
pere-turns now act on the magnetic circuit?

218. What is the magnetomotive force in gilberts in problem 215 (a) and
(&)? In problem 217?

219. In a certain iron-clad solenoid, Fig. 219A, the reluctance of the yoke
is negligible compared with that of the plunger. When the plunger is
inserted the lines of induction passing through the central core are observed

Fig. 219A.

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450

DIRECT CURRENTS

to increase from 350 to 52,000. What is the permeability of the plunger
at this flux density?

220. A steel field core of a dynamo is 4 in. in diameter and carries a mag-
netic flux of 1,280,000 lines. What is the flux density in lines per sq. in.
and per sq. cm.?

221. A magnet plunger, of circular cross-section and 1.5 in. diameter,
carries a flux of 200,000 lines. What is the flux density in lines per sq. in.
and per sq. cm.?

Fig. 226A.

At
re-

222. The field core of a dynamo is 3 in. long and 4 in. in diameter.
80,000 lines per sq. in. it has a permeability of 700. What is the
luctahce between opposite ends of this field core at this flux density?

223. The two iron pole pieces of an electromagnet are 6 in. in diameter
and are spaced K in. apart, forming the air-gap. What is the reluctance
of this gap? Neglect fringing.

yt^sMO

At- 900-

Fig. 226 a.

224. If the iron pole pieces of problem 223 are cylindrical and have axial
lengths of 1 in., what is their reluctance if their permeability is 1,200?

226. Fig. 225A shows three portions of a magnetic circuit. Compute
the reluctance of each portion and the total reluctance of the combination.
Each portion is circular in cross-section.

226. Compute the reluctance of the magnetic circuit shown in Fig. 226A.

227. Fig. 227 A shows a magnetic circuit composed of two branches,
which are similar and which are in parallel. Compute the reluctance of
each half and the total reluctance of the circuit. The iron has a permea-
bility of 6Q0 throughout.

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QUESTIONS AND PROBLEMS

451

228. Two coils, each having 1,600 turns, are placed on the magnetic cir-
cuit shown in Fig. 226 A, and are connected in series in such a way that they
act in conjunction. What is the magnetomotive force acting on the circuit
vv^hen 1 amp. flows in each coil? What is the resulting flux? What is the
flux density in the gap, in lines per sq. cm. ?

229. If a field coil of 1,000 turns is placed upon the central core of the mag-
netic circuit shown in Fig. 227 A, and 5 amp. flow through the coil, what is
the resulting magnetomotive force? What are the total flux and the gap
density in lines per sq. in. ?

230 A.

230. An anchor ring is shown in Fig. 230 A. Determine its reluctance
when the permeability of the iron is 800. What is the magnetomotive
force if there are 200 turns wound on this ring and 1.5 amp. flow through the
winding? What are the flux and the flux density?

231. A gap 1 mm. long is cut in the anchor ring (Fig. 230 A). The cor-
responding reduction of flux density in the iron increases its permeability
to 1,000. Determine the magnetomotive force, the reluctance, the flux
and the flux density.

232. Assume that the ring. Fig: 230 A, is made of cast steel whose mag-
netization and permeability curves are shown in Figs. 151 and 152 (pages 173
and 174) respectively. Determine the flux and the flux density in the steel
and air-gap when 1 amp. flows in the winding of 200 turns. (Use the trial
and error method of Par. 138.)

233. Repeat problem 232, assuming the ring is made of .cast iron and that
1 amp. flows in the winding. (See Fig. 154, page 177.)

234. Determine the ampere-turns necessary to send a total flux of 6,000
lines through the ring and the air-gap of problem 233.

235. Assuming that the magnet of Fig. 227 A is made of cast steel whose
permeability curve is shown in Fig. 152, page 174, determine the number of
ampere-turns on the central core necessary to send a flux of 400,000 lines
through each gap. Neglect fringing and leakage.

236. Repeat problem 235 using the cast steel magnetization curve of
Fig. 154, page 177.

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452

DIRECT CURRENTS

237. The magnet shown in Fig. 237A has a yoke of cast steel and pole
pieces of cast iron. Using the magnetization curves of Fig. 154. page 177,
determine the ampere-turns necessary to send 120,000 lines through the air-
gap. Neglect fringing.

ClrcnUr
CroM-

FiG. 238A.

238. Fig. 238A shows the magnetic circuit of a 2-pole d3mamo. The
field cores are of cast steel and are 4 in. square. The armature is of O.H.
sheet steel and has a net axial length of 3.6 in. over the iron; the yoke is
of cast iron and has a cross-section of 2 X 6 in. Using the magnetiza-
tion curves of Fig. 154, page 177, determine the necessary field ampere-turns
for an average flux density of 30,000 lines per sq. in. in the air-gap.

239. Repeat problem 238 assuming the air-gap to be 0.075 in. and that
only 80 per cent, of the flux in the yoke and field cores enters the armature.

(Leakage factor = -^ = 1-25. j

^' 240. Determine the hysteresis loss in ergs per cu. cm. per cycle for cast
iron operating at densities between 30,000 lines per sq. in. positive and nega-
tive. (Use data of Par. 143.)

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QUESTIONS AND PROBLEMS 453

241. A transformer yoke of silicon steel has a volume of 600 cu. in.
What is the hysteresis loss in ergs per cu. in. per cycle if the maximum flux
density is 40,000 lines per sq. in.?

242. In a certain electromagnet having 800 turns, a current of 5 amp. .
produces 1,200,000 lines of induction. What are the total linkages? What
are the linkages per ampere? What is the inductance of the circuit?

243. When a current of 12 amp. flows in a certain exciting coil of 2,000
turns, 2,000,000 lines of induction link the coil. What are the linkages?
What is the inductance in henrys?

244. Assuming that the permeability of the magnetic circuit of problem
243 remains constant, determine its inductance when the current is doubled.
Determine the inductance when the turns are doubled.

246. In a closed magnetic circuit of cast steel the net ampere-turns per in.
are 20. The cross-section of the magnetic path is 12 sq. in. and its net
length is 30 in. If 1 amp. flows in the exciting coil, determine the
inductance of the circuit using the curve of Fig. 154, page 177.

246. Repeat problem 245 for an exciting current of 2 amp., or double the
value of that in problem 245. To what is the change of inductance due?

247. When the exciting current of an electromagnet is flowing, there are
1,800,000 lines of induction linking the circuit. The exciting coil has 2,400
turns. If the exciting current is interrupted, requiring 0.5 second to
completely rupture the arc, what is the average induced voltage across
the ends of the exciting coil?

248. Re-compute problem 247, assuming that the circuit is interrupted in
0.2 second.

249. A certain electromagnet has an inductance of 2.8 henrys and a resist-
ance of 5 ohms and is connected across 110-volt mains. What is the time
constant of the magnet? How long will it take the current to reach 63
per cent, of its ultimate value? What will be the value of the current at
this instant? Illustrate the rise of the current by a sketch, marking the
values involved in the problem.

250. If the resistance of the electromagnet of problem 249 be doubled,
what does the time constant of the circuit become? How long does it take
the current to reach 63 per cent, of its ultimate value? What is the
ultimate value? Illustrate by a sketch and compare with problem 249.
In which problem is a given value of current first reached?

261. Six amperes flow in the exciting circuit of problem 247. Compute
the induced electromotive force when the circuit is opened in 0.5 .second,
using equation (75), page 191.

262. If the exciting current of problem 245 is reversed in 0.1 second,
determine the voltage induced across the ends of the exciting coil.

V 253. A certain generator field circuit has an inductance of 2 henrys and
carries 100 amp. The induced voltage across the field terminals must
not exceed 1,000 volts. What is the minimum time which can be allowed
for opening the field? How much energy is liberated in opening this field?
What is the average power during the opening period?

864. Repeat problem 253 with the total field resistance doubled by
means of the field rheostat. . t

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454

DIRECT CURRENTS

255. Two coils A and B, Fig. 255A, are insulated electrically but are so
placed that 80 per cent, of the flux produced by one of the coils links the
other. Coil A has 120 turns and coil B has 200 turns. When 2 amp.
flow in coil A, 220,000 lines link the coil. How many lines link B? What is
the coefficient of coupling of the two circuits? If the current in A is in-
terrupted in }^io second, what induced voltage results in 5? In A?

Fig. 255 a.

256. The same flux that was produced in A by 2.0 amp. is produced in B
by 1.2 amp. What voltage is induced in A upon interrupting the 1.2 amp.
of 5 in 0.1 second? What voltage is induced in 5?

257. Determine the mutual inductance in henrys of coils A and B, in
problems 255 and 256. What is the self inductance of A? Of B?

.1 B
Fig. 258^.

268. Coils A and B of problem 255 are now linked magnetically by an
iron core as shown in Fig. 258^4, so that practically all the flux of one links
with the other. 0.1 amp. in A now produces 200,000 lines in the joint

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QUESTIONS AND PROBLEMS 455

magnetic circuit. How many amperes in B will produce this same flux?
What is the self inductance of A ? Oi B?

269. If the 0.1 amp. in A of problem 258 is interrupted in 0.05 second,
what electromotive force is induced in A? In 5? What is the mutual
inductance of the circuits? The self inductance of A? Of B?

At what rate must the current in B be interrupted to induce 10 volts in
A? If the current in B is 0.05 amp., in what time should the circuit be
opened?

260. The flat pole pieces of an electromagnet are in contact with each
other and a total flux of 2,000,000 lines passes from one to the other. If
each cross-section is 4 in. X 5 in., what force in pounds is necessary to pull
these pole pieces apart?

QUESTIONS ON CHAPTER IX

1. If two insulated ellipsoids near each other are connected to the terminals
of an electrostatic machine, upon what portions of the ellipsoids will the
density of charge be greatest? Would any considerable change be observed
in these charges if the wires to the machine were disconnected? How can it
be shown that charges are "bound.'*

What force exists between two charges of unlike sign? What is its direc-
tion? For charges of like sign?

2. If a positive charge is brought into the neighborhood of an insulated
and uncharged ellipsoid or sphere, what phenomenon occurs? What is the
relation of the induced charge to the inducing charge? Distinguish between
free and bound charges. How may it be shown experimentally that free and
bound charges behave differently?

3. How does a small positive electrostatic charge act when placed near
two conducting bodies between which a difference of potential exists? Can
the distribution of electrical stress in the air between such bodies be repre-
sented by lines in a manner similar to that used in showing magnetic distri-
bution? Where do electrostatic lines originate and terminate? Compare
them with lines of induction and lines of force in this respect.

4. What is the effect in a dielectric medium of increasing the density of
electrostatic lines beyond a certain value? Is this same effect noted in the
magnetic circuit and in the electric circuit?

5. If a needle or other sharp projection be raised to a high potential, what
effect is noted at this projection? What is the condition of the air in this
region? What is the effect of a further increase of potentisd?

Distinguish between an insulator and a dielectric. Name two substances
that are good insulators but poor dielectrics; good dielectrics but poor in-
sulators. In what terms is dielectric strength expressed?

6. What is the effect of applying a voltage to an electric condenser?
What is the order of magnitude of the time required by a current to charge
such a condenser? Why does the current cease to flow? To what hydraulic
phenomenon can this be compared?

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456 DIRECT CURRENTS

7. How can it be shown that electricity is actually stored in a condenser?
How does the quantity which can be stored in a condenser vary with the
voltage? What simple relation does this give between charge, capacitance
and voltage?

8. What is the usual effect of inserting some dielectric medium other than
air between condenser plates? What is "specific inductive capacity" and
to what magnetic property is it analogous? What is the dielectric constant
of glass? Of mica? Of rubber?

9. How is the equivalent capacitance of condensers connected in parallel
determined? To what electric circuit condition is this analogous?

10. How is the equivalent capacitance of condensers connected in series
determined? What is the relation among the electric charges on each of a
number of condensers connected in series? To what equation in the electric
circuit is the equation relating to the equivalent capacitance of condensers
in series similar?

11. How can the voltage across each of a number of condensers in series be
calculated if the line voltage and the individual capacitances be known?
Are these voltage relations dependent at all upon the insulating properties
of the dielectrics? In the case of leaky condensers, upon what does the
ultimate voltage distribution depend?

12. How may it be shown that electric energy can be stored in a con-
denser? Upon what factors does this energy depend?

13. Upon what factors does the capacitance of a parallel plate condenser
depend? What is the effect upon the capacitance of changing the area of
the plates? Of decreasing the distance between them? Of substituting
hard rubber or glass for air?

14. What two methods are commonly employed in the measurement of
capacitance? Upon what fact does the ballistic galvanometer method de-
pend? What relation exists between the quantity passing through the gal-
vanometer and its m'aximum ballistic throw?

Should the measurement be made upon "charge" or upon "discharge? "
Explain. How is the galvanometer calibrated?

16. Describe the bridge method of capacitance measurement. Compare
it with the Wheatstone bridge method of resistance measurement. How
does the bridge formula for capacitance differ from the formula employed
when resistance is measured? What is the source of power and what simple
detector is used in the capacitance bridge?

16. How may a disconnection in a cable be located ? Upon what principle
does this method of measurement depend? Is this method applicable if
the fault is grounded?

PROBLEMS ON CHAPTER IX

261. A condenser has a capacitance of 12 m.f . What is its charge when
the potential between its plates is (a) 220 volts? (h) 440 volts? (c) What
current flowing at a uniform rate is necessary in order that the condenser
may be charged in 0.2 second in each case?

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QUESTIONS AND PROBLEMS 457

262. It is desired to store 70 microcoulombs in a condenser at a potential
of 760 volts. What should be the capacitance of the condenser?

263. What is the potential across a 40-m.f . condenser in which the charge
is 0.002 coulomb? How long must a current of 1 milliampere flow in
order to charge this condenser to the above potential?

264. A certain condenser consisting of two parallel plates, with air as
dielectric, has a capacitance of 0.00012 m.f. A slab of glass is placed be-
tween the plates occupying the entire space. The capacitance is now
found to be 0.00072 m.f. What is the specific inductive capacity of the
glass?

266. The condenser of problem 264 is charged to a potential of 300 volts
between plates and the supply then disconnected. Glass is then inserted
between the plates completely filling the space. This insertion of the glass
in no way changes the value of the electric quantity on the plates. What is
the condenser voltage after the insertion of the glass?

266. A plate condenser, with air as dielectric, has a capacitance of 0.0012
ni.f. and 300 volts is impressed across its terminals. The condenser is then
immersed in a bath of transformer oil having a dielectric constant of 2.5,
the voltage supply remaining connected. What is the charge on this con-
denser before and after immersion in the oil?

267. Four condensers having capacitances 12, 16, 20 and 30 m.f. respec-
tively are connected in parallel across 220-volt mains. What is the charge
on each and what must be the capacitance of a single condenser to replace
the four?

268. Three condensers connected in parallel across 400-volt mains have
charges of 600, 800 and 1,000 microcoulombs. What is the capacitance of
each and what single capacitance would replace the three?

269. The four condensers of problem 267 are connected in series across
these same mains. What is the voltage across each of them and what single
condenser would replace the four? What is the charge on each condenser?

270. Four condensers are connected in series. The voltages of the con-
densers are 50, 70, 80 and 100 volts respectively. This combination of con-
densers can be replaced by a single condenser having a capacitance of 6 m.f.
What is the capacitance of each condenser?

271. A condenser has a capacitance of 20 m.f. What is the stored energy
in the condenser when the voltage across it is 100 volts? 200 volts? In
what ratio is the energy increased if the voltage is doubled?

272. Three condensers having capacitances of 20, 40 and 60 m.f. respec-
tively are connected in series across a 600-volt supply, (o) What is the
voltage across each? (6) What is the energy of each? (c) What is the
energy of the system ?

273. Determine the stored energy of the system when the three condensers
of problem 272 are connected in parallel across the same voltage.

274. An air condenser consists of three plates. The two outer ones are
connected together as one terminal and the other terminal is formed by the
intermediate plate between the two outers. The dimensions of each plate
are 12 in. X 12 in. and the plates are spaced Ke in. apart. What is the
capacitance of this condenser? Digitized by CjOOQ Ic

458 DIRECT CURRENTS

275. If the space between the plates of the condenser of problem 274 Is
filled with paraffin, having a dielectric constant of 2,\, what does the capaci-
tance become?

276. A high voltage condenser is to be made of alternate layers of glass
and tin foil, the glass having a dielectric constant of 8. The glass is ^4
in. thick and the tin foil is 2 mils thick and its dimensions are 3 in. X 4 in.
How many plates and sheets of tin foil are necessary to make a condenser
having a capacitance of 0.02 m.f.? If the glass plates are 5 in. X 6 in.,
what is the siae of the completed condenser?

277. In a bridge measurement of condenser capacitance the bridge is
connected as shown in Fig. 183 (6), page 213. When a balance is obtained
Ri = 100, Rt = 1,242, Cj = 0.4 m.f. What is the value of C», the unknown
capacitance?

278. In a test for a cable fault the apparatus is connected as shown in
Fig. 184, page 213. In the capacitance measurement of the part x the
galvanometer has a ballistic throw of 4.2 cm. In the measurement of the
capacitance of the perfect cable plus the looped end of the faulty cable
the deflection is found to be 16.4 cm. If the length of each conductor is 1,800
ft., how far from the point of test is the cable broken?

QUESTIONS ON CHAPTER X

1. In what way is the flux linking the coil of a generator armature varied?
How does this induce voltage? How does this voltage vary with the speed?
The flux? The number of turns in the coil?

2. If instead of regarding this voltage as due to the change of flux linking
a coil, it is considered as being due to the individual conductors cutting
flux, is the ultimate result in any way affected? If the voltage is considered
as being due to the cutting of lines by individual conductors how does this
voltage vary with the length of conductor? The flux density ? The velocity
of the conductor?

3. What definite relation exists among the direction of the induced
emf ., the direction in which the conductor moves and the direction of the
flux? What simple rule enables one to determine these relations?

4. What is the value of the emf. induced in a rotating coil, (a) when
the coil is in the plane perpendicular to the flux? (6) When its plane lies
parallel to the flux? Does the voltage ever reverse its sign? Explain.

5. How may the alternating current produced in a coil be changed to
direct current? What is the effect of adding coils to the rotating member?
To what are the "ripples" in a voltage wave due?

6. In what way is the open coil type of armature different from* the closed
coil type? Which type is the gramme-ring armature (Fig. 192)? Show that
the resultant electromotive force is different in the two types, even though the
number of coils and turns be the same.

7. Name two serious objections to the ring winding. How are these
objections overcome in the drum winding? What two methods are used to
fasten conductors on the surfaces of armatures? Which is the better method
and why?

8. What is meant by "coil pitch" and what is its relat^^Si&CP^l® pitch?

QUESTIONS AND PROBLEMS 459

What relative positions in the slots do the two sides of any one coil occupy?
Why? What is meant by a ** winding element?" May it consist of more
than one conductor? Explain.

9. What is "front pitch?" "Back pitch?" Average pitch? What is
the relation between the number of winding elements and the number of
coils? The number of commutator segments?

10. In a simplex lap winding how many commutator segments does the
winding advance each time that a coil is added? What three fundamental
conditions must be fulfilled by a winding? What is a winding table and
what is its practical value?

11. Why is it sometimes desirable to place more than two winding ele-
ments in a slot? In what type of generator is this necessary? Is the con-
ductor numbering and are the winding relations in any way affected?
What one condition should be imposed upon this type of winding and why?

12. What is meant by "paths through an armature?" How is the cur-
rent output of- a machine affected by increasing the number of paths?
How is the voltage affected? The power output? How many paths are
there in all simplex lap windings?

13. What is meant by a duplex winding? Show that such a winding may
be composed of two simplex windings each lying in alternate slots. How
many closures may such a winding have and what is its degree of re-entrancy
in each case?

14. If a duplex winding does not close after one passage around the arma-
ture, is the number of segments even or odd? When does such a winding
close? How many times does it close and therefore what is its degree of re-
entrancy?

16. In a winding whose multiplicity is m, how many winding elements
separate a given element from the next returning element? How many
armature paths are there in a 6-pole machine having a simplex lap winding?
A duplex lap winding? A triplex lap winding?

16. To what causes are unequal voltages in different paths of an armature
winding due? Do equalizing connections do away with these inequalities?
What is the purpose of equalizing connections? What care should be
taken regarding the number of slots per pole when equalizing connections
are used? Why?

17. What is the fundamental difference between a lap and a wave winding?
Does the direction of induced emf . in opposite sides of a coil differ in the
two types? Explain.

18. After a wave winding has passed under every pole, in passing around
an armature, what relation should it bear to its starting position if the wind-
ing is simplex? What would a closure after one passage around the arma-
ture mean?

19. Show that the definitions of front pitch and back pitch in a wave
winding do not differ from the similar definitions in a lap winding? Can
the front pitch be even? Odd? Can the back pitch be even? Odd?
Can the two be equal? Can the average pitch be even? Odd? When
is a winding progressive? Retrogressive? Explain.

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460 DIRECT CURRENTS

20. Is it always possible to fit a wave winding to an armature having a
fixed number of slots if all the slots are utilized? Explain? What make-
shift may be used to accomplish the desired result?

21. If the number of pairs of poles is even, is the number of commutator
segments even or odd? Answer if the number of pairs of poles is odd?

22. What is the minimum number of brush sets that can be used in a wave
winding? What is the maximum number that it is possible to use? When
would two sets be used and why? Why is the maximum number usually
desirable?

23. How many paths are there in a simplex wave winding? In what
way is the number of such paths affected by the number of poles? How
many paths in a duplex wave winding? A triplex wave winding?

24. When is it desirable to use a wave winding and why? A lap winding?
Give specific reasons.

26. In addition to forming a part of the magnetic circuit, what other
function does the yoke of a generator perform? Of what two materials is it
made and why? Describe a process whereby the yoke is made without
casting.

26. Of what materials are the field cores made? The pole shoes? What
are the two general shapes of the core sections? Where is each used?

27. Is the armature a solid casting? If not, how is it built up? By what
two methods are the stampings produced? How are they held in position
when placed upon the armature? What is the purpose of the ventilating
ducts?

28. Sketch two general types of slot. Where is each used? What two
methods are used to prevent the conductors from being affected by centrif-
ugal forces?

29. Of what is the commutator made? What insulation is used between
segments? How are the segments clamped together? How are the coil
connections made?

30. What is the purpose of the brushes? Of what material are brushes
usually made? What pressure is used to hold the brush on the commutator?
What is the purpose of the plating on the brush? What is the purpose of
the pig-tail?

PROBLEMS ON CHAPTER X

279. A coil 20 cm. square, having 50 turns, rotates at a speed of 600 r.p.m.
in a uniform magnetic field having a density of 200 lines per sq. cm. (a)
What is the average voltage induced in the coil?

(&) If the flux and the speed are both doubled, what average voltage is
obtained?

280. A wire 40 cm. long moves at a speed of 2,000 cm. per second through
a magnetic field having an intensity of 6,000 lines per sq. cm. How many
volts are induced between the ends of this conductor?

281. A uniform magnetic field is just sufficient in cross-section to pass
perpendicularly through a coil 40 in. X 12 in. The coil has 80 turns. If

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QUESTIONS AND PROBLEMS 461

the coil slides out from this field in 0.001 second and at a uniform rate,
what voltage is induced due to the change in flux linking the coil? What
voltage is generated by the cutting of the flux by the individual conductors?
(Work with the coil sliding in the two directions, one parallel to the 12-in.
side and one parallel to the 40-in. side.)

282. An armature has 40 slots. Design a 2-layer, 4-pole, simplex lap
winding, in which the back pitch is 21 and the front pitch is 19. Make a
winding table.

283. Repeat problem 282 making the front pitch 21 and the back pitch 19.
Which winding is progressive and which is retrogressive?

284. Design a 2-layer, simplex lap winding for a 6-pole, 40-slot machine,
choosing the proper pitches.

285. An 8-pole armature has 128 slots and 6 winding elements per slot.
Determine a correct value of back and front pitch for a simplex lap winding.
Sketch a few slots with their winding elements and connections. How
many commutator segments are necessary?

286. Repeat problem 286 for a winding with 8 elements per slot.

287. A 6-pole, simplex, lap-wound armature delivers a total current of
228 amp. at 220 volts. How many amperes per path through the arma-
ture? How many volts per path? What is the kilowatt rating of the
machine?

288. If the machine of problem 287 had a duplex lap winding, what would
be the amperes per path? Per brush?

289. Repeat problems 287 and 288 for an 8-pole, 200-kw., 220-volt
generator.

290. Make a winding table for a 60-slot, 4-pole armature, the winding to
be a duplex, doubly re-entrant winding. There are 2 winding elements per
slot.

291. Repeat problem 290 using 61 slots and making the winding singly
re-entrant.

292. A 4-pole armature has 33 slots and 2 elements per slot. Design a
simplex wave winding for this armature, having a back pitch of 17 and a
front pitch of 17. Make a winding table. (Check the pitch, using equa-
tion 100.)

293. Repeat problem 292 making yi, = 19, and y/ = 16.

294. Attempt to place a similar winding upon a 34-slot armature. Then
omit one slot, using a dummy coil, and repeat.

295. An 8-pole, 660-volt, 60-kw. generator has a simplex wave-wound
armature. How many amperes per path?

296. Repeat problem 296 using a duplex wave winding.

QUESTIONS ON CHAPTER XI

1. A certain armature has a fixed number of conductors on its surface.
What are the separate effects on the induced voltage of (1) doubling the
speed of the armature; (2) doubling the fiux entering the armature; (3)
reconnecting the armature so that the number of paths through the armature
is doubled?

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462 DIRECT CURRENTS

2. In a given generator, upon what two factors does the induced voltage
depend? If the speed of the generator be maintained constant, upon what
factor does the induced voltage depend?

3. Show that a similarity should exist between two curves plotted as
follows :

1. The field ampere-turns of a generator as abscissas and the flux leaving
one of its north poles as ordinates.

2. The field current of the same generator as abscissas and the induced
armature voltage at constant speed as ordinates.

4. In the curve relating ampere-turns of the field and the flux of one north
pole, why does not the flux start at zero value? Why is the first part of the
curve practically a straight line? At the higher values of field current why-
does the induced voltage increase less and less rapidly for any given increase
in field current?

5. Is there any difference between the saturation curve obtained with
increasing values of field current and that obtained with decreasing values?
Explain any difference.

6. Sketch the connections used in determining a saturation curve. (1)
Using a simple field rheostat. (2) Using a drop wire with the field. Give
two reasons why the generator should be separately excited.

7. Show that Ohm^s Law can be expressed graphically. What two quan-
tities are plotted when expressing Ohm's Law in this manner?

8. Sketch the connections of a shunt generator. Is the field of compara-
tively low resistance or of high resistance? Explain.

9. Explain in detail how ashunt generator "builds up." What limits
the voltage to which a machine can build up?

10. What is the critical field resistance? Give three causes, each of which
may prevent the generator building up. What tests and remedies should be
used for each?

11. What is the general direction of the flux produced by the current in
the armature conductors? What effect does this have upon the resultant
flux in a machine? How does it affect the position of the neutral plane?
What effect does the change in position of the neutral plane have upon the
brush position?

12. What is the relation of the direction of the armature field to the brush
axis? When the brushes are moved forward in a generator what is the
resulting direction of the armature field? Into what two components can
this be resolved? What is the effect of each component?

13. Which conductors on an armature produce a demagnetizing effect?
Which produce a cross-magnetizing effect?

14. Sketch the conductors on the armature, together with the poles, for
a loaded multi-polar machine, indicating the current directions in the vari-
ous conductors. Sketch a curve showing the values of armature magneto-
motive force along the armature surface. Show the flux produced by this
magnetomotive force when acting alone.

15. Show the effect of the above flux on the distribution of the total flux
along the armature surface. How is the neutral zone affected? What
change must be made in the brush position?

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QUESTIONS AND PROBLEMS 463

16. Name four methods by which armature reaction is either eliminated
or reduced. State the principle of each method.

17. Sketch an ideal commutation curve assuming uniform current dis-
tribution over the brush.

18. What is the effect of having voltages induced in a coil during the
time that it is being short-circuited by the brush? What limits the current
in such a coil? How do these currents affect the uniform distribution of
current over the brush?

19. Sketch commutation curves for the following conditions: (a) Brush
advanced too far; (6) brush too far back; (c) brush too wide.

20. Why does an armature coil have self inductance? What is the effect
of this self inductance during the commutation period? What effect does
the voltage of self induction have upon the relation of the brush position to
the neutral zone?

21. What is the order of magnitude of the voltages induced in a coil
undergoing short circuit? If the voltages are low what makes them so
objectionable?

22. What is the advantage of copper brushes over carbon? Why are
carbon brushes used almost universally?

23. What evidence points to the fact that the taking of current
from the commutator by the brushes is not pure conduction? To
what is "high mica" due? How may it be reduced or even eliminated?
Name two methods.

24. In general, what is the effect of arcing on the commutator? Why
should any appearance of arcing be a reason for eliminating the cause of the
arcing as soon as possible? Why is it not desirable to use emery paper or
cloth in grinding brushes or smoothing the commutator?

26. What changes occur in the flux at the geometrical neutral of a gene-
rator as load is applied? What is the effect of these changes upon the brush
position? Why do the brushes have to be moved ahead of the load neutral
plane?

26. Show that instead of moving the brushes forward in order to obtain
the proper commutating flux, the same result may be obtained by the use
of a commutating pole.

27. Why is the commutating pole connected in series with the armature?
Why has it an unusually long air-gap?

' 28. What is the relation of the polarities of the main poles and of the
commutating poles to the direction of rotation, in a generator? In practice
how are the commutating poles adjusted to the proper strength?

29. Sketch the connections used in obtaining the shunt characteristic.
Sketch the characteristic. Why does the machine finally "break down?"
Why does the return curve from short-circuit not follow the curve obtained
with increasing values of armature current?

30. Give three reasons why the voltage of a shunt generator drops as
load is applied. Why are these three reactions cumulative? What pre-
vents a generator from "unbuilding" as load is applied?

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464 DIRECT CURRENTS

31. What effect does running a generator at higher than rated speed have
upon its characteristic, provided that the field current is so adjusted that
the no-load volts are the same in each case?

32. What is meant by generator regulation? Does a large value of the
per cent, regulation indicate that a generator is a desirable one for supplying
lamp loads? Explain.

83. What is meant by the "total characteristic" of a generator? What
is its relation to the shunt characteristic? How may the total power de-
veloped within an armature be determined?

34. How may the objectionable drooping characteristic of the shunt
generator be improved? How are the additional turns connected and
in what way do they differ from the shunt field turns?

36. Show the difference between "long shunt" and "short shunt" con-
nection. What is the effect of the connection upon the characteristic?
Sketch the characteristics of an over-compounded, a flat-compounded and an
imder-compounded generator. Where is each used and why?

36. How is the degree of compounding in a generator adjusted? When do
generators have two separate series fields?

37. What is the effect of speed upon the degree of compounding, if the
no-load voltage is the same in each case? Compare this with the effect
of speed upon the shunt characteristic and explain.

38. Show how the number of series turns for a desired degree of com-
pounding may be determined. What is the armature characteristic and how
may it be utilized?

39. How does the series generator differ fundamentally from the shunt
generator in construction? In the type of load that it supplies?

40. Describe the external characteristic of the series generator and show
its relation to the saturation curve.

41. In what way does the machine "build up?" What is meant by the
critical external resistance? Why is it desirable to operate upon the right-
hand side of the external characteristic?

42. Name a very common use of the series generator. Name two common
types of machines. Why are special commutators necessary?

43. What is the " Thury system " of power transmission? Where is it
used?

44. How may series generators be used to control the voltage at the end
of a feeder? Upon what portion of the characteristic does such a generator
operate? Sketch the connections. What precautions must be taken in the
installation and operation of such a booster?

46. How may the speed of a prime mover affect the generator characteris-
tic? Is such a drop in speed chargeable to the generator? How may it be
taken into consideration?

46. State one essential difference between a unipolar generator and the
ordinary type. What design is necessary to prevent the armature being
short-circuited on itself? What is the advantage of this type of machine
over the ordinary type and for what type of work is it best adapted ?.a^ What
are its disadvantages?

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QUESTIONS AND PROBLEMS 465

47. What is the basic principle of the Tirrill regulator? What is the
function of the main control magnet? Of the relay magnet? Why cannot
this regulator be applied directly to the fields of machines of large capacity?
How may it be applied to these machines?

PROBLEMS ON CHAPTER XI

297. The pole faces of a shunt generator are 8 in. square and the average
flux density under the pole is 45,000 lines per sq. in. The machine has
4 poles and there are 300 surface conductors on the armature. The
machine is wave wound making two parallel paths through the armature.
What is the induced voltage when the armature rotates at 800 r.p.m.?
1,000 r.p.m.?

298. If the current per path in problem 297 is 20 amp., what is the rating
of the machine in kilowatts?

299. Repeat problems 297 and 298 for a simplex lap winding, the number
of conductors, the speed, etc. remaining the same.

300. In an 8-pole, 220-volt generator, the pole faces are 12 in. square.
The flux density under the poles at no load is 47,600 lines per sq. in.
There are 16 slots per pole on the machine. The speed of the machine is
750 r.p.m. If the armature is lap wound, how many conductors per slot
are necessary to give the rated voltage at no load?

301. The following data were taken for the saturation curve of a 20-kw.,
220-volt generator, running at 600 r.p.m.

Field current 0

0.5

1.0

1.5

2.0

2.5

3.0

Volts 10

62.6

126

178

212.6

235

245

Plot this saturation curve and then replot it for 550 r.p.m.

302. The generator of problem 301 is a 4-pole, lap-wound machine and
has 440 conductors on the armature surface. There are 400 shunt field
turns per pole. Plot a curve between flux per pole and ampere-turns per
pole.

303. Determine the approximate number of ampere-turns required for
the gap and for the iron at 220 volts, when the generator of problem 301 is
operating at 550 r.p.m.

304. Determine the critical field resistance for both speeds in problem 301.
Determine the field resistance necessary for the generator to build up to 220
volts at each speed.

305. When the generator of problem 301 is operating at 600 r.p.m. and
the field resistance is adjusted so that the machine builds up to 220 volts,
what current flows through the field due to the residual magnetism? What
induced voltage results from this field current? What field current results
from this last voltage?

306. A generator fails to build up. When the shunt field is connected
across the armature, a voltmeter across the armature shows 4 volts. When
this field circuit is opened the voltmeter reads 7 volts. What is the probable
reason that the machine does not build up and what remedy is suggested?

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466 DIRECT CURRENTS

807. The no-load flux of a bi-polar generator is 3,000,000 lines. When
the generator is carrying its rated load and the brushes are in the no-load
neutral plane, the armature itself produces a flux of 1,000,000 lines. Ne-
glecting the effect of s&turation, what is the resultant flux?

308. Repeat problem 307 when the brushes are advanced 30**.

809. There are 240 conductors on the surface of the armature of a bi-polar
generator. The generator deUvers 50 amp., making 25 amp. flowing
in each conductor. If the brushes are advanced 15**, how many demagnet-
izing and cross magnetizing ampere-conductors are there? How many de-
magnetizing and cross-magnetizing ampere-turns are there?

310. The brushes of a 4-pole generator are advanced 10 space de-
grees. The armature is lap wound and has 496 surface conductors. How
many demagnetizing and cross-magnetizing ajnpeTe-tums are there on the
armature when the generator delivers 120 amp.?

311. Repeat problem 310 for a generator having the same number of poles
and armature conductors and deUvering the same current, but with a wave-
wound armature. What is the ratio of the kilowatt capacities of the two
machines?

312. A conmiutating-pole circuit has a resistance of 0.08 ohm. The rated
full-load current of the generator is 80 amp. The most satisfactory con-
dition of commutation is obtained when 60 amp. flow in the commutating-
pole circuit. What must be the resistance of a shunt to be connected
across this commutating-pole circuit?

313. The terminal voltage of a shunt generator is 550 volts when the
armature delivers 100 amp. If the armature resistance is 0.3 ohm, what
voltage is being induced in the armature?

314. A 75-kw., 220-volt shunt generator has 228 volts induced in its arma-
ture when it is delivering its rated load at 220 volts. At the same time 12
amp. are taken by the shimt field. What is the armature resistance?

316. The no-load voltage of the generator of problem 314 is 234 volts.
What is its per cent, regulation? Why is the induced voltage at rated load
not equal to the no-load volts?

316. A shunt generator has a no-load voltage of 119 volts. It is specified
that it shall regulate to within 6 per cent. What should be the terminal
voltage when it delivers its rated load?

317. What is the total power being developed in the armature of the gene-
rator in problem 314? How much of this power is lost in the armature and
how much is lost in the field? How much is available for delivery to the
external circuit?

318. The terminal voltage of a generator is 600 volts when delivering 50
amp. The armature resistance is 0.8 ohm and the shunt field resistance
is 250 ohms. What power is being generated in the armature? What is the
electrical efficiency of the armature?

319. A compound generator has a no-load voltage of 230 volts. It sup-
plies a 200-kw. load, situated 800 ft. distant, over a 1,000,000 CM. cable.
It is desired to maintain the voltage at the load constant at 230 volts from

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QUESTIONS AND PROBLEMS 467

no load to full load of 200 kw. What must be the no-load and the full-load
voltage rating of the generator?

320. Repeat problem 319 for the condition in which it is desired that the
voltage at the load rise from 230 to 240 volts from no load to full load.

321. The generator of problem 319 is connected long shunt. Its armature
resistance is 0.007 ohm, the shunt field resistance is 15 ohms, and the series
field resistance is 0.0016 ohm. What is the voltage induced in the armature
with the 200-kw. load of problem 319? How much power is lost in the
armature, in the shunt field, in the series field and in the cable? Of the power
generated, how much reaches the load?

322. Repeat problem 321 for the 200-kw. load of problem 320. Due
to the removal of the series field diverter the resistance of the series field is
now 0.003 ohm.

323. It Is desired to add series turns to a shunt generator, so that its rated
load voltage is the same as the no-load voltage. There are 260 shunt turns
per pole. It is found necessary to increase the shunt field current from .8 to
11 amp. in order to keep the rated load voltage equal to the no-load
voltage. The rated current of the machine is 300 amp. How many
series turns per pole should be added?

324. It is desired that the voltage of a 250-kw., 560-volt generator in-
crease from 650 volts at no load to 600 volts at rated load. With the series
field out of circuit and the shunt field excited from an external source it
is found that this increase of voltage may be obtained by increasing the
shunt field current from 6 amp. to 9 amp. There are 640 shunt turns per
pole and 6 series turns per pole. The total series field resistance is 0.03
ohm. What must be the resistance of a shunt or diverter to be connected
across the series field?

326. The terminal voltage of a series generator is 2,840 volts. The arma-
ture resistance is 20 ohms and the field resistance 26 ohms. What is the
voltage induced in the armature when the machine delivers 6.6 amp.?

326. A lOO-kw. load is situated 2,000 ft. distant from the 230-volt bus-
bars of a station. The load is supplied over a 600,000 CM. feeder. It is
desired that when the load is 100 kw. the load voltage shall not be less than
225 volts. What must be the current and voltage rating of a series booster
designed to maintain this voltage at the above value?

327. The booster in problem 326 has an efficiency of 80 per cent. It is
driven by a shunt motor connected across the bus-bars, the motor efficiency
being 80 per cent. What is the efficiency of transmission over the feeder?

QUESTIONS ON CHAPTER XH

1. In what way does a motor differ from a generator in the work which it
performs? In general construction?

2. What effect is noted when a conductor carrying a current is placed in a
magnetic field? How can this action be explained by two elementary laws
of magnetism? What is the effect of reversing the current in the conductor?

3. To what three factors is this force proportional? If the flux is doubled
how is the force affected? If the current is doubled?

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468 DIRECT CURRENTS

4. State a convenient rule by which the relation among the direction of
the current, the direction of the field and the direction of the force can be
determined. What other simple method enables one to determine this
relation?

6. What is torque? In what units is it expressed? In the British sys-
tem? In the metric system?

6. Show that a coil carrying current when placed in a magnetic field
develops a torque. In what position of the coil is the torque a maximum ?
When is it zero? What change in the connection to the armature coil
should be made when the torque reaches its zero value?

7. Why are a large number of conductors upon the armature desirable?
To what three factors is the torque of an armature proportional? In any
one machine, to what two factors b the torque proportional?

d. How can it be shown that resistance alone does not determine the
amount of current taken by a motor armature? Why must a motor of
necessity be generating a voltage when it is rotating? What is the relation
of this voltage to the direction of the current? To the direction of the
applied voltage?

9. Is the coimter electromotive force greater or less than the applied vol-
tage? Why? By what quantity do the two voltages differ from, each
other?

10. Fundamentally, upon what two quantities does the speed of a motor
depend?

11. In what direction is the flux of a motor distorted by armature reac-
tion? In what direction should the brushes be moved as the load is applied
to a motor? What general effect on the field flux does this movement of the
brushes have? What is the effect upon the speed?

12. What is the relation between the main poles, the interpoles and the
direction of rotation of a motor? How does this relation compare with the
similar one for a generator?

13. When load is appUed to a motor what is its first tendency? In the
case of the shunt motor, how does this tendency affect the back electro-
motive force? The current flowing into the armature?

14. What two characteristics are very important in considering the suit-
ability of a motor for commercial work?

16. How does the torque of the shunt motor vary with the load? Why?
How does the speed vary with the load? Demonstrate. Ordinarily is its
change of speed with load excessive? What effect does armature reaction
have upon the speed? What is meant by "speed regulation?" Does the
per cent, speed regulation have any significance as regards a motor's per-
formance? To what general type of work is a shunt motor adapted and
why?

16. How does the flux in a series motor vary with the load? How does
this affect the variation of torque with load?

17. To what extent is the speed of a series motor affected by the applica-
tion of load? By the removal of load? What precautions should be
taken when the series motor is being installed for industrial purposes?

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QUESTIONS AND PROBLEMS 469

18. To what general types of load is a series motor adapted and why?
For what reasons is it especially adapted to street -railway work?

19. What factors are plotted in giving the characteristics of a street
car motor? Why?

20. In what way do the windings of a compound motor differ from those
of a shunt motor? A series motor? In what two ways, with respect to the
shunt winding, may the series winding be connected?

21. What is the speed characteristic of the cumulative compound motor?
The torque characteristic? What advantage has it over the series motor?
For what general type of work is it best adapted?

22. What is the nature of the speed and torque characteristics of the dif-
ferentially-compounded motor? Is this type of motor in general use? Ex-
plain. What precaution is necessary in starting this type of motor?

23. How may a motor be reversed? What is the effect of reversing the
line terminals?

24. Why is a starting rheostat necessary for direct-current motors?
In what circuit is the starting resistance connected? Why should it not be
connected in the line?

26. What two additions to the starting resistance of Fig. 299, page 330,
incorporated in a 3-point starting box? Why? Sketch the connections of a
3-point box. Show that the starting resistance which is in series with the
shunt field when the arm is in the running position has little effect upon the
field current.

26. Under what conditions of motor operation is a 3-point box unde-
sirable? Why? Show that this objection is overcome by the use of a 4-
point box. Sketch the connections of a 4-point box. What is the principle
advantage of having the hold-up magnet in series with the shunt field?

27. Sketch the connections of a starting box containing the field resistance.
Why is it necessary to short-circuit this resistance on starting? How is
this accomplished?

28. How should a shunt motor be stopped? Give reasons. What is the
effect of stopping the motor by throwing back the starting arm?

29. Sketch the connections of series motor starters. What is the advan-
tage of the no-load release over the no- voltage release?

30. When are controllers used and why? What two functions may a
controller perform outside actual starting duty?

31. What are two advantages of automatic starters in medium sizes of
motors? In the larger sizes of motors? What limits the rate of cutting
out resistance in the type shown in Fig. 304, page 335? How is this starter
operated?

32. Upon what principle do the plungers and solenoids of the E.G. & M.
controller operate? Why do the plungers remain down when the current
is large? Why do they rise and close the contacts when the current
decreases?

33. What is the principle of the magnetic blow-out? When is it used?

34. Of what material are resistance units for the smaller types of starting
boxes made? The larger types?

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470 DIRECT CURRENTS

36. What two factors only can be varied in obtaining speed control of a
motor? In the armature resistance control, which of these factors is varied?
What are the advantages of this method of control? Name two serious
disadvantages.

36. What is the principle of the multi-voltage system? How are coarse
adjustments of speed obtained? Fine adjustments? What is the objec-
tion to this system ?

87. What factor in the speed equation is varied in the Ward-Leonard
system of speed control? How many machines are necessary in this system ?
What is its chief advantage and where has it been used extensively? Name
two disadvantages.

38. What factor in the speed equation is varied in the field control
method? Name two distinct advantages of this method. What limits the
range of speed obtainable? What type of motor is especially adapted to
this type of speed control?

39. What principle is involved in the speed control of the Stow motor?
Why can a wide range of speed, w^ith good commutation, be obtained with
this motor?

40. Upon what principle does the Lincoln motor operate? What are its
advantages?

41. What is meant by series-parallel control of railway motors? Why
is such control desirable? Sketch the half speed and the maximum speed
connections in a 2-motor car. In a 4-motor car.

42. Give three reasons why it is objectionable to place the main con-
troller on the platform in the larger sizes of electric car equipment. How
are these objections overcome? Give two other reasons why automatic
control is desirable.

43. What is the general principle underlying the multiple-unit control?
What is the train line?

44. Name briefly the sequence of closing of the contactors in starting a
train.

46. What is meant by "dynamic braking?" Where is it used? Can a
motor armature be brought to a standstill by this method of braking?
Explain. What is regenerative braking and where is it used?

46. Give two occasions where it is desirable to know the efficiency of a
motor. What type of brake is often used for loading motors? Does this
type lend itself to ready calculation of torque and power output of the
motor? Explain. What is meant by the dead weight of the brake arm and
how can it be determined and correction be made?

47. Describe a simple type of rope brake. How many balances are neces-
sary in this type? What is a common method of cooling prony brakes?

48. In what way does a speed counter differ from a tachometer? Upon
what principle is the magneto-voltmeter method of measuring speed based?

PROBLEMS ON CHAPTER XH
328. A bundle of 16 wires lies perpendicular to a magnetic field whose den-
sity is 800 lines per sq. cm. That part of the bundle of wires which lies

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QUESTIONS AND PROBLEMS 471

in this field is 25 cm. long. What force in kilograms is acting on the entire
bundle when a current of 12 amp. flows in each wire, the direction of cur-
rent being the same in each?

329. A gear having 130 teeth drives another having 60 teeth. The dis-
tance from the center of the first gear to the point of contact of the teeth is
6.5 in., the pitch circle having a diameter of 13 in. The pressure between
the teeth at the point of contact is 400 lb. What is the torque in pound-
feet developed by each of the gears?

330. A pulley having a diameter of 14 in. drives a 50-in. pulley with
a 6-in. belt. The respective tensions in the tight and loose sides of the
belt are 1,500 and 300 lb. respectively. What net torque in poimd-feet is
developed by each pulley?

331. A coil consisting of 16 turns of wire lies parallel to a magnetic field
having a strength of 30,000 lines per sq. in. (See Fig. 286(o), page 313.)
The distance across this coil parallel to the field is 12 in., and 14 in. of active
conductor lie in the magnetic field. What torque in kilogram-meters is de-
veloped by the coil when the current per conductor is 5 amp.? Sketch
the coil and the magnetic field, indicating the directions of the forces acting.

332. Repeat problem 331 for a similar coil in which the current in each
conductor is 8 amp. and the strength of field is 40,000 lines per sq. in.
Obtain the result in pound-feet.

333. When the flux density in the air-gap of a shunt motor is 45,000 lines
per sq. in. and the armature current is 60 amp., the motor develops 80 lb. -ft.
torque. What is the torque developed when the motor takes 30 amp.,
the flux remaining constant? 50 amp.?

334. When the load is entirely removed from the armature of problem 333,
the motor armature requires 8 amp. to keep it running. What torque is
required to overcome the motor losses? What is the torque available at
the pulley in each case of problem 333, assuming that the no-load torque re-
mains constant?

336. When the current (60 amp.) of problem 333 is halved, the flux is
also halved. What torque is developed?

336. The armature of a shunt motor has a resistance of 0.04 ohm. When
this motor is connected across 110-volt mains, it develops a counter electro-
motive force of 105 volts. What current does the armature take? What
current would it take if it were connected across the same mains while
stationary?

337. What counter electromotive force does this motor armature develop
when it is taking 80 amp. from the mains? If this same machine were
running as a generator what would be its internal electromotive force when
the armature is delivering 80 amp. at 110 volts?

338. The armature of a 4-pole shunt motor has 420 surface conductors
and is wave wound. What back electromotive force does it develop when
rotating at 1,400 r.p.m. ? The flux is 2,500,000 lines per pole. Its armature
resistance is 0.2 ohm. What is its terminal voltage when the motor takes
50 amp. from the line?

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472 DIRECT CURRENTS

339. What current does the armature of the motor in problem 338 take
from the hne when its speed is 1,360 r.p.m. if the terminal voltage and flux
remain constant?

340. A shimt motor has an armature resistance of 0.1 ohm. When con-
nected across 220-volt mains it takes 5 amp. and runs at 1,100 r.p-m. At
what speed will it nm when its armature current is 40 amp.? Neglect
armature reaction.

341. A compoimd winding having a resistance of 0.05 ohm is added to the
motor of problem 340. This increases the motor flux 20 per cent, between
zero current and 40 amp. Assuming that the increase of flux is propor-
tional to the current, find the speed at 40 amp. When running at 5
amp. the speed is adjusted to 1, 100 r.p.m. Neglect armature reaction.

342. By what percentage should the flux of problem 340 be decreased in
order that the speed at 40 amp. be the same as it is at 6 amp? Neglect
armature reaction.

343. A 560-volt series motor has a series field resistance of 0.05 ohm and
an armature resistance of 0.2 ohm. When taking 90 amp. from the line
its speed is 480 r.p.m. What is its speed when it takes 40 amp. from the
line? Assume that the saturation curve is a straight line and neglect arma-
ture reaction.

344. A 220-volt shunt motor has a field resistance of 100 ohms and its
armature has a resistance of 0.15 ohm. The total line current is 50 amp.
What is the back electromotive force of the armature?

346. The motor of problem 344 develops 65-lb.-ft. internal torque when
taking 50 amp. from the line. What internal torque does it develop when
taking 15 amp. from the line? Neglect armature reaction.

346. When the motor of problem 344 is running without load it takes 7.0
amp. from the hne and runs at 1,000 r.p.m. What is its speed when
taking 50 amp. and when taking 15 amp. from the line? What is its speed
regulation in each case? What is the torque at the pulley in problem 345?

347. A motor runs at 800 r.p.m. when running light. It has a speed
regulation of 3.5 per cent. What is its speed at its rated load?

348. When a series motor takes 40 amp. from the line it develops 220
Ib.-ft. torque. What torque does it develop at 60 amp.? At 90 amp.?
Assume that the saturation curve of the iron is a straight line.

349. The motor of problem 348 has an armature resistance of 0.2 ohm and
a series field resistance of 0.04 ohm. If it runs from 600-volt mains and
runs at 700 r.p.m. when taking 40 amp., what is its speed at 60 and at
90 amp.?

350. What is the motor speed of problem 349 when it takes 10 amp.
from the line?

351. A shimt motor is rated at 44 amp. at 230 volts. Its field current
is 1.0 amp. and its armature resistance is 0.2 ohm. It is desired that the
motor start with 125 per cent, rated current. What should be the initial
starting resistance?

352. The motor of problem 351 reaches 25 per cent, of its rated speed
before the second contact of the starting resistance is reached. When this

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QUESTIONS AND PROBLEMS

473

contact IB reached it is desired that the armature current be 43 amp. What
should be the resistance between the first two contacts?

363. The motor of problem 351 reaches half speed when the starting arm
reaches the third contact. Find the resistance between the second and third
contacts. The current should again be 43 amp. when the arm touches the
third contact.

354. A 220-volt shunt motor has an armature resistance of 0.15 ohm.
When the armature takes 4 amp. from the line it runs at 1,000 r.p.m. It
is desired to obtain 600 r.p.m. at 44 amp.
by inserting resistance in the armature cir-
cuit. What external resistance is necessary?
With this external resistance in circuit, at
what speed will the motor run when the
armature takes 22 amp.?

366. Repeat problem 354 for 300 r.p.m.

366. What percentage of the line power
is delivered to the armature at 44 amp. in
problems 354 and 355?

367. A motor when connected across the
110-volt mains of Fig. 309, page 341 runs
at 600 r.p.m. What speeds can be obtained
by the use of this system if the shunt field
is kept constant? Neglect the laRa drop in
the motor armature.

368. In a Ward Leonard system of speed
control the efficiencies of the machines are
as follows: Afi (Fig. 310, page 342), 85 per
cent. ; (r, 83 per cent. ; M2, 80 per cent. The
line voltage is 220 volts. When Af 2 delivers
7 hp. how much current is being supplied by
the line? What is the over-all eflBciency of ,
the system?

369. In a brake similar to that shown in
Fig. 317, page 349, the length L is 2 ft. The

balance reading is 32 lb. ; the dead weight of the arm is 2.8 lb. ; the speed of the
armature is 1,120 r.p.m. (o) What horsepower does this motor develop?
(6) The motor input is 49 amp. at 220 volts. What is its efficiency at this
load?

360. Repeat problem 359 for a balance reading of 23 lb. and a speed of
1,130 r.p.m. The motor input is now 36.2 amp. at 220 volts. The dead
weight of the arm remains unchanged.

361« In a brake similar to that shown in Fig. 319, page 352, the diameter
of the drum is 10 in. The speed is 1,400 r.p.m. One balance reads 19.8 lb.
and the other reads 4.3 lb. (a) What torque does the motor develop at this
load? (6) What is the horsepower output? (c) If the input is 18 amp. at
110 volts, what is its efficiency at this load?

362. Calculate the horsepower output developed by the rope brake shown
in Fig. 362A. It is running at 1,500 r.p.m. ^-^^-^^^^^^ ^^ CjOOgle

PiQ. 362A.

474 DIRECT CURRENTS

QUESTIONS ON CHAPTER Xm

L What becomes of the energy which is lost within electrical apparatus?
Is it useful or otherwise? Explain in detail its effect on the apparatus.

2. Into what three groups can the losses in either a motor or a gejierator
be classified? Name the losses under the first group, indicating how they
are determined. Are they readily determinable?

3. What constitutes the losses of the second group? How are the losses
supplied, electrically or mechanically? Upon what do they depend? How
is the eddy current loss made small? What is meant by pole face loss and
to what is it due? How is it reduced?

4. Why are all the losses except the copper loss grouped as one? Upon
what do they all depend? If it is desired to duplicate stray power losses
under two different conditions of load, what two factors must be maintained
constant?

6. Show that if the losses in a machine are known at any particular load
its efficiency can be calculated. Why is the formula for a generator differ-
ent from that of a motor?

6. How may the efficiency of a generator be measured directly? What
practical conditions make such measurements difficult ? What effect do errors
in the measurements have upon the precision of the results? What other
objections are there to direct measurements of efficiency?

7. How is a machine ordinarily operated in order to measure its stray
power? What measurements are made? To wliat is the stray power
then equal?

8. In stray power measurements, how is the flux adjusted to the proper
value? How is the speed adjusted? Does the flux adjustm^it have any
effect upon the speed and if so how are any readjustments made?

9. For what purpose is a set of stray power curves desirable? Why can-
not the stray power over the entire operating range of the machine be shown
with one curve? What errors are introduced by using the field current as a
measure of the flux and how is one of these errors partially neutralized?

10. For determining losses what is the advantage of the opposition
method over the stray power method? Upon what principle does this
method depend?

11. What assumption is it necessary to make in the opposition method?
Does this assumption introduce appreciable error? In this method how
are the two machines started and then adjusted? What instruments are
used and what measurements are necessary? State the disadvantages of
this method.

12. What determines the rating of a steam engine? A steam turbine?
A gas engine? An electric machine? Give reasons in each case.

13. State the effects of excessive temperatures upon the insulation of
electric machinery. What insulating materials can withstand the highest
temperatures?

14. What is the "hot spot" temperature and what difficulties accom-
pany its measurement ? Name one method by which an approximation of the
hot spot temperature may be reached.

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QUESTIONS AND PROBLEMS 475

16. What other well-known principle is utilized to determine the tempera-
ture rise? What is meant by "ambient" temperature?

16. For what length of time should a temperature test be run? How may
the temperature rise be accelerated? In what way may a machine's ap-
proach to constant temperature be determined? Why does the tempera-
ture of a machine rise more rapidly at the beginning of a test than at the
end? What relations exist between the heat supplied and the heat dis-
sipated when a constant temperature is reached?

17. Why must care be taken not to include the brush and contact resist-
ance when measuring the armature resistance for temperature determina-
tions? Where must the voltmeter leads be held?

18. What difficulties arise when the resistances of multi-polar armatures
are measured for temperature determinations? How may these difficulties
be eliminated? What precautions should be taken when the field tempera-
ture is being determined by resistance measurements? Why must the
temperature measured in this way be still further increased?

19. Give five reasons why it is either necessary or desirable to operate
shunt generators in parallel. What, in their characteristic, makes them
especially adapted to parallel operation?

20. If one generator starts to take more than its share of the load, what is
the resulting effect upon its voltage ? Does this effect oppose the generator's
taking additional load ? Explain. What is meant by * * stable equilibrium. ' '

21. State in detail the steps necessary to connect a machine in service. If
the generator is connected in service with its voltage equal to that of the bus,
why does it not take load? What must be done in order that it may take its
share of the load?

22. Describe the steps necessary to remove a machine from service.
Why is it undesirable to open the generator switch when the machine is
delivering load? What is necessary as regards the generator character-
istics in order that the machines may properly divide the load over their
entire range of operation?

23. Show that over-compounded generators in parallel are in unstable
equilibrium. What simple connection makes their operation stable?

24. What two conditions are necessary for two compound generators
to divide the load proportionately over their entire range of operation?

26. Why does not a diverter change the division of load between two
compound generators in parallel? What adjustment can be made to
change the load division?

26. How many equalizers may be necessary in certain types of compound
generators?' How many poles must the switch have for such a generator?

27. Compare circuit breakers and fuses, stating the advantages and dis-
advantages of each. Which has the higher first cost? Which ordinarily
has the lower maintenance cost? Which requires the more space? Which
operates the faster? Under what conditions should each be used?

28. Upon what principle do circuit breakers in general operate? How is
a "wiping contact" secured? What is the purpose of the carbon contacts?
Why should breakers always be mounted at the top of a switchboard?

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476 DIRECT CURRENTS

PROBLEMS ON CHAPTER Xm

363. The eddy current loss in a generator is 300 watts when it is running
at 800 r.p.m. and with a flux of 1,000,000 lines per pole, (a) What is
the eddy current loss when the speed is raised to 1,000 r.p.m., the flux re-
maining unchanged? (&) What is the loss at 800 r.p.m. with a flux of
1,200,000 lines? (c) With this same flux what is the loss at 1,000 r.p.m.?

364. If the hysteresis loss in a generator is 600 watts with a speed of
800 r.p.m., what is the loss when the speed is increased to 1,000 r.p.m. ? To
1,200 r.p.m.? The flux is the same in each case.

366. A shimt generator delivers 100 amp. at 115 volts. The total losses
in the machine are 1,200 watts. What is its efficiency?

366. A shunt motor takes 35 amp. at 220 volts. Its field resistance is
180 ohms, its armature resistance is 0.3 ohm and its stray power loss is
400 watts. What is its output in horsepower and what is its efldciency at
this load?

367. Assiune that the stray power loss of problem 366 is in error by 10
per cent. What error does this introduce into the efficiency ?

368. A shunt generator delivers 250 amp. at 220 volts. The shunt
field current is 5 amp., the armature resistance 0.035 ohm and the stray
power loss is 1,800 watts, (a) What is the horsepower input to the gen-
erator? (b) What is its efficiency? (c) If the generator speed is 400 r.p.m.
what is the torque applied to its shaft?

369. A shunt generator is running light as a motor. Its armature takes Ih
amp. from 220-volt mains. The armature resistance is 0.08 ohm. What
is the stray power loss of the machine under these conditions?

370. A motor running light takes 25 amp. from 110-volt mains. Its
shunt field resistance is 12 ohms and its armature resistance is 0.016 ohm.
What is the stray power loss in this motor?

371. It is desired to measure the stray power loss in the generator of
problem 368 by running it light as a motor. To what values should the
terminal volts and speed be adjusted? Make a diagram of connections
showing the instruments and methods of adjustment.

372. Two similar 10-kw. 220-volt generators are connected as shown in
Fig. 330, page 366, for the purpose of having their losses measured. When
the machine operating as a motor is taking its rated current of 48 amp. the
line current / is found to be 75 amp. The generator field current is 2.4 amp.
and the motor field current is 1.8 amp. Each machine has an armature
resistance of 0.2 ohm. What is the stray power of each machine at this
load? What is the efficiency of each machine at this load?

373. When the current in the generator armature of problem 372 is
24 amp. (half load) the line current / is 5 amp. The generator
field current is now 2.2 amp. and the motor field current is 1.8 amp.
Determine the stray power and the efficiency of each machine at this load.
Why are the field currents different in the two machines?

374. The armature and field resistances of a 550-volt shunt generator are
measured after the machine had been standing idle for some time in an engine

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QUESTIONS AND PROBLEMS 477

room whose temperature is 30° C. The voltage across the field wmding ex-
clusive of the rheostat is found to be 420 volts and the field current 4.8 amp.
Xhe armature resistance between two marked commutator seg-
ixients is found to be 0.21 ohm. After the machine had been running under
load for 2 hours these same measurements were repeated. The field
voltage is now 460 volts and the field current 4.8 amp. The armature
resistance is now 0.225 ohm. What is the temperature rise of each? Are
these maximum temperatures safe for untreated cotton insulation?

376. Two 60-kw. 220-volt generators are operating in parallel. They
both are adjusted to 230 volts at no load and are then paralleled. In gen-
erator No. 1 the voltage drops 8 volts from no load to full load and in No. 2
it drops 12 volts. When the aggregate load on the system is 360 amp.,
what current does each generator deliver? What kilowatt load does each
deliver? Assume that the voltage drops in a straight line in each case.

376. Repeat problem 375 for an aggregate load of 400 amp.

377. It is desired to operate a 100-kw. 220-volt shunt generator and a 60-
kw. 220-volt shunt generator in parallel. The voltage of the first drops 8
volts from a no-load voltage of 230 when its rated load is applied. What
should be the voltage drop of the second generator in order that each may
take its proportionate share of the load at all times? Assume that the
voltage drops follow a straight line in each case. How much current does
each deliver when the system demand is 700 amp.?

378. Two compound generators are operating in parallel. One has a
rating of 100 kw. and the other a rating of 75 kw. The resistance of the
series field of the first is 0.002 ohm. What should be the resistance of the
series field of the second machine for proper division of load?

QUESTIONS ON CHAPTER XIV

1. Why cannot the ordinary direct-current voltages be used for transmit-
ting considerable amounts of power over long distances? Where is direct
current most commonly utilized? What are its advantages under these
conditions?

2. What is the general scheme for transmitting large amounts of power
from a remotely situated power station to the consumers' premises? What
are the ranges of transmission voltages? Of distribution voltages? What
part does the sub-station play in the system?

3. How does the weight of conductor vary with the transmission voltage?
If the transmission voltage were doubled how would the weight of copper be
affected the other factors remaining unchanged?

4. What five conditions in general determine the size of conductor to be
used ? For what conditions does the question of heating particularly apply ?
How may the economics of the problem determine the size of conductor?
What is the disadvantage of having too large a conductor? Too small a
conductor?

6. Why is 110 volts most convenient for incandescent lighting? Why
is a higher voltage undesirable ? What are the advantages and disadvantages
of a lower voltage for this purpose?

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478 DIRECT CURRENTS

6. What are the common trolley voltages? Why are these voltages so
chosen?

7. What is meant by distributed loads? Where do such loads occur?
Where are conductors of uniform cross-section throughout most commonly
used?

8. Theoretically, what type of conductor is most economical for uni-
formly distributed loads? What is the practical condition that most
nearly approaches this theoretical condition?

9. Why is the "return loop" system of distribution used? What is its
one disadvantage?

10. What system overeomes the disadvantage of the return loop system?
Make a sketch and show how this system may be further modified to form a
still more efficient system.

11. What advantage is gained by connecting llQ-volt loads in series
groups of two and utilizing 22(>-volt supply? What are the disadvantages
of so grouping the loads?

12. How are the objections to the series-parallel system overcome?
What are the relations existing among the voltages of the Edison 3-
wire system?

13. If the neutral wire be of the same size as the two outers, what are
the relative weights of copper in the 3-wire system with 220 volts across
outers and in the simple llQ-volt system, other conditions being the same?

14. What is meant by balanced loads? Under this condition how much
current flows through the neutral?

16. In what direction does the neutral current flow if the positive load
is the greater? The negative load? What relation does the neutral current
bear to the current in the outer wires? What type of ammeter should be
used in the neutral? What is the commercial limit of unbalancing?

16. State briefly the effect of opening the neutral with (a) balanced loads
and (6) unbalanced loads. Why is the neutral usually grounded?

17. What in general is the effect of putting too heavy a load on one side
of a 3-wire system upon the voltage on that side of the system? Upon
the voltage on the other side of the system?

18. Sketch a method of obtaining a neutral by the use of two shunt
generators? What is the principal disadvantage of this method?

19. How may a storage battery be used for obtaining a neutral? In
general how does the current in the neutral wire divide when it reaches the
center of the battery?

20. Upon what principle does the balancer set operate? What deter-
mines which machine shall operate as a motor? As a generator? What
two methods are used to accentuate the motor and the generator actions?

21. Upon what principle does the 3-wire generator operate? Where
does the alternating current flow? The returning direct current from the
neutral? How is the direct current able to pass so readily back into the
armature?

22. How in general is power supplied to direct-current loads in the more
congested districts? What is the function of the feeders? The mains?

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QUESTIONS AND PROBLEMS 479

The junction boxes? Where are the house services connected? How are
the voltages at feeding points generally determined?

23. What type of generator is most conmionly used to supply power for
railways? How are such generators connected to the system?

24. Under what conditions does a single trolley suffice for transmitting
the power to the car? If a single trolley of the ordinary size is of insufficient
cross-section, what means can be taken to assist it in supplying the required
power? Why is the size of trolley not increased?

26. Under what conditions are multiple feeders employed? What is
the disadvantage of their use? How may this disadvantage be overcome?

26. Why does the return current from a trolley car leave the track?
What determines the paths which it follows? What damage, if any, occurs
at the point where the current enters a pipe? Where it leaves the pipe?

27. Name two methods by which electrolysis may be reduced. What
measurements give a good idea of the magnitude of stray currents between
pipes and track?

28. Sketch a typical central station load curve. Show how the habits of
a community determine the general shape of such a curve. Why is such a
load curve far more undesirable than a uniform load curve having the same
total kilowatt-hours?

29. What is meant by load factor? Is a high or a low load factor de-
sirable? Why?

30. How may a storage battery smooth out a station load curve? When
should the battery be charged? Discharged? ,Why are storage batteries
not more generally used for this purpose?

31. Where can storage batteries be used efficiently to carry the load
in off-peak times? For what purposes are they now commonty used by
central stations? Where should they be located? Under what conditions
is a battery very useful to a central station?

32. What difficulty is met when an attempt is made to operate storage
batteries in conjunction with a power plant? What simple method may be
used to control the battery load? What is the objection to this method?

33. Upon what simple principle do the counter-electromotive force cells
operate? What is the chief advantage of this method of control over the
resistance method?

34. What is meant by end cell control? How is such a battery charged?
In what manner is the connection changed from one cell to the next without
opening the circuit or dead-short circuiting the batteries?

36. What is meant by a "floating" battery? What is the purpose of
such a battery? Why is it often necessary to install auxiliary means for
accentuating the battery charge and discharge with change of load ? Sketch
the connections of one simple method for accomplishing this purpose.

36. Why will a battery placed at the end of a long feeder tend to equalize
the station load without auxiliary apparatus for charging and discharging?
Under what conditions does such a battery "float? "

37. What is the essential difference between the series system and the
parallel system of distribution? In the series system what is the effect of

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480 DIRECT CURRENTS

attempting to remove a load by opening the circuit? How is a load cut
out in a series system 7

38. By what devices is a series system supplied? What are the advan-
tages of the series system ? Where does its field of application lie ? Sketch
the layout of two different systems of series arc distribution. Name the
advantages of each.

PROBLEMS ON CHAPTER XIV

879. 140 kw. are transmitted a distance of 1,000 ft. over a cable of such
size that there is a potential difference of 215 volts at the load with 225
volts at the bus-bars, (a) What size of feeder is used, assuming that a
mil-foot of copper has a resistance of 10 ohms? (h) What is the weight
of copper if a<subic inch weighs 0.32 lb.?

880. Repeat problem 379 with the same power, the same loss, the same
distance and the same percentage line drop, but with 550 volts at the
load. How do the weights of copper compare with the respective voltages
in the two cases?

881. A 10-hp. motor is fed from a switchboard the bus-bars of which are
maintained at 115 volts. The motor is located at a distance of 500 ft. from
the switchboard and it is desired to have a voltage of 110 at the motor
terminals when the motor is carrying its full load of 10 hp. What must be
the diameter (mils)?

(a) Of the copper wire used to connect the motor to the switchboard?
Assume a temperature of 50® C. 1 hp. = 746 watts. The efficiency of the
motor is 86 per cent.

(6) If copper weighs 0.32 lb. per cu. in., what will be the weight of the
wire in (a)?

(d) Repeat (a) and (6) for a switchboard voltage of 550 and the same per
cent, drop to the motor.

882. A certain street is 2,000 ft. long. It is illuminated by eleven 20O-
watt multiple-connected lamps placed 200 ft. apart. No. 4 A.W.G. con-
ductors are used to supply this system. The voltage at the feeding end of
the street is 120 volts. What is the voltage drop between each two adjacent
lamps? What is the voltage at the last lamp? Assume that each lamp
takes 2.0 amp.

883. If the lamps of problem 382 are fed by the anti-parallel system (see
Fig. 343 (a), page 386), No. 4 wire still being used, determine the voltage
at the lamps on the two ends of the street. Compare their absolute
voltage and their difference of voltage with the results of problem 382.

884. A load of 100 amps, is situated 800 ft. from 600-volt bus-bars. 1,000
ft; farther on a second load of 65 amps, is located. A 4/0 annealed copper
feeder runs from the bus-bars to the first load. No. 1 wires run from the 100
amp. load to the 65 amp. load, (a) What is the voltage at each load?
(6) What is the weight of copper used?

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QUESTIONS AND PROBLEMS

481

386. (a) Determine the size of a uniform feeder which will have the same
weight as the two feeders of problem 384. (6) Determine the voltage at each
load with this uniform feeder, (c) Under which condition is the copper
most effectively utilized?

386. It is desired to operate 40 75-watt lamps on one circuit. Compare
the sizes of wire necessary to feed these lamps when all are connected in
parallel across 110 volts and when the lamps are connected in series groups
of two across 220 volts. (Use Table Appendix D, page 410.)

ii» V.

8a5

^-^

*lBa

1:5 v.

Fig. 388 a.

387. Repeat problem 386 for an Edison 3-wire system, assuming that
the neutral is the same size as the outer wires.

388. Fig. 388A shows an Edison 3-wire system with various loads. Indi-
cate the current and its direction at each of the points a-k inclusive.

389. If the neutral is cut at point X, Fig. 388A, find the voltages across
the two sides of the system, assuming that the load resistances do
not change. Neglect the drop in the mains themselves.

390. Find the voltages across loads A and B, Fig. 390 A, if loads A and B
are each 40 amp.

1 5 v.

z: jg-0.2 n

llfiV.

— 7?=0.lO

FlQ. 390 A.

391. Repeat problem 390 when load A is 60 amp. and load B is 20
amp.

892. Find the voltage across each load, problem 391, which would occur
if the neutral were opened.

393. Determine the voltage across each of the loads AB and BCj Fig.
393 A, and also the voltage across the motor.

394. Repeat problem 393 with the motor connected between the neutral
and the negative conductor. Owing to the fact that the voltage is halved
the motor must now take approximately 200 amp. to develop its former
power.

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482

DIRECT CURRENTS

895. Find the current in each machine of the balancer set of Fig. 395^ and
indicate which machine is the motor and which is the generator. There ia
110 volts across each machine and the efficiency of each machine is 80 per
cent. How much current does the main generator deliver?

121

-3000 ttr-

■h 500.000 P.M.

la V.

Z 2BO.00O O.M.

~ 500.000 0. M.

<-300ftr>-

150a;

100a:

Motor

Fig. 393A.

896. Solve problem 395 when there is a total load of 100 amp. connected
between the neutral and negative main and there is no load on the positive
side.

897. A 4/0 hard-drawn copper trolley wire runs from 600-volt bus-bais
to a station 6 miles out. For 4 miles it is paralleled by a 350,000 CM. feeder
which feeds it every quarter mile. (See Fig. 357(6), page 397.) The 4/0
wire has a resistance of 0.26 ohm per mile and the 350,000 CM. feeder has a
resistance of 0.163 ohm per mile. The resistance of the track and ground
return is 0.05 ohm per mile. Find the voltage at a car 5 miles out and
taking 60 amp. What is the voltage at the end of the line?

^aoo.ooo c.M.-N^

Fig. 395 a.

898. (a) Find the voltage at the car in problem 397, when the car is 4
miles from the power station, (b) When the car is 2 miles from the station.

899. Fig. 399A shows a 5-mile length of hard-drawn 4/0 copper trolley
wire. This is fed by three 300,000 CM. multiple feeders each feeding at
points l}^ miles apart. Find the equivalent resistance of the trolley and
feeders to the end of the line.

400. Find the voltage at a car, Fig. 399A, when it is at the end of the line
and taking 100 amp. The station voltage is 600 volts and the ground
and track resistance is 0.04 ohm per mile.

401. Find the voltage at the car of problem 400 when the car is 3 mile?
from the station and taking 100 amp.

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QUESTIONS AND PROBLEMS 483

402. Repeat problems 400 and 401 for a sectionalized trolley, the insu-
lated sections being at a, 6, c, Fig. 399 A. Which system of feeding is the
more economical of copper?

403. A central station has a peak load of 6,200 kw. It delivers 52,000
kw.-hr. over a 24-hour period. What is its daily load factor?

404. A storage battery helps the station of problem 403 by taking 1,600
kw. off the peak, it being necessary to use this battery for an hour and a half.
If the battery efficiency is 85 per cent, and it is charged off peak, what is
the new load factor of the station?

406. A storage battery consists of 64 cells connected in series, each cell
having an electromotive force of 2.1 volts and a resistance of 0.004 ohm.
It is desired to allow this battery to discharge at a 40-amp. rate into 115-
volt bus-bars, (a) What series resistance is necessary? (6) How much
power is developed within the battery? (c) How much is lost in the battery-
resistance? (d) How much is lost in the series resistance?

406. If counter-electromotive force cells, each having an electromotive
force of 2.08 volts, were used in problem 405, how many would be necessary?
Neglect the internal resistance of the counter-electromotive force cells.

407. If end cell control were used in problem 405 how many cells would be
cut off the end of the battery?

408. A 4/0 hard-drawn copper trolley, having a resistance of 0.26 ohm
per mile, extends 5 miles from 600-volt bus-bars. The track return has a
resistance of 0.05 ohm per mile. A storage battery consisting of 260 cells
floats at the end of the line. Each cell has an electromotive force of 2.08
volts and a resistance of 0.002 ohm. For what value of current at the
battery terminals will the battery "float?"

409. What current must a car 3 miles from the station of problem 408
taJte in order that the battery may "float?"

410. At what rate does the battery charge when there is no car at all on
the system?

411. When a car at the battery takes 120 amp., how much current
does the station supply and how much does the battery supply ? How much
power is supplied by each?

412. Repeat problem 411 for a car taking 120 amp., 3 miles from the
station.

413. A series arc generator supplies 60 500-watt 6.6-amp. magnetite
arcs over a No. 6 cable, which has a resistance of 0.395 ohm per 1,000 ft.
The length of the arc circuit is 1© miles. What is the terminal voltage of
the generator and what is the efficiency of transmission?

"' 414. Repeat problem 413 for an 80-lamp circuit in which the length of
the circuit is 15 miles.

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INDEX

Absolute potential, 52
Accumulator (see Storage Battery),

96
Alloys, 43

Aluminum conductors, 47
American Wire Gage (A.W.G.), 44
Ammeter, definition of, 128
hot-wire tjrpe, 136
shunts, 131
solenoid type, 128
Weston, 129
Ampere, definition of, 48
Ampere-turn, definition of, 170

determination of, 176
Annealed Copper Standard, 45
Anode, definition of, 85
Anti-parallel feeder system, 384
Armature, characteristic, 300
coils, 225

construction of, 251
electromotive force of, 257, 316
paths through, 230
reaction, 267
calculations of, 271
components of, 270
compensation of, 274
laminated pole cores, 274
slotted pole faces, 275
Thompson-Ryan method,
275
in multipolar machines, 272
of a motor, 319
resistance, 355

measurement of, 355
losses due to, 355
windings (see Windings), 222
Astatic watt-hour meter, 168

485

Automatic starting box, 335

Cutler-Hammer, 335

Electric Controller and Mfg. Co.,
336
Ayrton shunt, 128

Back electromotive force of motor

armature, 316
Back pitch of windings, 225, 239
Balancer set, 391

Bar magnet, effect of breaking a, 3
Battery, 84

anode, definition of, 85
cathode, definition of, 85
cells, in parallel, 73

series, 73

series-parallel, 75
charging, 71, 111
Clark cell, 93
Daniell cell, 89
definition of, 86 -
dry cell, 94

Edison-Lalande cell, 91
electrodes, definition of, 85
electrolyte, definition of, 85
electromotive force of, 68
floating, 403
gravity cell, 90
grouping, for best economy, 76

for maximum current, 76

for quick action, 76
internal, resistance of, 69, 87

voltage of, 68
lead cell, 97
Le Clanch^ cell, 91
nickel-iron-alkaline, 115
polarization of, 88

remedies for, 89

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486

INDEX

Battery, primary cell, 86
principles of, 84
secondary cell, 86
storage (see Storage Battery),

96
terminal voltage of, 68
voltage drop in, 68
Weston Standard Cell, 92
Boosters, regulation of storage bat-
tery discharge with, 403
series generators as, 304
Brakes, cradle dynamometer, 360
Prony, 348
rope, 351
Braking, dynamic, 347
British Thermal Unit (B.t.u.), 62,

407
Browne and Sharpe Wire Gage, 44
Brush Arc machine, 303, 405
Brush, construction, 255

position, in a dynamo, 221, 228,
244
in a generator, 281
in a motor, 319
rocker ring for holding, 256
Building up of generator, 264

Cable testing, 147

Murray loop for locating a

ground, 147
total disconnecion, location of,

213
Varley loop for locating
ground, 148
Calibration curve of ammeter, 159
Capacitance, calculation of, 209
of co-axial cylinders, 211
of parallel plates, 209
definition of, 202
measurement of, 211

ballistic galvanometer

method, 211
bridge method, 213
of parallel condensers, 205
of series condensers, 206

Capacities of storage batteries, 115
Cathode, definition of, 85
Characteristics of generators, 257

armature electromotive force,
257

armature reaction, 267

commutation, 276

compound (see Compound Gen-
erator), 295

effect of speed on, 305

regulation, 292

saturation curve, 258

series (see Series Generator),
301

shimt (see Shimt Generator),
264

total, 293
Characteristics of motors, 309

compoimd, 328

series, 324

shunt, 321
Charge, electrostatic, 198
Charging of storage batteries, 71,
111

booster method, 112

constant current method. 111

constant potential method, 112
Chemical reaction, of lead cell, 99
of nickel-iron-alkaline battery,
116
Circuit breakers, 377
Circular mil, definition of, 38

foot, definition of, 39
Clark cell, 93

Closed loop feeder system, 385
Coefficient of coupling, 195
Coercive force, 181
Coils, dummy, 242

formed, 225
Commutating, of a motor, 321

poles, 285
Commutation, 276

high mica, 283

sparking due to, 281

undercut mica, 283

with commutating poles, 285
(Commutator construction, 253

Digitized by VjOOQIC

INDEX

487

Compass, magnetic, 8
Compensation of amature reaction,

274
Compound generator, 295

characteristics of, 296
efifect of speed on, 299
over compounding, 297
under compounding, 297

long shunt, connection for,
296

parallel operation of, 374

series field diverter for, 298

short shunt, connection for,
296
Compound motor, 328

characteristics of, 328

cumulative, 328

differential, 328
Condensers, charge of, 202

definition of, 202

energy stored in, 208

parallel connection of, 205

series connection of, 206
Conductance, definition of, 36

specific, 36
Conductivity, definition of, 36

per cent., 36
Conductors, 32, 46

aluminum, 47

copper, 46

field around, 17

iron, 47

silver, 46

steel, 47
Consequent poles, 5
Constant potential, battery charg-
ing, 112

distribution system, 383

feeder system, 384
Copper, Standaid Annealed, 45
Corkscrew rule, 19
Coiona, 201

Coulomb, definition of, 48
Counter electromotive force, of
motor, 316

demonstration of, 318
Coupling, coefficient of, 195

Cradle dynamometer, 360
Creeping in winding, 243
Critical field resistance, 265
Cumulative compound motor, 328
Current, decay in inductive circuit,
189
measurement, 53

with potentiometer, 168
rise in inductive circuit, 187
unit of, 48
Cutler-Hammer automatic starter,
335

Damping of galvanometers, 125
Daniell cell, 89.
D'Arsonval galvanometer, 123
Decade bridge, 144
Development of a winding, 227
Dielectric, constants, 204
table of, 205

materials, 202

strength, 202
Differential compound motor, 328
Direct current, definition of, 220

production of, 220
Disc dynamo, 305
Discharge switch, 190
Distribution systems, 380

constant voltage, 383

electric railway, 396

series, 405

storage battery, 399

three-wire, 385

Thury System of, 303, 380
Diverter, series field, 298
Dobrowolsky method, 394
*Doubly re-entrant winding, 235
Drop-wire, 157
Drum winding, 223
Dry cell, 94
Dummy coil, 242
Duplex winding, 235
Dynamic, braking, 347

electricity, 198
Dynamo construction, 249

Digitized by VjOO^IC

488

INDEX

Dynamo, construction, 249
armature, 251
brushes, 255
commutator, 253
cores, 249
field coils, 254

cores, 250
frame, 249
shoes, 250
efficiency of, 359
heating of, 369
losses in, 355
copper, 355

armature, 355
determination of, 365
friction, 358
iron, 356

eddy currents, 356
hysteresis, 357
pole face, 358
series field, 355
shunt field, 355
stray power, 359
measurement of, 361
magnetic calculations in, 179
rating of, 368

windings (see Windings), 222
Dynamometer, cradle, 360

Earth's magnetism, 15

intensity of, 16
Eddy current losses, 356
Edison battery, 115
applications of, 118
charging of, 117
chemical reaction of, 116
Edison three-wire system, 375
advantages of, 385
effect of open neutral on, 387
methods of obtaining neutral
for, 390
balancer set, 391
storage battery, 390
three-wire generator, 394
two-generator, 390
Toltage unbalancing in, 388

Edison-Lalande cell, 91
Efficiency of dynamos, 359
Electric batteries, 84
Electric railway distribution system,

396
Electric Controller and Mfg. CJo.

automatic starter, 336
Electrical units, . definition of, 48

ampere, 48

coulomb, 48

farad, 204

henry, 184

joule, 60, 407

kilowatt, 59

kilowatt-hour, 60

ohm, 32

volt, 48

watt, 58

watt-second, 60, 407
Electrode, definition of, 85
Electrolysis, 397

Electrolyte, defintion of, 85, 105
Electromagnet, plunger type, 23
Electromagnetism, 17
Electromotive force, 48

generated in armature, 215, 257

induced, 184

in motor armature, 316

of battery, 68

of self induction, 186
calculation of, 190
Electroplating, 120
Electrostatic, charges, 198

field,200

induction, 199

lines, 200
of force, 210
Electrotyping, 121
End cells, 402
Energy, efficiency of conversion. 61

of magnetic field, 191

stored in condenser, 208

units of electrical, 60
Equalizing connections in windings,

236
Exide Vehicle battery, 109
Extension coils, 135

Digitized by VjOOQ IC

INDEX

489

Farad, definition of, 204
Faraday disc dynamo, 305
Feeders, 395

estimation of, 65

potential diop in, 63

power loss in, 67

systems of, 384
Field, around a conductor, 17

coil construction, 254

control of speed by, 342

discharge switch, 190

intensity, unit of, 7

resistance line, 262
Fleming's Left Hand Rule, 311

Right Hand Rule, 218
Floating battery, 403
Flux density, 7, 171
Force, acting on a conductor, 309

coercive, 181

lines of, 6

magnetic, 5
Forced winding, 243
Four-point starting box, 332
Fractional pitch winding, 224
Friction losses, 358
Fringing, of electromagnetic lines, 171

of electrostatic lines, 210 .
Front pitch, 226, 239

G
Gage, American Wire (A.W.G.), 44
Galvanometer, 123

Ayrton shunt for, 128

damping of, 125

D'Arsonval, 123

methods of reading, 124

shunts, 126

Weston portable, 130
Gauss, definition of, 7, 171
Generated electromotive force, 215

equation of, 216

in armature, 215, 257

light hand rule for, 218
Generator, armature characteristic
of, 300

armature reaction of, 267

Generator, characteristics of, 257
effect of speed on, 305
commutation, 276
compound (see Compound Gen-
erator), 295
definition of, 215
electromotive force of, 257
homopolar, 305
regulation of, 292
saturation curve of, 258
determination of, 261
field resistance line, 262
hysteresis, 260
series (see Series Generator),

301
shunt (see Shunt Generator),

264
total characteristic of, 293
unipolar, 305

windings (see Windings), 222
Gilbert, definition of, 170
Gould ploughed plates, 100
Gradient, potential, 202
Gram-calorie, 62, 407
Gramme-ring winding, 222
Gravity cell, 90

Hand rule, 19

Heat, mechanical equivalent of, 62

Heating of dynamos, 369

measurement of, 370

Standardization Rules for, 368
Henry, definition of, 184
High mica, 283
Homopolar generator, 305
Horseshoe, magnet, 13

solenoid, 24
Hot-wire instruments, 136
Hydrometer, 105
Hysteresis, 181, 260

coefficients, 183

losses due to, 182, 357

Induced electromotive force, 184
in generator armature, 215

Digitized by VjOOQ IC

490

INDEX

Induced electromotive force, in
motor armature, 316
rule for direction of, 218
Inductance, 183
mutual, 193
self, 183
Induction, coil, 196

electromotive force of self, 186

calculation of, 190
electrostatic, 199
lines of, 2, 171, 200
magnetic, 11
Inductive circuit, 186

decay of current !n, 189
rise of current in, 187
Instruments, 122

ammeters (see Ammeter), 128
galvanometers, 123
damping of, 125
shunts for, 126
hot-wire, 136
voltmeters, 134
wattmeter, 161
Insulation testing, 150
Insulators, 32
International ohm, 49

volt, 48
Interpoles, 285
Iron, as a conductor, 256
losses, 356

eddy current, 356
hysteresis, 357
pole face, 358
Iron-clad, Exide battery, 102
solenoid, 22

Jagabi tachoscope, 353
Joule, 60, 407
Joule's Law, 62
Junction boxes, 395

Kapp opposition test, 365
Kilowatt, definition of, 69
Kilowatt-hour, definition of, 60

Kirchhoff's Laws, 77

applications of, 78, 82

Ladder sjrstem of distribution, 396
Laminated, magnets, 14
pole cores, 250, 274
Lap winding, 224

development of, 227
equalizing connections in, 236
number of paths in, 233
requirements of, 228
simplex, 226
uses of, 246
Lead cell, 97

chemical reaction of, 99
Leakage, magnetic, 27
LeClanch6 cell, 91
Leeds & Northrup, dial bridge, 145
low resistance potentiometer,
155
Left hand rule, Fleming's, 311
Lenz's Law, 186
Lifting magnet, 26
Lincoln motor, 343
Linkages, definition of, 183
Load, curve, 399

factor, 399
Lodestone, 1
Losses, dynamo, 355
armature, 355
determination of, 365
friction, 358
iron, 356
eddy current, 356
hysteresis, 357
pole face, 358
series field, 356
shunt field, 356
stray power, 359
measurement of, 361
hysteresis, 182

Magnet, artificial, 1
electro-, 23
exploration of field around a 6,

Digitized by VjOOQIC

INDEX

491

Magnet, fonns of, 13
laminated, 14
lifting, 26
natural, 1
neutral zone of, 3
wire, 409
Magnetic, blowout, 338

calculations for dynamos, 179
circuit, definition of, 3, 169
circuit of dynamos, 27
compass, 8
field, 2

around ccJnductor, 17

due to parallel conductors, 19

energy of, 191

flux of, 171

intensity of, 7

law of, 12, 174

magnetomotive force of, 170

of solenoid, 20
figures, 10
flux density, 7, 171
force, 5
induction, 11
leakage, 27
materials, 1
permeability, 171
poles, 2, 5,
pull, 197
reluctance, 170
screens, 14
separator, 27
units, 170

ampere-turn, 170

gauss, 171

gilbert, 170

maxwell, 171

oersted, 170
Magnetizing, 15
Magnetism, 1
earth's, 16

intensity of, 16
Magnetization curves, 173
of dynamos, 268
tjrpical, 177
uses of, 178
Magneto, 14, 363

Magnetomotive force, 170
Manchester plate, 100
Maxwell, definition of, 171
Measurement, cuirent, 63, 168
power, 160
resistance, 137
voltage, 167
Mechanical equivalent of heat, 62
Megohm, definition of, 32
Mho, definition of, 36
Mica, high, 283

undercut, 283
Microfarad, 204
Microhm, definition, 32
Mil, circular, 38

foot, 38
Motor, 309

compound (see Compound

Motor), 328
counter electromotive force of,

316
Lincoln, 343
principle of, 309
series (see Series Motor), 324
shunt (see Shunt Motor), 321
speed control, 339

armature resistance method,

339
field control, 342
Lincoln motor, 343
multivoltage, 341
of railway motor, 346
Stow method, 343
Ward Leonard system, 341
starters (see Starting Box), 329
Stow, 343
testing, 348

cradle dynamometer, 360
Prony brake, 348
rope brake, 361
torque, 313
Multiple unit control, 345
Multiplex winding, 233
doubly re-entrant, 236
duplex, 235
singly re-entrant, 236
Multipliers, 136

Digitized by VjOO^IC

492

INDEX

MuUi-voltage speed control, 341
Murray loop, 147
Mutual inductance, 193

coefficient of coupling, 195

effect of iion on, 196

Neutral zone of magnets, 3
Nickei-iron-alkaline battery, 115
applications of, 118
charging of, 117
chemical reaction of, 116

Oersted, definition of, 170

Ohm, definition of, 32

Ohm's Law, 53

Open loop series distribution, 406

Open spiral feeder system, 384

Opposition test, 365

Parallel, batteries, 73

circuits, 55

conductors, fielii due to, 19

loop feeder system, 406

operation, 372

compound generators, 374
shunt generators, 372
Pasted plate, 101
Per cent, conductivity, 36
Permeability, curve for cast steel,
174

definition of, 171

of iron and steel, 173
Permeance, definition of, 170
Permittivity, 205
Pilot cell, 106
Pitch of winding, 225

back, 225, 239

front, 226, 239
Plants plate, 100
Plunger electromagnet, 23
Poggendorf method, 155
Polarization, 88

remedies for, 89

Pole, commutating, 285, 321
consequent, 5
-face losses, 358
interpole, 285, 321
magnetic, 2

strength, definition of, 5
Potential, absolute, 52
difference, 48, 51
drop in feeders, 63
measurement of, 53
Potentiometer, 153

current measurement with, 158

standard resistances, 158
Leeds & Northrup low resist-
ance, 155
voltage measurement with, 157
drop wire, 157
volt box, 157
Power, distribution systems, 380
constant potential, 383
Edison three-wire, 385
electric railway, 396
feeder systems, 384
anti-parallel, 384
closed loop, 385
open loop, 406
open spiral, 384
parallel loop, 406
return loop, 384
series-parallel, 385
series, 405

size of conductor for, 382
storage battery, 399
three-wire, 385
Thury, 303, 380
voltage of, 381
weight of conductor for,
381
electrical imit of, 58
loss, in dynamos, 355

in feeders, 67
measurement, 160
Primary cell, definition of, 86
requirements of, 86
Weston, 92
Production of direct current, 220
Pfeogressive winding, 226, 239

Digitized by VjOOQIC

INDEX

493

Prony brake, 348

cooling of, 350

power of, 350

zero reading of, 350
Pull due to magnetic field, 197

Q
Quantity of electricity, definition of,
48

R
Railway motors, 328

multiple unit control, 345
speed control, 345
Rating, of dynamos, 368

of storage battery, 110
Reaction, armature, 267, 319

chemical, 99, 116
Regulation, speed, 323

voltage, 292
Regulator, Tirrill, 306
Relay, telegraph, 24
Reluctance, definition of, 170

unit of, 171
Remanence (magnetic induction) ,

181
Resistance, definition of, 31, 40
insulation, 150

International Standard of, 49
measurement of, 137
voltmeter method, 139
voltmeter-ammeter method,

137
Wheatstone Bridge, 141
parallel connection of, 37
relation to direction of current,

32
series connection of, 37
standard, 158
temperature coefiicient of, 41

table of, 43
unit of, 32, 40
units for starting boxes, 338
Resistivity, 34
table of, 40
volume, 35
Retrogressive winding, 227, 239, 240
Return loop feeder system, 384

Right hand rule, Fleming's, 218
Ring winding, 222
Rocker ring, 256
Rope brake, 351

Saturation curve, 258
determination of, 261
effect of hysteresis on, 260
field resistance line, 262
Screens, magnetic, 14
Secondary cell, 86

(see Storage Battery), 96
Self-induction, electromotive force
of, 186
calculation of, 190
Separator, in batteries, 104

magnetic, 27
Series, batteries in, 73 ^
circuits, 54

condensers, 206 ^^

distribution, 405
field, calculation of turns for,
300
diverter for, 298
loss in, 356
uses of, 295
generator, 301

Brush Arc machine, 303
characteristics of, 302
Thompson-Houston, 303
Thury system, 303
used as booster, 304
motor, 324

characteristics of, 325
railway, 328
speed equation of, 325
starting boxes for, 334
no load release, 334
no voltage release, 334
torque of, 324
uses of, 326
parallel system, 385
resistances, 37
turns, determination of, 300
Shunt, ammeter, 131
Ayrton, 128

Digitized by VjOO^IC

494

INDEX

Shunt, field, loss in, 366
resistance line, 262
for galvanometer, 126
generator, 264

armature reaction of, 267
building up of, 265
characteristics of, 288
commutation oi, 276
critical field resistance, 265
failure to build up, 266
parallel operation, 372
regulation, 292
motor, 321
characteristics of, 323
speed, 322

regulation, 323
starting torque, 324
uses of, 324
Silver conductors, 46
Simplex winding, 226
Sine wave, 219
Slide wire bridge, 144
Slotted pole faces, 275
Solenoid, ammeter, 128
commercial, 22
definition of, 21
horseshoe, 24
iron-clad, 22
magnetic field of, 20
plunger, 22
Spark coil, 192
Sparking at commutator, 281

effect of brush position on, 282
high mica, 283
undercut mica, 283
Specific, conductance, 36
gravity, 106

table of, 408
inductive capacity, 204

table of, 205
resistance, 34 .
Speed, control of motors, 339

armature resistance method,

339
field, 342

Lincoln method, 343
multi-voltage, 341

Speed, control of motors, railway, 345
Stow method, 343
Ward Leonard system, 341
equation for determining, 319 '
measurement of, 353
Jagabi tachoscope, 353
magneto and voltmeter, 353
revolution counter, 353
tachometer, 353
regulation, 323
Standard, Annealed Copper, 45
Clark cell, 93
resistances, 158
Weston cell, 92
Starting boxes, 329
* automatic, 335

Cutler-Hammer, 335
Electric Controller & Mfg.
Co., 336
four-point, 332
magnetic blowouts for, 338
resistance units for, 338
series motor, 334
no load release, 334
no voltage release, 334
speed adjustment, 333
three-point, 331
Static electricity, 198
Stationary battery, 103
Steel conductors, 47
Storage battery, 96
capacity of, 116
charging. 111
booster method, 112
constant current method, 111
constant potential method,
112
distribution systems, 399

counter electromotive force,

control of, 401
floating battery, 403
resistance control, 401
Edison, 115
efficiency, 118
electrolytes, 105
Gould ploughed plates, 100
installation, 107, 114

Digitized by VjOOQIC

INDEX

495

Storage battery, Iron-clad Exide, 102

lead cell, 97

chemical reaction of, 99

Manchester plate, 100

nickel-iron-alkaline, 115

pasted plate, 101

pilot cell, 106

Plants plate, 100

rating, 110

separators, 104

specific gravity, 106

stationary, 103

tanks, 103

temperature, 114

vehicle, 108
Stow motor, 343
Stray power, 369

curves of, 363

measurement of, 361
Switch, discharge, 190
Syringe hydrometer, 106
Systems of feeders, 384

anti-parallel, .384

closed loop, 385

Edison three-wire, 385

open loop, 406

open spiral, 384

parallel loop, 406

return loop, 384

series-parallel, 385

Tables, American Wire Gage, 44

current capacity of wires, 410

layer windings, 409

relations of units, 407

resistivity, 40

specific gravities, 408

temperature coefficients of re-
sistance, 43
Tachometer, 353
Tachoscope, Jagabi, 353
Tanks for batteries, 103
Telegraph relay, 24
Temperature coefficient of resist-
ance, 41

table of, 43

Thermal units, 62, 407
Thompson-Houston generator, 303,
405
-Ryan method, 276
Thomson watthour meter, 163
Three-point starting box, 332
-wire generator, 394
-wire system, Edison, 385
advantages, 386
effect of open neutral on, 387
methods of obtaining neutral,
390
balancer set, 391
storage battery, 380
three-wire generator, 394 »
two generator, 390
voltage unbalancing, 388
-wire watt-hour meter, 167
Thury system, 303, 380
Time constant, 187
Tirrill voltage regulator, 306
Torque, definition of, 312
developed by motor, 313
series motor, 324
shunt motor starting, 324
imits of, 312
Total characteristic of generator,

293
Types of generators, 263
compound, 295
series, 301
shunt, 264
Typical magnetization curves, 177

Undercut mica, 283
Unipolar generator, 305
Units, magnetic, 170
relations of, 407

Varley loop, 148
Vehicle battery, 108
Exide, 109

Digitized by

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496

INDEX

Volt, box, 157

definition of, 48

International, 48
Voltage, generated by rotating coil,
219

gradient, 202

measurement, 52

with potentiometer, 157

regulation of generator, 202

regulator, Tirrill, 306
Voltmeter, 134

extension coils, 135

multipliers, 135

Ward-Leonard system, 341
Watt, definition of, 58
-hour meter, 162
astatic, 168
Thomson, 163

adjustments of, 165
three-wire, 167
meter, 161

-second, definition of, 60
Weber's theory of magnets, 3
Weston, ammeter, 129

portable galvanometer, 130
Standard Cell, 92
normal, 94
secondary, 94
Wheatstone bridge, 141

method of using, 143
Winding, closed circuit, 223

Winding, comparisons of lap and
wave, 245
creeping, 243
development of, 227
drum, 223
dummy coil, 242
forced, 243
formed coils for, 225
fractional pitch, 224
Gramme-dng, 222
lap, 224

development of, 227

equalizer connection, 236

multiplex, 233

paths through armature, 230

requirement for, 228

simplex, 226

uses of, 246
multiplex, 233

doubly re-entrant, 235

duplex, 235

singly re-entrant, 236
numbering slots for, 226
open circuit, 221
pitch of, 225
progressive, 226
table, 228
wave, 238

brushes requiried for, 244

paths through armature, 244

progressive, 239

retrogressive, 239

uses, 246
Wire gage, American, 44

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