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£A«t
\ i v ,4*,'
\
An Electrical Library*
By PROF. T. O'CONOR SLOANE.
How to become a Successful Electrician.
PRICE tl.OO.
Electricity Simplified.
PRICE, $1.00.
Electric Toy Making, Dynamo Building, etc.
PRICE, tl.OO.
Arithmetic of Electricity.
PRICE, tl.OO.
Standard Electrical Dictionary.
price, ta.oo.
NORMAN W. HENLEY & CO., Publishers,
132 Naeeau Street, New Tork.
Arithmetic of .electricity
A PRACTICAL TREATISE ON ELECTRICAL CALCULATIONS OF
AU, KINDS REDUCED TO A SERIES OF RULES, ALL OF THE
SIMPLEST FORMS, AND INVOLVING ONLY ORDINARY ARITH-
METIC, EACH RULE ILLUSTRATED BY ONE OR MORE PRAC-
TICAL PROBLEMS, WITH DETAILED SOLUTION OF EACH,
FOLLOWED BY AN EXTENSIVE SERIES OF TABLES.
BY
T. O'CONOR SLOANE, A.M., E.M., Ph.D.
AUTHOR OF
Standard Electrical Dictionary, Electricity Simplified,
Electric Toy Making, etc.
TWENTIETH EDITION
NEW YORK :
THE NORMAN W. HENLEY PUBLISHING COMPANY
132 Nassau Street
1909
Copyrighted, 1891,
BY
NORMAN W. HENLEY & CO.
Copyrighted, 1903,
BY
NORMAN W. HENLEY & CO.
Copyrighted, 1909,
BY
THE NORMAN W. HENLEY PUBLISHING CO.
4v.\v. .-«m.^r
X
PREFACE.
The solution of a problem by arithmetic, although
In some cases more laborious than the algebraic method,
gives the better comprehension of the subject. Arith-
metic is analysis and bears the same relation to algebra
that plane geometry does to analytical geometry. Its
power is comparatively limited, but it is exceedingly
instructive in its treatment of questions to which it
applies.
In the following work the problems of electrical en-
gineering and practical operations are investigated on
an arithmetical basis. It is believed that such treatment
gives the work actual value in the analytical sense, as
it necessitates an explanation of each problem, while
the adaptability of arithmetic to readers who do not
care to use algebra will make this volume more widely
available.
In electricity there is much debatable ground, which
has been as far as possible avoided. Some points seem
quite outside of the scope of this book, such as the intro-
duction of the time-constant in battery calculations.
Again the variation in constants as determined by dif-
ferent authorities made a selection embarrassing. It is
believed that some success has been attained in over-
coming or compromising difficulties such as those sug-
383491
iv PREFACE.
Enough tables have ^een introduced to fill the limits
of the subject as here treated.
The full development of electrical laws involves the
higher mathematics. One who would keep up with the
progress of the day in theory has a severe course of
study before him. In practical work it is believed that
such a volume as the Arithmetic of Electricity will
always have a place. We hope that it will be favorably
received by our readers and that their indulgence will
give it a more extended field of usefulness than it can
pretend to deserve.
PREFACE TO TWENTIETH EDITION.
The steady progress of electrical science in conjunc-
tion with a continued demand for this work have made
advisable a revision and extension of this book.
The author feels that in the matter which has been
added much more could have been said on the subjects
treated of, but, since a full exposition of each theme
would alone fill a volume, it is hoped that the practical
value of the rules, etc., will atone for the brevity of the
text.
In the preparation of this edition the author -would
express his indebtedness to A. A. Atkinson's excellent
work on Electrical and Magnetic Calculations and also
to the instruction papers of the Electrical Engineering
Course of the International Correspondence School of
Scranton, Pa. He would also express his thanks to
Henry V. A. Parsell, for his valued advice and assistance
in the preparation of the manuscript.
The Authob.
CONTENTS.
CHAPTER L
INTBODUCTOBY.
Space, Time, Force, Resistance, Work, Energy, Mass
and Weight.— The Fundamental Units of Dimension,
and Derived Units, Geometrical, Mechanical, and Elec-
trical.— C. G. S. and Practical Electrical Units.— Nomen-
clature.— Examples from Actual Practice 9
CHAPTER IL
ohm's law.
General Statement. — Six Rules Derived by Transpos-
ition from the Law. — Single Conductor Closed Circuits.
— Batteries in Opposition. — Portions of Circuits. — Di-
vided Circuits, with Calculation of Currents Passed by
Each Branch, and of their Combined Resistance 13
CHAPTER III.
BESISTANCE AND CONDUCTANCE.
Resistance of Different Conductors of the same Mater-
ial. — Relations of Wires of Equal Resistance. — Ratio of
Resistance of Two Conductors. — Specific Resistance. —
Universal Rule for Resistances. — Resistance of Wires
Referred to Weight. — Conductance. — Ohm's Law ex-
pressed in Conductance 26
Yi CONTENTS.
CHAPTER IV.
POTEOTIAL DIFFERENCE.
Drop of Potential in Leads, and Size of Same for Mul-
tiple Arc Connections. — Diminishing Size of Leads
Progressively
CHAPTER V.
CIBCT7LAB MILS.
The Mil.— The Circular Mil as a Unit of Area.— Cir-
cular Mil Rules for Resistance and Size of Leads.-
CHAPTER VL
SPECIAL SYSTEMS.
Three Wire System.— Rules for Calculating Leads in
Same. — Alternating Current System. — Ratio of Conver-
sion.— Size of Primary Wire.— Converter Winding. 46
CHAPTER VH
WOBK AND ENBBGY.
Energy and Heating Effect of the Current. — Differ-
ent Rules Based on Joule's Law. — The Joule or Gram-
Calorie. — Quantity of Heat Developed in an Active Cir-
cuit in a Unit of Time. — Watts and Amperes in Rela-
tion to Time.— Specific Heat. — Heating of Wire by a
Current.— Safety Fuses. — Work of a Current. — Elec-
trical Horse-Power.— Duty and Efficiency of Electrical
Generators 60
CONTENTS. Vil
CHAPTER VIII.
BATTERIES.
Arrangement of Battery Cells. — General Calcula-
tions of Current. — Rules for Arrangement of Cells
in a Battery. — Battery Calculations for Specified
Electromotive Force and Current — Efficiency of
Batteries. — Chemistry of Batteries. — Calculation
of Voltage. — Work of Batteries. — Efficiency of Bat-
teries, to Calculate. — Chemicals Consumed in a
Battery. — Decomposition of Compounds by a Bat-
tery. — Electroplating 66
CHAPTER IX.
ELECTRO-MAGNETS, DYNAMOS AND MOTORS.
The Magnetic Field and Lines of Force. — Per-
meance and Reluctance.— Magnetizing Force and
the Magnetic Circuit. — General Rules for Electro-
Magnets and Ampere-Turns for Given Magnetic
Mux. — Magnetic Circuit Calculations. — Leakage of
Lines of Force. — Example of Calculation of a Mag-
netic Circuit. — Dynamo Armatures. — Voltage and
Capacity of Armatures. — Drum Type Closed Cir-
cuit Armatures. — Field Magnets of Dynamos. — The
Kapp Line 82
CHAPTER X.
ELECTRIC RAILWAYS.
Sizes of Feeders. — Power to Move Cars 109
Till CONTENTS.
CHAPTER XI.
ALTERNATING CURRENTS.
Self-induction 115
CHAPTER XII.
CONDENSERS.
Page 121
CHAPTER XIII.
DEMONSTRATION OF RULES.
Some of the Principal Rules in the Work Demon-
strated 127
CHAPTER XIV.
NOTATION IN POWERS OF TEN.
The Four Fundamental Operations of Addition,
Subtraction, Multiplication, and Division in Powers
of Ten 136
TABLES.
A Collection of Tables Needed for the Operations
and Problems Given in the Work 139
ARITHMETIC OF ELECTRICITY.
CHAPTER L
INTRODUCTORY.
Space is the lineal distance from one point to
another.
Time is the measure of duration.
Force is any cause of change of motion of matter.
It is expressed 'practically by grams, volts, pounds
or other unit.
Resistance is a counter-force or whatever opposes
the action of a force.
Work is force exercised in traversing a space
against a resistance or counter-force. Force multi-
plied by space denotes work as foot-pounds.
Energy is the capacity for doing work and is
measurable by the work units.
Mass is quantity of matter.
Weight is the force apparent when gravity acts
upon mass. When the latter is prevented from
moving under the stress of gravity its weight can
be appreciated.
\Q ARITHMETIC OF ELECTRICITY.
Physical and Mechanical calculation, are based on
three fundamental units of dimension, as follows:
the unit of time — the second, T; the unit of length
— the centimeter, L; the unit of mass — the gram,
M. Concerning the latter it is to be distinguished
from weight. The gram is equal to one cubic centi-
meter of water under standard conditions and is
invariable; the weight of a gram varies slightly with
the latitude and with other conditions.
* Upon these three fundamental units are based the
derived units, geometrical, mechanical and electrical.
The derived units are named from the initials of
their units of dimension, the C. G. S. units, indi-
cating centimeter-gram-second units.
In practical electric calculations we deal with
certain quantities selected as of "convenient size
and as bearing an easily defined relation to the~
fundamental units. They are called practical
units.
The cause of a manifestation of energy is force;
if of electromotive energy, that is to say of electric
energy in the current form, it is called electromotive
force, E. M. F. or simply E. or difference of poten-
tial D. P. What this condition of excitation may
be is a profound mystery, like gravitation and much
else in the physical world. The practical unit of
E. M. F. is the yolt, equal to one hundred mil-
lions (100,000,000) C. G-. S. units of E. M. F. The
last numeral is expressed more briefly as the eighth
ZNTBODUCTOBY. 11
power of 10 or 10 8 . Thus the volt is defined as
equal to 10 8 0. G. S. units of E. M. F.
This notation in powers of 10 is used throughout
C. G. S. calculations. Division by a power of 10 is
expressed by using a negative exponent, thus 10~*
means loooiooos . The exponent indicates the number
of ciphers to be placed after 1.
When electromotive foroe does work a current is
produced. The practical unit of current is the
ampebe, equal to & 0. G. S. unit, or 10" 1 C. G. S.
unit, ib being expressed by 10~*.
A current of one ampere passing for one second
gives a quantity of electricity. It is called the
coulomb and is equal to 10" 1 C. G. S. units.
A coulomb of electricty if stored in a recipient
tends to escape with a definite E. M. F. If the
recipient is of such character that this definite E.
M. F. is one volt, it has a capacity of one Fabad
equal to t oooo^oooo or 10"* C. G. S. unit.
A current of electricity passes through some
substances more easily than through others. The
relative ease of passage is termed conductance. In
calculations its reciprocal, which is resistance, is
almost universally used. A current of one ampere
is maintained by one volt through a resistance of
one practical unit. This unit is called the Ohm and
is equal to 10* 0. Gr. S. units.
Sometimes, where larger units are wanted, the pre-
fix deka, ten times, heka* one hundred times, kilo, one
12 ARITHMETIC OF ELECTRICITY.
thousand times, or mega, one million times are used,
as dekalitre, ten liters, kilowatt, one thousand watts,
megohm, one million ohms.
Sometimes, where smaller units are wanted, the
prefixes, deci, one tenth, cenii, one hundredth, milli,
one thousandth, micro, one millionth, are used. A
microfarad is one millionth of a farad.
For the concrete conception of the principal units
the following data are submitted.
A DanielFs battery maintains an E. M. P. of 1.07
volt. A current which in each second deposits
.00033 grams copper (by electro-plating) is of one
ampere intensity and from what has been said the
copper deposited by that current in one second cor-
responds to one coulomb. A column of mercury one
millimeter square and 106.24 centimeters long has
a resistance of one ohm at 0° O. The capacity of the
earth is 1088800 farad. A Leyden jar with a total
coated surface of one square meter and glass one
mm. thick has a capacity of A microfarad. The
last is the more generally used unit of capacity.
These practical units are derived from the C. G.
S. units by substituting for the centimeter (C.) one
thousand million (10 9 ) centimeters and for the gram,
the one hundred thousand millionth (10~ u ) part of a
gram.
CHAPTER H.
ohm's law.
This law expresses the relation in an active
electric circuit (circuit through which a current of
electricity is forced) of current, electromotive force,
and resistance. These three factors are always pres-
ent in such a circuit. Its general statement is as
follows :
In an active electric circuit the current is equal to
the electromotive force divided by the resistance.
This law can be expressed in various ways as it is
transposed. It may be given as a group of rules, to
be referred to under the general title of ohm's law.
Rule 1. The current is equal to the electromotive
force divided by the resistance. c = —
R
Rule 2. The electromotive force is equal to the cur-
rent multiplied by the resistance. E = C R
Rule 3. The resistance is equal to the electromotive
force divided by the current. r = —
C
Rule 4. The current varies directly with the electro*
motive force and inversely with the resistance.
14 ARITHMETIC OF ELECTRICITY.
Rule 5. The resistance varies directly with the elee*
tromottve foree and Inversely with the current.
Rale 6. The electromotive force varies directly with
the current and with the resistance.
This law is the fundamental principle in most
electric calculations. If thoroughly understood it
will apply in some shape to almost all engineering
problems. The forms 1, 2, and 3 are applicable to
integral or single conductor circuits; when two or
more circuits are to be compared the 4th, 5th and 6th
are useful. The law will be illustrated by examples.
Single Conductor Closed Circuits.
These are circuits embracing a continuous con*
ducting path with a source of electromotive force
included in it and hence with a current continually
circulating through them.
Examples.
A battery of resistance 3 ohms and B. M. P. 1.07
volts sends a current through a line of wire of 55
ohms resistance ; what is the current?
Solution : The resistance is 3 + 55 = 58 ohms.
By rule 1 we have for the current 1 #- giving .01845
Ampere.
Note. — A point to be noticed here is that whatever is
included in a circuit forms a portion of it and its resistance
must be included therein. Hence the resistance of the
battery has to be taken into account. The resistance of a
battery or generator is sometimes called internal resistance
OEM'S LAW. 15
to distinguish it from the resistance of the outer circuit,
called external resistance. Resistance in general is
denoted by R, electromotive force by E, and current by
C.
A battery of B 2 ohms; sends a current of .035
ampere through a wire of E 48 ohms; what is the
E. M. F. of the battery?
Solution: The resistance is 48 + 2 = 50 ohms. By
Rule 2 we have as the E. M. P. 50 X .035 - 1.75
volts.
A maximum difference of potential E. M. F. of
30 volts is maintained in a circuit and a current of
191 amperes is the result; what is the resistance of
the circuit?
Solution: By Eule 3 the resistance is equal to
ffc= .157 ohms.
In the same circuit several generators or gal-
vanic couples may be included, some opposing the
others, i. e. connected in opposition. All such can
be conceived of as arranged in two sets, distrib-
uted according to the direction of current produced
by the constituent elements, in other words, so as
to put together all the generators of like polarity.
The voltages of each set are to be added together to
get the total E. M. F. of each set.
Rule 7. Where batteries or generators are | n opposi-
tions add together the E. M. F of all generators of like
polarity, thus obtaining two opposed E. M. F.s. Sub-
tract the smaller E. M. F. from the larger E. IS. F. to
16 ARITHMETIC OF ELECTRICITY.
obtain the effective E. M. F. Then apply Ohm's law on
this basis of E. M. F.
It will be understood that the resistances of all
batteries or generators in series are added to give
the internal resistance.
Examples.
There are four batteries in a circuit: Battery No,
1 of 2 volts, y 2 ohm; Battery No. 2 of 1.75 volts, 2
ohms; Battery No. 3 of 1 volt, 1 ohm; Battery No.
4 of 1 volt, 4 ohms constants; Batteries 1 and 4 are
in opposition to 2 and 3. What are the effective
lattery constants?
■-AAAAAAA/WW
Solution: Voltage = (2 + 1) — (1.75 + 1) = .25 volt.
Eesistance = J^ + 2+l + 4 = 7^ ohms, or .25
volt, 7j4 ohms constants.
What current will such a combination produce in
a circuit of 5 ohms resistance?
Solution: By Ohm's law, Rule 1, the current =*
.25 h- (1}4 + 5) = .02 amperes.
A battery of 51 volts E. M. F. and 20 ohms resist-
OHM'S LAW. 17
ance has opposed to it in the same circuit a battery
of 26 volts E. M. F. and 25 ohms resistance. A
current of }£ ampere is maintained in the circuit.
What is the resistance of the wire leads and con-
nections?
Solution: The effective B. M. P. is 51 — 26 = 25
volts. By Eule 3 we have 25 ■*- 1 = 200 ohms, as the
total resistance. But the resistance of the batteries
(internal resistance) is 20 4- 25 = 45 ohms. The re-
sistance of leads, etc. (external resistance), is there-
fore 200 — 45 = 155 ohms.
POBTIONS OF OlECUITS.
All portions of a circuit receive the same current,
but the E. M. P., in this case termed preferably dif-
ference of potential, or drop or fall of potential,
and the resistance may vary to any extent in differ-
ent sections or fractions of the circuit. Ohm's Law
applies to these cases also.
Examples.
An electric generator of unknown resistance main-
tains a difference of potential of 10 volts between its
terminals connected as described. The terminals
are connected to and the circuit is closed through a
series of three coils, one of 100 ohms, one of 50
ohms, and one of 25 ohms resistance. The connec-
tions between these parts are of negligibly low re-
18 ARITHMETIC OF ELECTRICITY.
sistance. What difference of potential exists be*
tween the two terminals of each coil respectively?
Solution: The solution is most clearly reached by
a statement of the proportion expressed in Rule 6,
viz. : The electromotive force varies directly with the
resistance. The resistance of the three coils is 175
ohms; calling them 1, 2, and 3, and their differences
of potential E 1 , E a , and E 1 , we have the continued
proportion, 175 : 100 : 50 : 25 :: 10 volts : E 1 : E a : E 1 .
because by the conditions of the problem the total
E. M. P. «= 10. Solving the proportion by the regu-
lar rule, we find that E 1 — 5.7, E* — 2.8 and E» — 1.4
volts.
The same external circuit is connected to a
battery of 30 ohms resistance. The difference of
potential of the 100 ohm coil is found to be 30 volts.
What is the difference of potential between the ter-
minals of the battery, and what is the E. M. F. of
the battery on open circuit, known as its voltage or
E. M. P. (one of the battery constants)?
Solution: The total external resistance is 100 + 50
+ 25 = 175 ohms. By Rule 6, we have 100 : 175 :: 30
volts : x = 52 j4 volts, difference of potential between
the terminals of the battery. The current is found
by dividing (Rule 1), the difference of potential of
the 100 ohm coil by its resistance. This E. M. F. is
30. The current therefore is -Aftr amperes. The to-
tal resistance of the circuit is that of the three coils
or 175 ohms plus that of the battery or 30 ohms, a
0HW8 LAW. 19
total of 205 ohms. To maintain a current of tffo
amperes through 205 ohms (Rule 2), an E. M. F. is
required equal to i% X 205 volts or 61^ volts.
Divided, Branched or Shukt Circuits.
A single conductor, from one terminal of a gener-
ator may be divided into one or more branches
which may reunite before reaching the other ter-
minal. Such branches may vary widely in resist-
ance.
Rule 8. In divided circuits, eacn brancn passes a
portion of a current Inversely proportional to its re-
sistance.
Examples.
A portion of a circuit consists of two conductors,
A and B, in parallel of A = 50, and B = 75 ohms,
respectively; what will be the ratio of the currents
passing through the circuit, which will go through
each conductor?
Solution: The ratio will be current through A :
current through B :: 75 : 50, which may be ex-
pressed fractionally, is : tV.
Where more than two resistances are in parallel,
the fractional method is most easily applied.
Three conductors of A = 25, B = 50, and = 75
ohms are in parallel. What will be the ratio of cur-
rents passing through each one?
Solution: Fractionally A : B : C :: A : is : iV.
20 ARITHMETIC OF ELECTRICITY.
Rule 9. To determine the amount of a given cur-
rent that will pass through parallel circuits of dUTer-
ent resistances, proceed as follows s Take the resist*
ance of each branch for a denominator of a fraction
having 1 for its numerator. In other words, for each
branch write down the reciprocal of its resistance.
Then reduce the fractions to a common denominator,
and add together the numerators. Taking this sum
of the numerators for a new common denominator,
and the original single numerators as numerators,
the new fractions will express the proportional cur*
rents as fractions of one. If the total amperage is
given, it is to be multiplied by the fractions to give
the amperes passed by each branch. The solution <
also be done in decimals.
Examples.
A lead of wire divides into three branches; No.
1 has a resistance of 10,000 ohms, No. 2 of 39 ohms,
and No. 3 of 1 ohm. They unite at one point.
What proportion of a unitary current will pass each
branch?
Solution: The proportion of currents passed are
as irriw : A : g or 3. Reducing to a common de-
nominator, these become irSfinr : s tfoWo : WoWo . The
proportions of the numerators is the one sought for;
taking the sum of the numerators as a common de-
nominator, we have in common fractions the follow-
ing proportions of any current passed by the three
branches. No. 1, iri&Tv; No. 2, ri»; No. 3,
ttHTO.
Four parallel members of a circuit have resistances
respectively of 25, 85, 90, and 175 ohms; express
OHM'S LAW. 21
decimally the ratio of a unitary current that will
pass through them.
Solution: The ratio is as A : is : A : ih, or reduc*
ing to decimals (best by logarithms), .04 : .011765 :
.011111 : .0057. Adding, these together, we have
.068576, which must be multiplied by 14.582 to pro-
duce unity. Multiplying each decimal by 14.58
(best by logarithms), we get the unitary ratio as
.5832 : .17153 : .1620 : .08310, whose sum is 1.0000.
Unless logarithms are used, it is far better to work
by vulgar fractions.
A current of .71 amperes passes through two
branches of a circuit. One is a lamp with its con-
nections of 115 ohms resistance; another is a resist-
ance coil of 275 ohms resistance. What current
passes through each branch?
Solution: The proportions of the current are as
jh : *fo or reduced to a common denominator and to
their lowest terms tHt : irJi*. Proceeding as before,
and taking the sum of the numerators (55 + 23 =*
78), as a common denominator, we find that the lamp
passes H, and the resistance coil H of the whoie
current. Multiplying the whole current, .71 by H,
we get H$$ amperes, or i ampere for the lamp, leaving
.21 or a little over i ampere for the resistance coil.
Another problem in connection with parallel
branches of a circuit is the combined resistance of
parallel circuits. This is not a case of summa-
23 ABITHMETIC OF ELECTRICITY.
tion, for it is evident that the more parallel paths
there are provided for the current, the less will be
the resistance.
Rule 1 0. In shunt circuits, the resistance of the com-
bined shunts is expressed hy the reciprocal of the sum
or the reciprocals or the resistances.
Example.
Two leads of a 50 volt circuit (leads differing in
potential by 60 volts), are connected by a 20 ohm
motor. A 50 ohm lamp and 1000 ohm resistance
coil are connected in parallel or shunt circuit there-
with, what is the combined resistance? and the total
current?
Solution: The reciprocal of resistance is conduc-
tance, sometimes expressed as mhos. (Rule 19.)
The conductance of the three shunts is equal to
ft t ts t ioi)o mhos = i8oo t looo t TTnnr ass ioA©
mhos. The reciprocal of conductance is resistance.
The combined resistance is therefore m* ohms =
14.09 ohms. The current is ^ or 3.5 amperes.
Rule 11. The combined resistance or two parallel
circuits is round by multiplying the resistance* to-
gether, and dividing the product by the sum or the re-
sistances. Where there are several circuits, any two
can be treated thus, and the result combined in <he
same way with another circuit, and so on to get the
final resistance. B _ rXrI
Example. ^
Four conductors in parallel have resistances of
100 _ 50 — 27 — 19 ohms. What is their combined
resistance?
OHM' 8 LAW. 23
Solution: Combining the first and second, we
have lro*M = 33* ohms. Combining this with the
resistance of the third wire, we have |£qp| =■ i*.9
ohms. Combining this with the resistance of the
fourth wire, we have ji£pfS = 8.3 ohms. The
result is, of course, identical by whatever rule
obtained.
Rule 12. When all the parallel circuit* are of uni-
form resistance, as in multiple arc Incandescent light-
ing, the resistance of the combined circuits is found by
dividing the resistance of one circuit by the number of
circuits. o — —
R ~" n
Examples.
There are fifty lamps of 100 ohms resistance each
in multiple arc connection. What is their com-
bined resistance?
Solution: W = 2 ohms.
A motor can take 3 amperes of currents at
30 volts safely without burning out or heating
injuriously. A 110 volt incandescent circuit is at
hand. The motor is to be connected across the
leads so as to receive the above amperage. A shunt
or branch of some resistance is carried around it,
and a resistance coil intervenes between the united
branches and one of the main leads. The resistance
of the coil is 20 ohms. What should the resistance
of the shunt be?
24 ARITHMETIC OF ELECTRICITY.
\JfctvK
7
.Zeads.
A
Ster]*
VSAtfWwJ
. Jtksistetnee .
Solution: The resistance of the motor (Ohm's
Law, Rule 3), is found by dividing the E. M. F. by
the resistance — 30 -*- 3 = 10 ohms. By Eule 5 the
resistance of the coil in series ([20 ohms) must be to
the combined (not added) resistance of the motor
and shunt coil, as 110 — CO (total voltage minus
voltage for motor) : 30 (voltage for motor) or 20 :
x :: 80 : 30 . *. x = 7.5 combined resistance of parallel
or shunt coil and motor. The reciprocal of 7.5
(conductance, Rule 19), may be expressed as flHfths of
the combined (in this case added) conductances of
shunt coil and motor. The conductance of the
motor is equal to the reciprocal of 10 which may be
expressed as A or as ^. The conductance of the
shunt coil must therefore be Hi — t%t = ^fo = A
mho. The reciprocal of this gives the resistance of
the shunt coil which is 30 ohms. The total current
going through the system by Ohm's law is n+aQ = 4
amperes. The resistance of the shunt coil— 30 ohms
— is to that of the motor in parallel with it — 10
ohms — as the current received by the motor is to
OHM'S LAW. 25
that received by the coil, a ratio of 30 : 10 or 3:1
giving 3 amperes for the motor and 1 ampere for
the coil. This is a proof of the correctness of
operations.
Two conductors through which a current is
passing are in parallel circuit with each other.
One has a resistance of 600 ohms. The other has a
resistance of 3 ohms. A wire is carried across from
an intermediate point of one to a corresponding
point of the other. It is attached at such a point of
the first wire that there are 400 ohms resistance be-
fore it and 200 after it. Where must it be connected
to the other in order that no current may pass?
Solution: The E. M. F. up to the point of con-
nection of the bridge or cross wire is to the total
E. M. P. in the 600 ohm wire as 400: 600 or as 2: 3.
The other wire which by the conditions has the
same drop of potential in its full length must be
divided therefore in this ratio. The bridge wire
must therefore connect at 2 ohms from its begin-
ning, leaving 1 ohm to follow. The principle here
illustrated can be proved generally and is the Wheat-
stone Biidge principle.
CHAPTER IIL
besistakoe akd conductance.
Resistance of Different Conductors of the
same Material.
Conductors are generally circular in section.
Hence they vary in section with the square of their
diameters. The rule for the resistance of conduc-
tors is as follows:
Rule 13. The resistance of conductors of Identical
material varies Inversely as their section, or If or circu-
lar section Inversely as the squares of their diameters,
and directly as their lengths.
Example.
1. A wire a, is 30 mils in diameter and 320 feet
long; another b, is 28 mils in diameter and 315 feet
long. What are their relative resistances?
Solution: Calling the resistances R* : R b we would
have the inverse proportion if they were of equal
lengths R b : R* :: 30* : 28* or as 900 : 784. Were
they of equal diameter the direct proportion would
hold for their lengths: R b : R* :: 315 : 320. Com-
bining the two by multiplication we have the com-
pound proportion R b : R a :: 900 X 315 : 784 X 320 or
as 283,500 : 250,880, or as 28 : 25 nearly. The
RESISTANCE AND CONDUCTANCE. 27
combined proportions could have been originally
expressed as a compound proportion thus: E b : E* ::
30 a X 315 :28 a X 320.
For^ wires of equal resistance the following is
given.
Rule 14* The length of one wire multiplied by the
square of the diameter of the other wire must equal the
square of Its own diameter multiplied by the length of
the other If their resistances are equal* Or multiply the
length of the first wire by the square of the diameter of
the second. This divided by the length of the second
will give the square of the diameter of the first wire;
or divided by the square of the diameter of the first will
glTC the length of the second. Id* — Fd*
Examples.
1. There are three wires, a is 2 mils, 5 is 3 mils,
and c is 4 mils in diameter; what length must b
and c have to be equal in resistance to ten feet of at
Solution: Take a and c first and apply the rule,
10 X 4* +2* = 40 feet; then take a and J 10 X
3* -*- 2 a = 22 j£ feet. To prove it compare a and c
directly by the same rule 22j£ X 4 a -*- 3 2 = 40.
As this gives the same result as the first operation,
we may regard it as proved.
A conductor is 75 mils in diameter and 79 feet
long; how thick must a wire 1264 feet long be to
equal it in resistance?
Solution: 75 a X 1264 •*- 79 = 7,110,000 •*- 79 =
90,000. The square root of this amount is 300 which
is the required diameter.
28 ARITHMETIC OF ELECTRICITY.
For problems involving the comparison of wires of
unequal resistance the rule may be thus stated:
Rule 1 5. Multiply the square of the diameter of eaeh
wire by the length of the other. Of the two products
divide the one by the other to get the ratio of resist-
ance of the dividend to that of the divisor taken «*t
unity* The term Including the length of a given wire
is the one expressing the relative resistance of such
wire*
Examples.
A wir9 is 40 mils in diameter, 3 miles long and
40 ohms resistance. A second wire is 50 mils in
diameter and 9 miles long. What is its resistance?
Solution: 9 X 40 2 = 14,400 relative resistance of
the first wire. 3 X 50 2 = 7,500 relative resistance
of second wire. 14,400 •*- 7,500 = 1.92 — ratio of
resistance of second wire to that of first taken at
unity. But the latter resistance really is 40 ohms.
Therefore the resistance of the second wire is 40 X
1.92 = 76.80 ohms.
The result may also be worked out thus:
40* X 9 = 14,400 = relative resistance of the 3 mile
wire.
50 a X 3 = 7500 = relative resistance of the 9 mile
wire.
14,400 -*- 7500 = 1.92 = ratio of 9 mile (dividend)
to 3 mile (divisor) wire.
.*. 40 ohms X 1.92 = 76.8 ohms.
A length of a thousand feet of wire 95 mils in
diameter has 1.15 ohms resistance; what is the di«
BE8I8TANCE AND CONDUCTANCE. 29
ameter of a wire of the same material of which the
resistance of 1000 feet is 10.09 ohms? (E. E. Day,
M. A.).
Solution: 10.09 -*• 1.15 = 8.77 ratio of resistances.
If we divide 1000 by 8. 77 we obtain a length of the
first wire which reduces the question to one of iden-
tical resistances. 1000 •*- 877 = 114 feet. Then
applying Rule 14, 114 X 95* •*- 1000 = 1037.88.
This is the square of the diameter of the other wire.
Its square root gives the answer: 32.2 mils.
Specific Resistance.
Specific resistance is .the resistance of a cube of
one centimeter diameter of the substance in ques-
tion between opposite sides. It is expressed in
ohms for solutions and in microhms for metals.
From it may be determined the resistance of all
volumes, generally prisms or cylinders, of substance.
Very full tables of Specific Resistance are given in
their place.
Rule 16. The resistance of any prism or cylinder of a
substance Is equal to Its specific resistance multiplied
by Its length In centimeters and divided by Its cross-
sectional area In square centimeters. If the dimensions
are given In Inches or other units of measurements
they must be reduced to centimeters by the table.
n _ Sp. R x I
Examples.
An electro-plater has a bath of sulphate of copper,
sp. resistance 40 ohms. His electrodes are each 1
80 ARITHMETIC OF ELECTRICITY.
foot square and 1 foot apart. What is the resist*
ance of such a bath?
Solution : By the table 1 square foot = 929 sq. cent,
and 1 foot = 30.4797 cent. .\ Eesistance = 40 X
30.4797 + 929 = 1.31 ohms.
Where the electrodes in a solution are of uneven
size take their average size per area. The facing
areas are usually the only ones calculated, as owing
to polarization the rear faces are of slight efficiency,
and where the electrodes are nearly as wide as the
bath or cell the active prism is practically of cross-
sectional area equal to the area of one side of a plate.
In a Bunsen battery the specific resistances of the
solutions in inner and outer cells were made alike,
each equalling 9 ohms. The central element was a
y 2 inch cylinder of electric light carbon. The outer
element was a plate of zinc 6 inches long bent into a
circle. When there were 2 inches of solution in the
cell what was the resistance?
Solution : Area of carbon = f X 2 = 3.14 square
inches. Area of zinc = 2 X 6 = 12 square inches.
This gives an average facing area of (12 + 3.14) ■*■
2 = 7.57 square inches = 48.38 sq. cent. The
distance apart = % inches (nearly) = 1.9049 cent.
.-. Eesistance = 9 X 1.9049 -*- 48.38 = .354 ohms.
For wires, the specific resistance of metals being
given in microhms, the calculation may be made in
microhms, or in ohms directly. As wire is cylindri-
RESISTANCE AND CONDUCTANCE. 31
cal a special calculation may be made in its case to
reduce area of cross section to diameter. This may
readily be taken from the table of wire factors, thus
avoiding all calculation.
Role 17. The resistance In microhms of a wire of given
diameter In centimeters Is equal to the product of the
specific resistance by 1.2737 by the length In centi-
meters dlylded by the square of the diameter In cen-
timeters. ^ n m __, .
_ Sp. Re* x «.2737 x L
* — -* 5»
BZAMPLES.
The Sp. Res. of copper being taken at 1.652 mi-
crohms what is the resistance of a meter and a half
of copper wire 1 millimeter thick?
Solution: The diameter of the wire (1 millimeter)
is .1 centimeter. The square of .1 is .01. The
length of the wire (1 y 2 meter) is 150 centimeters.
Its resistance therefore is 1.652 X 1.2737 X 150 «•
.01 = 31,561 microhms or .031,561 ohms.
TTNTVBB8AL RULE FOR RESISTANCES.
Into the problem of resistances of one or two wires
eight factors can enter, these are the lengths, sec-
tional areas, specific resistances and absolute resist-
ances of two wires. Their relation may be ex-
pressed by an algebraic equation, which by transposi-
tion may be made to fit any case. The rule is
arithmetically expressed by adopting the method of
cancellation, drawing a vertical line and placing on
82 ARITHMETIC OF ELECTRICITY.
the left side, factors to be multiplied together for
a divisor, and on the right side factors to be multi-
plied together for a dividend. In the expression of
the rule as below the quotient is 1, in other words
the product of all the factors on the left hand of
the line is equal to that of all the factors on the
right hand. Calling one wire a and the other b we
have the following expression:
^Resistance of b
Specific Resistance of a
Length of a
Cross-sectional area of b
Resistance of a
Specific Resistance of b
Length of b
Cross-sectional area of d
Rule 18. Substitute In tbe above expression tbe values
of any factors given. Substitute for factors not given
or required tbe figure 1 or unity. Sucb a value deter-
mined by division must be given to tbe required factor
and substituted In Its place as will make tbe product
of tbe left-band factors equal to tbat of tbe rlgbt-band
factors. Only one factor can be determined* and all
factors not given are assumed to be respectively equal
for botb conductors.
Examples.
If the resistance of 500 feet of a certain wire is
•09 ohms what is the resistance of 1050 feet of the
same wire?
Solution: The cross sectional areas and specific
resistance not being given are taken as equal. (This
of course follows from the identical wire being re-
ferred to.) The vertical line is drawn and the
values substituted :
RESISTANCE AND CONDUCTANCE. 33
Resistances: Eesistance of .09
required wire
Lengths : 500 1050
(Other factors omitted as unnecessary.)
1050 X .09 ■*- 500 = .189 ohms.
What is the diameter of a wire 2 miles long of
23 ohms resistance, if a mile of wire of similar ma-
terial of seventy mils diameter has a resistance of
10.82 ohms?
Solution. We use for simplicity the square of the
diameter in place of the cross sectional area of the
known wire, thus:
Resistances : 23
Lengths : 1
Areas : Unknown
10.82
2
70*
As the specific resistances are identical they are
not given.
2 X 70 2 X 10.82 ■*- 23 X 1 = 4610 square of diameter
required : 4610* — 68 mils.
What must be the length of an iron wire of cross-
sectional area 4 square millimeters to have the same
resistance as a wire of pure copper 1000 yards long,
of cross-sectional area 1 square millimeter, taking
the conductance of iron as } that of copper? (Day).
Solution:
34 ARITHMETIC OF ELECTRICITY.
Specific Resistances : 1
Lengths : 1000
Cross-sectional areas : 4
7 (i.e. the reciprocal of
conductance)
Unknown
1
As the resistances are identical they are not
given.
Solving we have 1000 X 4 ■* 7 = 571f yards.
There are two conductors, one of 35 ohms resist-
ance, 1728 feet long and 12 square millimetres cross-
sectional area and specific resistance 7: the other of
14 ohms resistance, 432 feet long and 8 square milli-
metres cross-sectional area. What is its specific
resistance?
Resistances: 35
14
Specific Resistances
Unknown
7
Lengths: 432
1728
Cross-Sectional areas: 12
8
By cancellation this reduces to 14X8-*-5X3=*
7.4 Specific Resistance.
In these cases it is well to call one wire a and the
other b, and to arrange the given factors in two
columns headed by these designations. Then the
formula can be applied with less chance of error.
Thus for the last two problems the columns should
be thus arranged.
RESISTANCE AND CONDUCTANCE.
35
Area, 4 sq. mils.
L. Unknown
Sp. Res. 7
Resist. 85 ohms
1,000 yards L. 1,728 feet
Area, 12 sq. mils.
Sp. Res 7
14 ohms
432 feet
8 sq. mils,
required
From such statements of known data the formula
can be conveniently filled up.
Besistance of Wires Beferred to Weight.
The weight of equal lengths of wire is in propor-
tion to their sections. The problems involving
weight therefore can be reduced to the Eules al-
ready given.
Problem. A wire, A, is 334 feet long and weighs 25
oz.; another, B, is 20 feet long and weighs 1 oz.
what are the relative resistances?
Solution: 20 feet of the wire " A" weigh fh x
25 = 1.50 oz. The weights of equal lengths of A
and B respectively are as 1.50 : 1.00 which is also
the inverse ratio of the resistances of equal lengths.
By compound proportion Eule we have E. of " A "
: E. of "B" :: 1 X 334 :: 1.50 X 20; reducing to
16.7 : 1.5 or 11.1 : 1.0 or the wire " A " has about
eleven times the resistance of the wire " B."
Solution: By general Eule for resistance (Eule
18). Taking 1.50 : 1.00 as the ratio of cross-sec-
tional areas and taking the resistance of the long
wire A as 1 we have :
36 ARITHMETIC OF ELECTRICITY.
Resistances : 1
Lengths : 20
Cross-sectional area : 1.50
Unknown
334
1
Resistance of B = 1.50 X 20 -*- 334 = .0899 oi
about it ae before.
Conductance.
Conductance is the reciprocal of resistance and is
sometimes expressed in units called mhos, which
is derived from the word ohm written backwards.
Rule 19. To reduce resistance In ohms to conductance
in mhos express Its reciprocal and the reverse*
Examples.
A wire has a resistance of ^ftr ohms, what is its
conductance?
Solution: 126 -*- 18 = 7 mhos.
Reduce a conductance of lit to ohms.
Solution: 1H = H mhos which gives tt ohms.
It is evident that the data for problems or that
constants could be given in mhos instead of ohms.
In some ways it is to be regretted that the positive
quality of conductance was not adopted at the out-
set instead of the negative quality of resistance.
One or two illustrations may be given in the form of
examples involving conductance.
Express Ohm's law in its three first forms in
conductance.
RESISTANCE AND CONDUCTANCE. 37
Solution: This is done by replacing the factor
" resistance " by its reciprocal. Thus, Rule 1
reads for conductance: "The current is equal to
the electromotive force multiplied by the conduc-
tance " (C = EK)— Eule 2 as " The electromotive
force is equal to the current divided by the conduc-
tance " (E = — )— Eule 3 as "The conductance is
K
equal to the current divided by the electromotive
force." (K = — )
E
A circuit has a resistance of .5 ohm and an E. M.
F. of 50 volts; determine the current, using con*
ductance method.
Solution: The conductance = .-£- = 2 mhos. The
current = 50 X 2 = 100 amperes.
In a circuit a current of 20 amperes is main-
tained through 2t ohms. Determine the E. M. F.
using conductance.
Solution: The conductance = 4 = H mhos.
E. M. F = 20 -*- 1* = 52 volts.
Assume a current of 30 amperes and an E. M.
F. of 50 volts, what is the conductance and resis-
tance?
Solution: Conductance = 30 -•• 50 = .6 mho.
Resistance = 1 •*- .6 = 1.667.
CHAPTER IV.
potential difference,
Dbop op Potential in Leads and Size op
Same for Multiple Abo Connections.
Subsidiary leads are leads taken from large sized
mains of constant E. M. F. or from terminals
of constant E. M. F. to supply one or more lamps,
motors, or other appliances. A constant voltage is
maintained in the mains or terminals. There is a
drop of potential in the leads so that the appliances
always have to work at a diminished E. M. F. The
E. M. F. of the leads is known, the requisite
E. M. F. and resistance of the appliance is known,
a rule is required to calculate the size of the wire to
secure the proper results. It is based on the princi-
ple that the drop or fall in potential in portions of
integral circuits varies with the resistance. (See
Ohm's law). A rule is required for a single appli-
ance or for several connected in parallel.
Rule 20* The resistance of the leads supplying any ap-
pliance or appliances for a desired drop In potential
within the leads Is equal to the reciprocal of the cur-
rent of the appliances multiplied by the desired drop
In volts*
POTENTIAL DIFFERENCE. 89
Example.
A lamp, 100 volts X 200 ohms, is placed 100 feet
from the mains, in which mains a constant E. M. F.
of 110 volts is maintained. What must be the resist-
ance of the line per foot of its length; and what
size copper wire must be used?
Solution: The lamp current is obtained (Ohm's
law) by dividing its voltage by its resistance, (Hi =
i ampere). The reciprocal of the current is f ; mul-
tiplied by the drop (i X 10 = 20) it gives the resist-
ance of the line as 20 ohms. As the lamp is 100
feet from the mains there are 200 feet of the wire.
Its resistance per foot is therefore ffo = A ohm or
it is No. 30 A. W. G. (about).
For several appliances in parallel on two leads a
similar rule may be applied. There is inevitably a
variation in E. M. F. supplied to the different appli-
ances unless resistances are intercalated between the
appliances and the leads.
Rule 21. The E. M. F. of the main leads or terminals
the factors of the lamps or other appliances, their num.
ber and the distance of their point of connection are
given. The combined resistance Is found by Rules 8 to
12. Then by Rule 20 the resistance of the leads Is cal-
culated.
Example.
A pair of house leads includes 260 feet of wire,
or 130 in each lead. Six 50 volt 100 ohm lamps are
connected thereto at the ends. The drop is to be 5
40 ARITHMETIC OF ELECTRICITY.
volts, giving 55 volts in the main leads. Eequired the
total resistance of and size of wire for the house leads.
Solution: The resistance of six 100 ohm lamps in
parallel is 100 + 6 = 16.66 ohms. The current re-
quired is by Ohm's law 50 -*- 16.66 or 3 amperes.
Its reciprocal multiplied by the drop, (tX5 = i =
1?3 ohms) gives the required resistance = lji ohms.
This, divided by 260 feet gives the resistance per
foot as .0064 ohm, corresponding by the table to
No. 18 A. W. G.
A rule for the above cases is sometimes expressed
otherwise, being based on the proportion: Eesist-
ance of appliances is to resistance of leads as 100
minus the drop expressed as a percentage is to the
drop expressed as a percentage. This gives the fol-
lowing:
Rale 22. The resistance of tbe leads is equal to the
combined resistance of the appliances multiplied by
the percentage of drop and divided by 100 minus the
percentage of drop*
Problem. Take the data of last problem and
solve.
Solution: The percentage of drop is A = 9%. The
resistance of the leads = 16g6 x 9 = U9M = \yi ohms
100 — 9 91 '
about.
JVote.— To obtain accurate results the figures of percen-
tage, etc., must be carried out to two or more decimal
places. Rules 20 and 21 are to be preferred to any percen-
tage rule. Also see Rule 23.
POTENTIAL DIFFERENCE. 41
Where groups of lamps are to be connected along
a pair of leads but at considerable intervals, the
succeeding sections of leads have to be of diminish-
ing size. The same problem arises in calculating
the sizes of street leads. The identical rule is ap-
plied, care being taken to express correctly the ex-
act current going through each section of the lead.
The calculation is begun at the outer end of the
leads. A diagram is very convenient; it may be
conventional as shown below.
Examples.
At three points on a pair of mains three groups of
fifty 220 ohm lamps in parallel are connected; a
total drop of 5 volts is to be divided among the
three groups, thus: 1.6 volts — 1.6 volts — 1.8 volts.
The initial E. M. P. is 115 volts; what must be the
resistances of the three sections of wire?
Solution: The following diagram gives the data
as detailed above:
1. 2. 3.
S&Tmmfm dttiZ+myu. f S&J*unj*.
-x+Ka,j, T - -/*K&j>~r- -tsnusDrvp-
Starting at group 3 we have 50 lamps in parallel
each of 220 ohms resistance, giving a combined re-
sistance (Eule 12) of 4.4 ohms and a total current
(Ohm's law) of 110 + 4.4 = 25 amperes. The re-
sistance of section 2 — 3 is by the present rule *V X
42
ARITHMETIC OF ELECTRICITY.
1.8 = .072 ohms. Taking group 2 the current
through this group of lamps is 111.8 -*- 4.4 = 25.41
amperes. The section 1 — 2 has to pass also the cur-
rent 25 amperes for group 3 giving a total current
of 25 + 25.41 =» 50.41 amperes. The resistance of
section 1 — 2 is therefore^ X 1.6 = .0317 ohm.
Taking group 1 the current for its lamps is 113.4 ■+-
4.4 — 25.7 amperes. The total current through
section 0—1 is therefore 25 + 25.4 + 25.7 — 76.1
amperes. The resistance of the section is ^ X 1.6
= .021 ohms. Arranging all these data upon a
diagram we have the full presentation of the calcu-
lation in brief as below:
b
. _ j££C4ta._» • -^S- -'SH^Vs* ^Jf±&& &*£!?!$■
<#»*fi^
tev*
Irv
4~e*
v J
CHAPTER V.
CIBCULAB MILS.
A mil is TtAnr of an inch. The area of a circle,
one mil in diameter, is termed a circular mil. The
area of the cross-section of wires is often expressed in
circular mils. Thus a wire, 3 mils in diameter, has
an area of 9 circular mils, as shown in the cut. A
*— .. -.--'jjgtgf. •
circular mil is .7854 square mil. Rules for the sizes
of wires for given resistances are often based on cir-
cular mils, and include a constant for the conduc-
tivity of copper. By the table of specific resistances,
the values found can be reduced to wires of iron or
other metals.
44 ARITHMETIC OF ELECTRICITY.
A commercial copper wire, one foot long, and one
circular mil in section, has a resistance of 10.79
ohms at 75° F. This is, of course, largely an as-
sumption, but is taken as representing a good aver-
age. No two samples of wire are exactly alike, and
many vary largely. From Eule 13, and from the
above constant, we derive the following rules:
Rule 23. The resistance of a commercial copper wire
Is equal to Its length divided by the cross-section In cir-
cular mils, and multiplied by 10.79.
Example.
A wire is 1050 feet long, and has a cross-section
of 8234 circular mils. What is its resistance?
Solution: 1050 X 10.79 -*- 8234 = 1.37/ ohms.
Rule 24. The cross-section of a wire In circular mils
Is equal to Its length divided by Us resistances and mul-
tiplied by 10.79.
Example.
A wire is 1050 feet long, and has a resistance of
.68795 ohms. What is its cross-section in circular
mils?
Solution: 1050 X 10.79 -*- .68795 = 16,468 circu-
lar mils.
Rule 25. The cross-section of the wire* of a pair of
leads In circular mils for a given drop expressed in per-
centage is equal to the product of the length of leads by
the number of lamps (in parallel), by 21.5 8, by the dif-
ference between 100 and the drop, the whole divided
by the resistance of a single lamp multiplied by the
drop. __ In x 81.58 X (100— e)
CIRCULAR MILS. 45
Example.
Fifty lamps are to be placed at the end of a double
lead 150 feet long (= 300 feet of wire). The resist-
ance of a lamp is 220 ohms. What size must the
wire be for 5% drop?
Solution: 150 X 50 X 21.58 X (100 — 5) + (220 X
5) = 13,977.9 circular mils.
In these calculations and in the calculations given
on page 48 it is important to bear in mind that the
percentage is based upon the difference of potential
at the beginning of the leads or portion thereof
under consideration; in other words upon the high-
est difference of potential within the system or the
portion of the system treated in the calculation.
CHAPTER VL
special systems.
Three Wire System,
As there are three wires in the three wire system,
where there are two in the ordinary system, and as
each of the three wires is one quarter the size of
each of the two ordinary system wires, the copper
used in the three wire system is three-eighths of
that used in the ordinary system.
In the three wire system the lamps are arranged
in sets of two in series. Hence but one-half the
current is required. The outer wires, it will be no-
ticed, have double the potential of the lamps.
Hence to carry one-half the current with double the
B. M. P., a wire of one quarter the size used in the
ordinary system suffices.
Rule 26. The calculations for plain multiple arc
work apply to the three wire system, as regards size of
each of the three leads. If divided by 4.
While the central or neutral wire will hare noth-
ing to do when an even number of lamps are burning
on each side, and may never be worked to its full
capacity, there is always a chance of its having to
carry a full current to supply half the lamps (all on
SPECIAL SYSTEMS. 47
one side). Hence it is made equal in size to the
others.
Altebnating Cubrent System.
The rules already given apply in practice to this
system also, although theoretically Ohm's law
and those deduced from it are not correct for this
current. A calculation has to be made to allow for
the conversion from primary to secondary current
in the converter as below.
Note. — The ratio of primary E. M. F. to secondary is ex-
pressed by dividing the primary by the secondary, and is
termed ratio of conversion. Thus in a 1000 volt system with
50 volt lamps in parallel the ratio of conversion is 1000 ■+■ 60
= 20.
Rule 27. The current in the primary is equal to the
current in the secondary divided by the ratio of conver-
sion.
Examples.
On an alternating current circuit whose ratio of
conversion = 20, there are 1000 lamps, each 50 volt;
50 ohms. When all are lighted what is the primary
current?
Solution : By Ohm's law the ^secondary current is
1000 x 1 (each lamp using $# — 1 ampere, Ohm's
law) = 1000 amperes. 1000 -*-i 20 = 50 amperes is
the primary current.
The current being thus determined the ordinary
rules all apply exactly as given for direct current
work.
48 ABITHMETIC OF ELECTRICITY.
Given 650 lamps, 50 volt 50 ohms each, 3600
feet from the station. The primary circuit pressure
is 1031 volts. A drop of 3% is to be allowed for in
the primary leads. What is the resistance of the
primary wire?
Solution: Current of a single lamp = 50 •+■ 50 = 1
ampere; current of 650 lamps = 650 amperes, cur-
rent of primary 650 •+■ 20 = 321 amperes, drop of
primary = 1031 X 3* = 30.9 volts, resistance of pri-
mary (Rule 20) 1X30.9 = .9516 ohm.
Rale 28. For obtaining the size of the primary wire
In circular mils, calculate by Rule 25, and divide the
result by the square of the ratio of conversion.
Example.
Take data of last problem and solve.
Solution: [3600 X 650 X 21.58 X (100—3) -h (50 X
3)] +■ 20 a = 81,637 circular mils.
The two last examples may be made to prove each
other, thus:
The total length of leads is 3600 X2«= 7200 feet.
If of 1 mil thickness its resistance would be 7200 X
10.79 = 77,688 ohms. As resistance varies inversely
as the cross sectional area we have the proportion
.9516 : 77,688 :: 1 : x which gives x = 81,639 cir-
cular mils corresponding within limits to the result
obtained by Rule 28.
In all cases of this sort where percentage is ex-
pressed the statement in the last paragraph on page
SPECIAL 8T8TEM8. 49
45 should be kept in mind. The ratio of conversion
must be based on the E. M. F. at the coil (in this
case on 1031 — 31 = 1000 volts) not on the E. M. P.
at the beginning of the leads or portion thereof con-
sidered in the calculation. The percentage of drop
must be subtracted before the ratio of conversion is
calculated.
For winding the converters, the following is the
rule :
Rule 29. The convolutions of tbe primary are equal
In number to tbe product of tbe convolutions of tbe sec-
ondary multiplied by tbe ratio of conversion, and vice
versa*
Examples.
The.ratio of conversion of a coil is 20; there are
1000 convolutions of tl\e secondary. How many
should there be of the primary?
Solution: 1000 X 20 = 20,000 convolutions.
There are in a coil 5000 convolutions of the pri-
mary; its ratio of conversion is 50. How many con-
volutions should the secondary have?
Solution: 5000 -*- 50 = 100 convolutions.
CHAPTER VBL
work and energy.
Ekbbgy and Heating Effect of the Current.
It has been shown experimentally by Joule that
the total quantity of heat developed in a circuit is
equal to the square of the current multiplied by the
resistance. This is equal, by Ohm's law, to the
square of the E. M. P. divided by the resistance,
which again reduces to the E. M. F. multiplied by
the current. Each of these expressions has its own
application, and they may be given as three rules.
Rule 30* The energy or heat developed in 'circuits to
In proportion to the square of the current multiplied by
the resistance.
Examples.
An electric lamp has a resistance of 50 ohms; it is
connected to a street main by leads of 24 ohms re-
sistance. What proportion of heat is wasted in the
house circuit?
Solution: The current being the same for all parts
of the circuit, the heat developed is in proportion to
the resistance, or as 2i : 50 equal to 1 : 20. The
WORK AND ENERGY. 51
heat developed in the wire is wasted, therefore the
waste is A of the total heat developed.
The internal resistance of a battery is equal to
that of 3 meters of a particular wire. Compare the
quantities of heat produced both inside and outside
the battery when its poles are connected with one
meter of this wire with the quantities produced in
the same time when they are connected by 37 meters
of the same wire. (Day.)
Solution: The relative currents produced in the
two cases are found (Ohm's law) by dividing the
E. M. F. of the battery (a constant quantity = E)
by the relative resistance. As the battery counts
for the resistance of 3 meters of wire, the relative
resistances are as 4 : 40. Were the same current de-
veloped in both cases these figures would give the
desired ratio. But as the current varies it has to be
taken into account. To determine the relative bat-
tery heat only, we may neglect resistance of the bat-
tery, as it is a constant for both cases, the battery
remaining identical. By Ohm's law the currents
are in the ratio of f : ^ and their squares are in pro-
portion to f£ : j^ = 100 : 1. By the rule this is
the proportion of the heats developed in the battery
alone, with the short wire 100, with the long wire
1. For the heating effects on the outside circuit,
as resistance and current both vary, the full for-
mula of the rule must be applied. The ratio of the
heat in the short wire connection to that in the long
52 ARITHMETIC OF ELECTRICITY.
wire connection is as (* ) f X 1 : (£)* X 37 = 100 X 1 •
37 X 1. The ratio therefore is as 100 is to 37 for
the total heat produced in the circuit which includes
battery and connections.
Owing to irregular working of a dynamo, an in-
candescent lamp receives sometimes the full amount
or i ampere of current; at other times as little as Jt
ampere. What proportion of heat is developed in it
in both cases, assuming its resistance to remain
sensibly the same?
Solution: By the rule the ratio is (i) : (tt) or i:
*$ftr = 2025 : 400. The diminution of current there-
fore cuts down the heat to I the proper amount.
Rule 3 1 • The energy or beat developed In a circuit is
In proportion to the square of the electromotive force
divided by the resistance. E*
H = -
K
Examples.
There are two Grove batteries, each developing
1.98 volts E. M. P. One has an internal resistance
of A ohm; the other of i ohm. They are placed in
succession on a circuit of 2 ohms resistance. What
is the ratio of heats developed by the batteries in '
each case?
Solution: As the E. M. F. is constant it may be
taken as unity. Then for the two cases we have
«~£ : «1 *** ^i : 2A as the ratio of heat produced.
A battery of one ohm resistance and two volts E.
WORK AND ENERGY. 53
M. P. is put in circuit with 4 ohms resistance.
Another battery of 4 ohms and 1 volt is connected
through 1 ohm resistance. What ratio of heat if
developed in each case?
Solution: if 8 : 1|I or 4:1.
Rale 32. The energy or heat developed In a circuit %m
In proportion to the K. m. F. multiplied by the current*
H = BP
Examples.
Take data of last problem and solve.
Solution: For first battery, by Ohm's law, current ■■
f ampere; for the second, current = I ampere. The
heat developed, is by the present rule, in the propor-
tion as 1 X 2 : i X 1 or 4 : 1.
Compare the heat developed in a 100 volt 200 ohm
Jamp and in a 35 volt 35 ohm lamp and in a 50 volt
50 ohm lamp.
Solution :The currents (Ohm's law) are : Hfcli and
U in amperes — i, 1, and 1 amperes. The heats devel-
oped are, by the rule, in the ratio 100 X i : 35 X 1
and 50 X lor50:35 :50.
The same problem can be done directly by Rule
31, thus: The three lamps develop heat in the
ratio m* : tt* : U* « 50 : 35 : 50. This is the direct
and preferable method of calculation.
Note, — For " heat," " rate of energy," or " rate of work"
can be read in these rules.
54 ARITHMETIC OF ELECTRICITY.
The Joule ob Gram-Calorie.
The last rules and problems only touch upon the
relative proportions of heat; they do not give any
actual quantity. If we can express in units of the
same class a standard quantity of heat, then by deter-
mining the relation of the standard to any other
quantity, we arrive at a real quantity. Such a stan-
dard is the joule, sometimes called the " calorie" or
" gram-calorie. " A joule is the quantity of heat
required to raise the temperature of 1 gram of
water 1 degree centigrade. It is often expressed
as a water-gram-degree C. or w. g. d. 0. or for
shortness g. d. 0., from the initials of the factors.
It is unfortunate that it is called the calorie
as the name is common to the water-kilogram-
degree C. or kg.d. C. The joule is equal to 4.16
X 10 7 or 41,600,000 ergs.
It will be remembered that practical electric units
are based on multiples of the C. G. S. units of which
the erg is one. The joule comes in the 0. G. S.
order. Therefore to determine quantities of heat
the following is a general rule when the practical
electric units are used.
Rule 33. Obtain the expression of rate of heat devel-
oped, or of rate of energy, or of rate of work In volt
amperes. Reduce to €. G. 8. units (ergs) by multiply-
ing by 10 7 and divide by the value of a joule In ergs
(4.16 X 10 7 ). The quotient Is joules or water-gram de-
grees C. per second*
Q _ E X C X 1Q T
H " 4.16 XlO*
WORK AND ENERGY. 55
Example.
A current of 20 amperes is flowing through a wire.
What heat is developed in a section of the wire
whose ends differ in potential by 110 volts?
Solution: The rate of energy in watts or volt-
amperes = 110 X 20 = 2200. In the 0. G. S. units
this is expressed by (110 X 10 8 ) X (20 X 10 1 ) = 2200
X 10 7 ergs, per second; . \ quantity of heat = 2200 X
10 7 -*- 4.16 X 10 7 = 528.8 joules or gram-degrees-
centigrade per second.
As 10 7 + 10 7 = 1 the rule can be more simply
stated thus:
Rule 34* The quantity of heat produced per second
In a circuit by a current Is equal to the product of the
watts by ^-j^ or by .24.
Q = 0.24 OE. or 0.24 ^
Examples.
A difference of potential of 5.5 volts is main-
tained at the terminals of a wire of A ohm resist-
ance. How many joules per second are developed?
Solution: By Ohm's law, current = 5.5 -*- tV = 55
amperes. By the rule 55 X 5.5 X 0.24 = 72.6
joules per second.
Note.— The energy of a current is given by Rules 30, 31
and 32 in watts, so that all cases are provided for by a com*
bination of one or the other of these rules with Rule 34
An example, will be given for each case.
56 ARITHMETIC OF ELECTRICITY.
A current of .8 ampere is sent through 50 lamps
in series, each of 137J4 ohms resistance. What
heat does it develop per second?
Answer: The resistance = 137J6 X 50. By rules
30 and 34 we have, rate of heat produced = .8* X
137# X 50 = 4400 watts. 4400 X 0.24 = 1056
joules per second.
Rules 31 and 34. Fifty incandescent lamps, 110
volt, 137# ohms, each are placed in parallel.
What heat per second do they develop?
Solution: By Ohm's law total resistance = 137J4
••- 50 = 2.75 ohms. By rules 31 and 34 rate of heat
produced = HO 8 + 2.75 « 4400 watts and 4400 X
0.24 == 1056 joules per second as before.
Eules 32 and 34. Through 50 incandeseent 110
volt lamps a current of .8 ampere is passed, the
lamps being in series. What heat per second do
they develop?
Solution: By rules 32 and 34 rate of heat «= 110 X
50 X .8 = 4400 watts and 1056 joules per second as
before.
These three examples are purposely made to refer
to the same set of lamps, to show that rules 30, 31,
and 32 are identical. Each fits one of the three
forms of statement of data. They also are designed
to bring out the fact that the unit "watts/' being
based partly on amperes, includes the id«m of rate,
not of absolute quantity. Hence watts "per sec-
ond " is not stated, as it would be meaningless o*
WORK AND ENERGY. 57
redundant, while the joule, denoting an absolute
quantity, has to be stated "per second " to indicate
the rate. There is such a unit as an "ampere-
second," viz., the "coulomb," but there is no such
thing as an "ampere per second," or if used it
means the same as an " ampere per hour," " ampere
per day" or "ampere." The same applies to watts
and to power units such as "horse-power."
Specific Heat.
The specific heat of a substance is the ratio of its
capacity for heat to that of an equal quantity of
water. It almost invariably is referred to equal
weights. Here it will be treated only in that con-
nection.
The coefficient of specific heat of any substance is
the factor by which the specific heat of water (= 1 or
unity) being multiplied the specific heat of the sub-
stance is produced.
Rule 35. The weight of any substance corresponding
to any number of joules multiplied by Its specific heat
gives the corresponding -weight of water, and vice versa.
Example.
A current of .75 amperes is sent for 5 minutes
through a column of mercury whose resistance was
0.47 ohm. The quantity of mercury was 20.25
grams, and its specific heat 0.0332. Find the rise
of temperature, assuming that no heat escapes by
radiation. (Day.)
58 ARITHMETIC OF ELECTRICITY.
Solution: By Eules 30 and 34, we find rate of heat
« .75* X .47 = .264375 watts; .264375 X .24 =>
.06345 joules per second. The current is to last for
300 seconds .-. total joules = .06345 X 300 = 19.035
joules. This must be divided by the specific heat
of mercury to get the corresponding weight of mer-
cury; 19.035 -*- .0332 — 573 gram degrees of mercury.
Dividing this by the weight of mercury, 20.25 grams,
we have 573 ■*■ 20.25 = 28° 0. ^
Rule 36. By radiation and conTectlon, 4000 Joole
about Is lost by any unpolished substance In the air for
each square centimeter of surface, and for each degree
that It Is heated above the atmosphere*
Example.
A conductor of resistance, 8 ohms, has a current
of 10 amperes passing through it. It is 1 centi-
meter in circumference, and 100 meters long. How
hot will it get in the air?
Solution: By Eule 30, etc., the heat developed per
second in joules is 10 2 X 8 ■*- 4.16 = 192.3 joules. The
surface of the conductor in centimeters is 10,000 X
1 = 10,000 sq. cent. It therefore develops heat at
the rate of 192.3 X 10" 4 = .01923 joules per second
per square centimeter of surface. When the loss by
radiation and convection is equal to this, it will re-
main constant in temperature. Therefore .01923
•+" 4TJW = 76.92, the number of degrees C. above the
air to which the conductor could be heated by such
a current.
WORK AND ENERGY. 69
Besults of this character are only approximate,
as the coefficient, dta, is not at all accurate.
Rule 37. The cube of the diameter In centimeters of
a wire of any given inateriai that will attain a given
temperature centigrade under a given current Is equal
to the product of the square of the current In amperes x
Sp a .Resistance In microhms of the substance of the
wire, multiplied by .000391, and divided by the tem-
perature in degrees centigrade.
d . 3> Q»xSp.Bes. x. 000391
Examples.
A lead safety catch is to be made for a current of
7.2 amperes. Its melting point is 335° 0., and its
specific resistance 19.85 microhms per cubic centi-
meter. What should its diameter be? (Day.)
Solution: By the rule, the cube of the diameter =
7.2 2 X 19.85 X .00039 + 335 = .001198. The cube
root of this gives the diameter in centimeters. It is
.10582 or .106 cm.
A copper wire is to act as safety catch for 500 am-
peres: melting point 1050° C— Sp. Eesistance 1.652
microhms. What should its diameter be? (Day.)
~ , . ^ , . j« i. 600»x 1.652 x. 000391
Solution: Cube of diameter = j^ ■■
1B0 » 418 = .1523. The cube root of this is .5341 centi-
1050 *
meter, the thickness of the wire sought for.
It will be observed that in this rule no attention
is paid to the length of the wire, as the effect of ^ a
current in melting a wire has no reference to its
60 ARITHMETIC OF ELECTRICITY.
length. The time of fusion will vary with the spe-
cific heat, but will, of course, be only momentary.
WOEK OF A CUBBENT.
Role 38* The work of a current In a giren circuit Is
proportional to the volt amperes. w — EC
Example.
A current A of 3.5 amperes with difference of
potential in the circuit of 20 volts is to be com-
pared to B, a 3 ampere current with a difference of
potential of 1000 volts in the circuit; what is the
ratio of work produced in a unit of time?
Solution: Work of A : work of B :: 3.5 X 20 :
3 X 1000 or as 70 : 3000 or as 1 : 42^-
Rule 39. The work or a current In a given ctrcnlt Is
equal to the volt -coulombs divided by the acceleration
of gravitation (9.81 meters). This gives the result In
kilogram-meters. (7.23 foot pounds.) _ Tr _E^C 1 t
w -~9^r
Example.
A current of 20 amperes is maintained in a circuit
by an E. M. F. of 20 volts. What work does it do
in one minute and a half (90 sec.)?
Solution : Work — 20 x 20 x 90 sec -4- 9.81 — 3670
kgmts. and 3670 X 7.23 — 26,534 foot pounds.
Note. — This is easily reduced to horse-power. 26,534 foot
pounds in 90 sec. = 17,688 foot pounds in 1 min. 1 H. P.
= 33,000 foot pounds per min. .\ £$$$$ = .536 H. P. of
above current and circuit. The same result can be ob-
tained by Rule 41 thus: ^ = .536 H. P.
WORK AND ENERGY. 61
Rule 40. To determine work done by a current In a
given clrcnit apply Rules 30, 31 or 32 as the case re-
quires. These give directly the watts. Multiply by sec-
onds and divide by 9.81. The result Is kilogram-me-
ters.
Examples.
10 amperes are maintained for 55 sec. through 15
ohms. What is the work done?
Solution : By rule 32, watts — 10 a X 15 — 1500.
Work = 1500 X 55 + 9.81 = 8409 kgmts.
1000 volts are maintained between terminals of a
lead of 20 ohms resistance. Calculate the work
done per hour.
Solution: By Eule 32 watts = 1000* -«- 20 = 50,000.
One hour = 3600 sec. Work = 50,000 X 3600 -*- 9.81
= 18,348,623 kgmts.
These rules give the basis for determining the ex-
pense of maintaining a current. The expense of
maintaining a horse-power or other unit of power or
work being known the cost of electric power is at
once calculable.
Electrical Horse-Power.
Power is the rate of doing work or of expending
energy. In an electric circuit one horse-power is
equal to such a product of the current in amperes,
by the E. M. F. in volts as will be equal to 746.
Rule 41. The electric horse power Is found by multi-
plying the total amperes of current by the volts or K. M.
F, of a circuit or given part of one and dividing: by 746.
MP -&■
63 ARITHMETIC OF ELECTBICITT.
Example.
260 incandescent lamps are in parallel or on mul-
tiple arc circuit. Each one is rated at 110 volts
and 220 ohms. What electric H. P. is expended on
their lighting?
Solution: The resistance of all the lamps in par-
allel is equal to Hi ohm. The current is equal to
110 + H» = 125 amperes. H. P. = 125 X 110 -*- 746
=» 18A H. P. or 13& lamps to the electrical H. P.
As it is a matter of indifference as regards absorp-
tion of energy how the lamps are arranged, a simpler
rule is the following, where horse-power required for
a number of lamps or other identical appliances is
required.
Rule 42. multiply together the voltage and amperage
of a single lamp or appliance ; multiply the prodnct by
the nnmber of lamps or appliances and divide by 746*
Example.
Take data of last problem and solve it.
Solution: Current of a single lamp = ffl = % am-
pere. H. P. = 110 X y 2 X 250 -*- 746 = 18A H. P.
When the voltage and amperage are not given
directly, the missing one can always be calculated
by Ohm's law and the above rules can then be ap-
plied. The same can be done by applying following:
Rale 49. To determine the electrical Horse-power
apply Rules 30, 3 1, or 32; these give directly the watts;
multiplying the result by -i^ or dividing by 746 gives
the horse-power. • 4S
WORK AND ENERGY. 63
Examples.
A current of 10 amperes is maintained through 50
ohms resistance. What is the electrical horse-
power?
Solution: By rules 30 and 43 we have watts =
10* X 50 = 5000 and electrical horse-power = 5000
-f- 746 or 6.7 H. P.
An electromotive force of 1500 volts is maintained
in a circuit of 200 ohms resistance. What is the
electrical horse-power?
Solution: By Bules 31 and 43 we have watts »
1500* -*- 200 = 11,250. Electrical horse-power =
11,250 «•• 746 or 15 H. P. (nearly).
Thus the volt-amperes or watts are units of rate
of heat energy or of rate of mechanical energy.
The ratio of joules per second to a horse-power is
746: 4.16 or 179.3 joules per second = 1 H. P.
Other ratios of power and heat units will be found
in the tables.
Duty and Efficiency of Electric Generators.
Rule 44. The duty of an electric generator Is the quo-
tient obtained by dividing the total electric energy by
the mechanical energy expended In turning the arma-
ture.
j. e. H. P. (total)
v " m.H.F.
Examples.
A dynamo is driven by the expenditure of 58
H. P. Its internal resistance is 10.7 ohm. Th*
64 ABITHMETIC OF ELECTRICITY.
resistance of the outer circuit is 150 ohms and it
maintains a current of 16 amperes. What is its
duty?
Solution: The total electrical H. P. is found by
Rules 30 and 43 to be 16* X 160.7 + 746 = 55.1
H. P. Duty = 55.1 * 58.0 = 95*.
The result must always be less than unity; if it
exceeded unity it would prove that there had been
an error in some of the determinations.
Rule 45. The commercial efficiency of a generator is
the quotient obtained by dividing the electric energy in
the outer circuit by the mechanical energy expended In
turning the armature.
O THff- - e * H * P * (Qjjter circuit)
m. H. P.
Examples.
What is the commercial efficiency of the dynamo
just cited?
Solution: The electrical H. P. of the outer circuit
is found by the same rules to be 16 2 X 150 -«-746 =
61.5 commercial efficiency = 51.5 ■+■ 58.0 = 88.8*.
Rule 46. The resistance of the outer circuit Is to the
total resistance, as the commercial efficiency Is to the
duty.
Examples.
Take the case of the generator last given and
from its duty calculate the commercial efficiency.
Solution: 150: 160.7: : x: 95.0. \ x = 88.8 or
88.8*.
OHAPTEE VIII.
batteries.
General Calculations of Current.
A battery is rated by the resistance and electro-
motive force of a single cell, which factors are
termed the cell constants. In the case of storage
batteries, whose susceptibility to polarization is very
slight, the resistance is often assumed to be neglible.
It is not so, and in practice is always knowingly or
otherwise allowed for.
From the cell constants its energy-constant may
be calculated by Eule 31, as equal to the square of
its electro-motive force divided by its resistance.
This expresses its energy in watts through a circuit
of no resistance.
There are two resistances ordinarily to be consid-
ered, the resistance of the battery which is desig-
nated by E or by n E if the number of cells is to be
implied and the resistance of the external circuit
which is designated by r.
Rule 47. The current given by a battery Is equal to
Its electro-motive force divided by tbe sum of the exter-
nal and Internal resistances.
o =_
u B + r
66 ARITHMETIC OF ELECTRICITY.
|H$) ($)($)<$)($)
Six cells in parallel.
Six cells in series.
Biz cells— two in parallel, three in series.
Six cells— three in parallel, two in series.
«
Abbangkment of Battery Cells.
BATTERIES. 67
Example.
A battery of 50 cells arranged to give 75 volts
E. M. P. with an internal resistance of 100 ohms
sends a current through a conductor of 122 ohms
resistance. What is the strength of the current?
Solution: Current = 75 -*- (100 + 122) = .338
ampere. This rule has already been alluded to
under Ohm's law (page 14).
Abbangement of Cells in Battery.
In practice the cells of a battery are arranged in
one of three ways, a: All may be in series; b: all
may be in parallel; c: some may be in series and some
in parallel, so as to represent a rectangle, s cells in
series by p cells in parallel, the total number of cells
being equal to the product of s and p.
Other arrangements are possible. Thus the cells
may represent a triangle, beginning with one cell,
followed by two in parallel and these by three in
parallel and so on. This and similar types of arrange-
ment are very unusual and little or nothing is to be
gained by them.
Rale 48. The electromotive force of a battery Is
equal to the E. M. F. of a single cell multiplied by the
number of cell* In series.
Rule 49. The resistance of a battery is equal to the
number of its cells in series, multiplied by the resist*
ance of a single cell and divided by the number of lta
cells in parallel.
•o battery _ s_B
68 ARITHMETIC OF ELECTRICITY.
Examples.
A battery of 50 gravity cells 1 volt, 3 ohms each
is arranged 10 in parallel and 5 in series. What is
its resistance and electromotive force?
Solution: Eesistance = 5 X 3 + 10 = 1.5 ohms.
E.M.F.=5X1 = 5 volts.
The same battery is arranged all in parallel; what
is its resistance and E. M. F. ?
Solution: This gives one cell in series.
Eesistance = 1 X 3 -*- 50 = .06 ohms.
E. M. F. =1X1 = 1 volt.
The same battery is arranged all in series; what is
its resistance?
Solution: This gives one cell in parallel.
Eesistance = ^x 8 = 150 ohms.
E. M. F. = 50 1 X 1 = 50 volts.
The current given by a battery is obtained from
these rules and from Ohm's law.
Example.
150 cells of a battery (cell constants 1.9 volts,
i ohm) are arranged 10 in series and 15 in parallel.
They are connected to a circuit of 1.7 ohms resist-
ance. What is the current?
Solution: The resistance of the battery = ^2Li =
.333 ohms. The E. M. F. = 10 X 1.9 = 19 volts.
Current = 19 ••- (.333 + 1.7) = 9.34 amperes.
BATTERIES. 69
Cells Kequired foe a Given Current.
To calculate the cells required to produce a given
current through a given resistance and the arrange-
ment of the cells proceed as follows.
Rule 50. Calculate the cell current through zero exter-
nal resistance. Case A. If It Is twice as great or more
than twice as great as the current required apply .Rule
51, Case B. IT less than twice as great and more than
equal or less than equal and more than one half as
great as the current required and so on apply Rule
52. Case C. If the cell current is equal to or Is a unit-
ary fraction <* * £, etc.) of the current required apply
Rule 53. TVS 9
Rule 51. Case A. Divide the required difference of po-
tential of the outer circuit by the voltage of a single cell
diminished by the product of the required current mul-
tiplied by the resistance of a single cell. Arrange the
cells In series.
Examples.
Five lamps in parallel, each of 100 volts 200 ohms,
are to be supplied by a battery whose cell constants
are 2 volts i ohm. How many cells and what
arrangement are required?
Solution: Cell current = | = 10 amperes. The
resistance of the five lamps in parallel (Rule 12) =
2p_o _ 4q Q h ms# Th e required current therefore =»
W = 2} amperes. As 10 exceeds 2i X 2 (Eule 50) it
falls under case A. By Eule 51 the number of cells
is 2 -(2ix*) = *f = 66.6 or 67 cells, as a cell cannot
be divided. The cells must be in series.
70 ARITHMETIC OF ELECTRICITY.
Proof: The E. M. F. of the 67 cells in series
= 67 X 2 = 134 volts; their resistance = 67 X j =
13.4 ohms. The resistance of the lamps in par-
allel is 40 ohms. Hence by Ohm's law the cur-
rent **" 40+13.4 = &51 amperes, the current required.
The same lamps are placed in series. Calculate
the cells of the same battery required. Cell current
■= 10 amperes. Current required = ^^| or | am-
pere. As 10 exceeds i X 2 (Rule 50) it falls again
under case A. By Eule 51 cells required = fe_^° xi)
= 263.
Proof: Current «= foe+iooo = i ampere the current
required. 526 is the number of cells multiplied by
the Toltage of one cell; 52.6 is the number of cells
multiplied by the resistance in ohms of a single cell;
1000 is the resistance of a single lamp, 200 ohms,
multiplied by the number, 5, of lamps in series.
Whenever the arrangement and number of cells of
a battery has been calculated the calculation should
be proved as above.
Rule 52. Case B. Group two or more cells In parallel
so as to obtain by calculation from them through no ex-
ternal resistance a current twice as great or more than
twice as great as the required current. Then treating
the group as If It was a single cell apply Rule 51 to de-
termine the number of groups In series.
Example.
Assume the same lamps in parallel, requiring the
BATTERLB8. 71
current already calculated of 2i amperes. Assume
a battery of constants 1 volt .25 ohm, giving a
cell current of 4= amperes. This is less than 2i X 3
and more than M X 1; therefore it falls under Case
B
Solution: A group of two cells in parallel gives
tH = 8 amperes. 8 exceeds 2i X 2 . •. applying Rule
49 we have number of groups = 100+ [1 - (2j X
.125)] = 146 groups in series. Total number of cells
= 2 in parallel, 146 in series = 292 cells.
Proof: Current = 146 + (40 + 18.25) = 2.5 am-
peres.
Rule 53. Case C. Place as many cells in series as will
give twice the required voltage. Place as many cells In
parallel as will give a resistance equal to that of the
external circuit.
Example.
Assume the same lamps in parallel. Assume a
battery of cell constants, 1 volt, 4 ohms. The lamp
current is 2J amperes. The cell current is } ampere.
The cell current therefore equals (i -*- 2i) A of the
required current. This falls under Case C. and is
solved by Rule 53.
Solution: Voltage required 100. By the rule
cells in series = 100 X 2 = 200. These have a re-
sistance of 800 ohms. To reduce this to the resist-
ance of the outer circuit, viz., 40 ohms, 800 -+• 40
= 20 cells must be placed in parallel. Total cell* -*•
20 x 200 = 4000 cells.
72 ARITHMETIC OF ELECTRICITY.
Proof: Current = 200 «*- (40 + 40) = 2.5.
Rule 54. All cases coming under Case C. may be slm*
ply solved for the total number of cells by dividing the
external energy by the cell energy and multiplying by 4.
This gives the number of cells.
Example.
Take as cell constants .75 volt 3 J ohm giving i am-
pere. Assume 20 lamps, each 50 volts, 50 ohms and
1 ampere. As J -*• 1 is a unitary fraction (J) Case
C. applies.
Solution: Cell energy = J X .75 = .375 watts. Ex-
ternal energy = 50 x 1 X 20 = 1000 watts. (1000 -*-
.375) X 4 =-10,666 cells.
Solution by Eule 53: Voltage required taking
lamps in series = 20 x 50 = 1000. To give twice
this voltage requires 2000 -*- .75 =• 2667 cells in
series whose resistance is 2667 X 1.5 = 4000 ohms.
To reduce this to 1000 ohms we need 4 such series
of cells in parallel giving 10,668 cells.
Proof: Current = ( syi^f^) + iooo = 1 am P er ©-
Slight discrepancies will be noticed in the current
strength given by different calculations. This is
unavoidable as a cell cannot be fractioned or di-
vided.
Efficiency of Batteries.
Rule 55 • The efficiency of a battery Is expressed by di-
viding the resistance of the external circuit by the total
resistance of the circuit.
Efficiency = B , ■
BATTERIES. 73
Example.
A battery consists of 67 cells in series of constants
2 volts i ohm. It supplies 5 lamps in parallel, each
100 volts 200 ohms constants. What is its effi-
ciency?
Solution: The resistance of the battery is 67 X *
= 13.4 ohms. The resistance of the lamps is (Rule
12) *P = 40 ohms. Therefore the efficiency of the
battery is 40.0 + (40 + 13.4) = .749 or 74. W.
Rale 5 6. To calculate the number of battery cells and
tbeir arrangement for a given efficiency : Express. tb©
efficiency as a decimal, multiply tbe resistance of tb©
external circuit by tbe complement of tbe efficiency
(1— efficiency) and divide tbe product by tbe efficiency;
this fives tbe resistance of tbe battery* Add tbe two
resistances and multiply tbeir sum by tbe current to be
maintained for tbe E. M. F. of tbe battery. Arrange tbe
cells accordingly as near as possible to tbese require*
ments.
Examples.
Five lamps, each 100 volts 200 ohms in parallel
are to be supplied by a battery of cell constants 2
volts .4 ohm. The efficiency of the battery is to be
as nearly as possible 75#. Calculate the number of
cells and their arrangement.
Solution: The constants of the external circuit are
40 ohms (Rule 12) and 100 volts. Applying the
rule we have [40 X (1— .75)] + .75 = X W = 13*
ohms, the resistance of the battery. By Ohm's law
the E. M. P. of the battery = (40 + 13*) X 2.5 =
74 ARITHMETIC OF ELECTRICITY.
133i volts. These constants, 13i ohms and 133i
volts, require 67 cells in series and 2 in parallel.
Proof: a. Of efficiency, by Kule 55, -^^ = .75
or 75*. b. Of number of cells and of their arrange-
ment 67 X 2 = 134 volts; (67 X .4) + 2 = 13.4
ohms; 134 -*- (13.4 + 40) = 2.5 amperes.
Rale 57* Where a fractional or mixed number of cells
in parallel are called for to produce a given efficiency,
take a group of the next highest Integral number of
cells In parallel and proceed as in Rule 51 •
Example.
Assume a current of 3i amperes to be supplied
through a resistance of 30 ohms, absorbing 100
volts E. M. F. Let the cell constants of a battery
to supply this circuit be 2 volts, I ohm. Calculate
the cells and their arrangement for 80 per cent.
efficiency.
Solution: By Rule 56 efficiency = .80 and
qq ' = 74 ohms, which is the required resist-
ance of the battery; 74+ 30 = 374 ohms are the total
resistance of the circuit. By Ohm's law, 37J X 34 =
125 volts, the required E. M. P. of the battery.
This requires 63 cells in series, with a resistance for
one series of 63 X 4 = 104 ohms. To reduce this to
7j ohms ~ = 1.4 cells in parallel are required. As
this is a mixed number we take the next highest in-
tegral number and place 2 cells in parallel. The
constants of this group of 2 cells are 2 volts, &
BATTERIES. 75
ohm. Applying Rule 51 we have for the number of
such groups in series; 2 -^ x A) = 58 g rou P s * n series.
As there are 2 cells in parallel the total cells are 116,
of resistance, 58 X A = 4.83 ohms, and of E. M. P.,
58 X 2 = 116 volts.
Proof: Of efficiency by Rule 55, g^j^ = 86.1*.
Of number and arrangement of cells 30 V^gg = 3.33
amperes.
It is to be observed that the efficiency thus
obtained is far from what is required. In most
cases accuracy can only be attained by arranging
the battery irregularly, which is unusual in prac-
tice. An example will be found in a later chapter.
Chemistry of Batteries.
One coulomb of electricity will set free .010384
milligrams of hydrogen. The corresponding weights
of other elements or compounds are found by multi-
plying this factor by the chemical equivalent, and
dividing by the valency of the element or metal of
the base of the compound in question.
An element or other substance in entering into
any chemical combination develops more or less
heat, always the same for the same weight and com-
bination. The atomic weight of an element or the
molecular weight of a compound divided by the val-
ency of the element or metal of its base gives the
original chemical equivalent.
76 ARITHMETIC OF ELECTRICITY.
The quantities of heat evolved by the combination
of quantities of substances expressed by their original
chemical equivalents multiplied by one gram are
termed the thermo-electric equivalents of the ele-
ments or substances in question. In the tables it is
expressed in kilogram degrees 0. of water (kilogram-
calories).
From the thermo-electric equivalent of a combi-
nation we find the volts evolved by it or absorbed by
the reciprocal action of decomposition.
Rale 58. The volts evolved by any chemical com-
bination or required for any chemical decomposition
are equal to the thermo-electric equivalent in kilo*
gram-calories multiplied by .043.
E=.043xH.
Examples.
What number of volts is required to decompose
water ?
Solution: From the table we find that the com-
bination of one gram of hydrogen with eight grams
of oxygen liberates 34.5 calories. Then 34.5 X
.043 = 1.48 volts.
Rule 59. To determine the voltage of a galvanic
couple subtract the kilogram calories corresponding to
decompositions In the cell from those corresponding to
combinations In the cell for effective energy and multi-
ply by .043 for volts*
Examples.
Calculate the voltage of the Smee couple.
Solution: In this battery zinc combines with
BATTERIES. 77
oxygen, giving out 43.2 calories and combines with
sulphuric acid, giving 11.7 more calories; a total of
54.9 calories. An equivalent amount of water is at
the same time decomposed acting as counter-energy
of 34.5 calories. The effective energy is 54.9 —
34.5 = 20.4 calories. The voltage = 20.4 X .043 =
.877 volts.
Calculate the voltage of the sulphate of copper
battery.
Solution: Here we have combination of zinc with
sulphuric acid as above 54.9 calories; decomposition
of copper sulphate 19.2 + 9.2 = 28.4 .-. 54.9 — 28.4 •
= 26.5 calories effective energy 26.5 X .034 = 1.14
volts.
It will be noticed that these results are approxi-
mate. Some combinations are omitted in them as
either of unknown energy, or of little importance.
Work of Batteries.
The rate of work of a battery is proportional to
the current multiplied by the electro-motive force.
The work is distributed between the battery and the
external circuit in the ratio of their resistances as by
Eule 55. The horse-power, and heating power are
calculated by Rules 30-43, care being taken to dis-
tribute the energy acording to the resistance by
the following rule:
Role 60. The effective rate of work or the rate of
work In the external circuit of a battery, Is equal to the
78 ARITHMETIC OF ELECTRICITY.
total rate multiplied by the efficiency of the battery ex-
pressed decimally.
Example.
25 cells of 2 volt 1 ohm battery are arranged in
series on an external circuit of 250 ohms resistance.
What work do they do in that circuit?
Solution: The current (Ohm's law) = —^ = 1
1.818 amperes. Total rate of work = 1.818 X 50
volts = 90.9 watts. Efficiency of battery = Wk = 90
per cent, (nearly). Effective rate of work = 1.818 X
50 X .90 = 81.81 watts.
Chemicals Consumed in a Battery.
Rule 61. The chemicals consumed In grams by a bat*
tery for one kilogram-meter (7.23 foot lbs.) of work are
found by multiplying the combining equivalent of the
chemical by the number of equivalents In the reaction
by the constant .000101867 and dividing by the pro-
duct of the E. M. F. by the valency of the element Id
question*
— ^ Equiv. x n x .000101867
B X valency
Examples.
What is the consumption of zinc and sulphate of
copper per kilogram-meter of work in a Daniel's
battery?
Solution: Take the E. M. F. as 1.07 volt. The
equivalent of zinc, a dyad, is 65 and one atom en-
ters into the reaction. The zinc consumed there-
fore = « x ^» = mQ9 g^
BATTERIES. 79
The equivalent of copper sulphate, is 159.4. One
equivalent enters into the reaction carrying with it
one atom of the dyad metal copper. The weight
consumed therefore = m4 ^ x x f )10t867 = .0076
grams. Add 56.46# for water of crystallization.
All these quantities are for one kilogram-meter of
work (7.23 foot lbs.) which may be more or less
Effective according to circumstances as developed in
Rules 44, 45, and 60.
Decomposition of Compounds by the Batteet.
In cases where a compound has to be decomposed
by a battery two resistances may be opposed to the
work. One is the ohmic resistance of the solution,
which is calculated by Eule 16. The other is the
electromotive force required to decompose the solu-
tion. This is best treated as a counter-electromotive
force. Then from the known data the current rate
is calculated, and from the electro-chemical equiva-
lents the quantity of any element deposited by a
given number of coulombs is determined.
Rule 62. To calculate the metal or other element lib-
erated by a given cnrrent per given time proceed as fol-
lows* Calculate the resistance. Determine the counter-
electromotive force of the solution by Rule 58 and sub-
tract It from the K. M. F. of the battery or generator.
Apply Ohm's law to the effective voltage thus deter-
mined and to the calculated resistances to find the cur-
rent. Multiply the electro-chemical equivalent of the
element by the coulombs or ampere-seconds.
80 ARITHMETIC OF ELECTRICITY.
Example.
A bath of sulphate of copper is of specific resis-
tance 4 ohms. The electrodes are supposed to be
10,000 sq. centimeters in area and 5 centimeters
apart. Two large Bunsen elements in series of
1.9 volts .12 ohms each are used. What weight
in milligrams of copper will be deposited per
hour?
Solution: By Rule 16 the resistance of the solu-
tion is ^| = 0.023. The electro-chemical equiva-
lent of copper is .00033 grams. The thermo-electric
equivalent for copper from sulphate of copper
is 19.2 + 9.2 = 28.4 calories. The E. M. F.
corresponding thereto = 28.4 X .043 = 1.22 volts
counter E. M. F. The E. M. F. of the bat-
tery = 1.9 X 2 = 3.8 volts, giving an effective E. M.
F. of 3.8 — 1.22 = 2.58 volts. The resistance of the
battery = .12 X 2 = .24 ohms. The current =
.24+028 = 9.8 amperes. This gives per hour 9.8 X
3,600 = 35,280 coulombs, and for copper deposited
.00033 X 35,280 = 11.64 grams.
In many cases one electrode is made of the mater-
ial to be deposited and being connected to the car-
bon end of the battery or generator is dissolved as
fast as the metal is deposited. In such case there
is no counter electro-motive force to be allowed
for.
BATTERIES. 81
Example.
Take the last case and assume one electrode (the
anode) to be of copper and to dissolve. Calculate
the deposit.
Solution: Current = 3.8 -*- (.24 + .023) = 14.4
amperes = 51,840 coulombs per hour = .00033 X
51,840 = 17.10 grams of copper.
CHAPTER IX.
ELECTRO-MAGNETS, DYNAMOS AND MOTORS.
The Magnetic Field and Lines of Force.
A current of electricity radiates electro-magnetic
wave systems, and establishes what is known as a
field of force. The field is more or less active or in-
tense according to the force establishing it. The
intensity of a field is for convenience expressed in
lines of force. These are the units of magnetic
intensity, often called units of magnetic flax, and
the line as a unit is comparable to the ampere X 10,
which is the C. G. S. unit of current. A line of
force is that quantity of magnetic flux which passes
through every square centimeter of normal cross-
section of a magnetic field of unit intensity. The
line is at right angles to the plane of normal cross-
section of such field. Such intensity of field exists
at the center of curvature of an arc of a circle of ra-
dius 1 centimeter, and whose length is 1 centimeter,
when a current of 10 amperes passes through this
arc. Practically it is the amount which passes
through an &rea of one square centimeter, situated
in the center of a circle 10 centimeters in diameter,
ELECTBO-MAGNETS, DYNAMOS AND MOTORS. 83
surrounded by a wire through which a current of
7.9578 amperes is passing. The plane of the circle
is a cross-sectional plane of the field; a line perpen-
dicular to such plane gives the direction of the lines
of force, or of the magnetic flux.
This cross-sectional area is often spoken of as the
field of force. As a field exists wherever there are
lines of force, there are in each magnetic circuit
either an infinite number of fields of force, or a
field of force is a volume and not an area.
The number of lines of force or of magnetic flux
per unit cross-sectional area of the magnetic cir-
cuit, i. e. per unit area of magnetic field, expresses
the intensity of the field. In soft iron, it may run as
high as 20,000 or more lines per square centimeter
of cross-section of the iron which is magnetized.
Just as we might speak of a bar of copper acting
as conductor for 20,000 0. G. S. units of current, or
2000 amperes, so we may speak of the iron core of a
magnet carrying 20,000 lines of force.
Permeance and Reluctance.
This action of centralizing in its own material
lines of force is analogous to " conductance. " It is
termed permeance. Its reciprocal is termed
reluctance, which is precisely analogous to
" resistance. " Iron, nickel, and cobalt possess high
permeance; the permeance of air is takten as unity.
At a low degree of magnetization, soft iron pos-
84 ARITHMETIC OF ELECTRICITY.
sesses 10,000 times the permeance of air. At high
degrees of magnetization, it possesses much less in
comparison with air, whose permeance is unchanged
under all conditions.
There is no substance of infinitely high reluctance,
which is the same as saying that there is no insula-
tor of magnetism.
Magnetizing Fobce and the Magnetic Cib-
cuit.
The producing cause of the magnetic flux or mag-
netization just described is in practice always a cur-
rent circulating around an iron core. The name of
magnetizing fobce is often given to it. It is the
analogue of electro-motive force, and is measured by
the lines of force it establishes in a field of air of
standard area.
A high value for the magnetic force is 585 lines
per square centimeter. It is proportional to the
amperes of current and to the number of turns the
conductor makes around it. Its intensity is often
given in ampere-turns.
Magnetization always implies a circuit. As far as
known, magnetic lines of force cannot exist without
a return circuit, exactly like electric currents. But
owing to the imperfect reluctance of all materials,
the lines of force can complete their circuit through
any substancer. In a bar magnet the return branch
of the circuit is through air.
ELECTBO-MAGNETS, DYNAMOS AND MOTOBS. 85
In the same magnetic circuit, the planes of nor-
mal cross-section lie at various angles with each
other.
The law of a magnetic circuit is exactly compar-
able to Ohm's law. It is as follows:
Rule 63* The magnetization expressed in lines of
force is equal to the magnetizing force divided by the
reluctance or multiplied by the permeance of the entire
circuit.
This rule would be of very simple application, ex-
cept for the fact that reluctance increases, or per-
meance decreases, with the magnetization, and the
rate of variation is different for different kinds of
iron.
Rule 64. Permeability is the ratio of magnetization
to magnetizing force, and is obtained by dividing mag*
netlzation by magnetizing force*
Permeability has to be determined experimentally
for each kind of iron. It is simply the expression
of a ratio of two systems of lines of force. It
always exceeds unity for iron, nickel, and cobalt.
The specific susceptibility of any particular iron to
magnetization is its permeability. The susceptibil-
ity of a portion, or of the whole of a magnetic cir-
cuit is its permeance.
General Rules for Electro-Magnets.
The traction of a magnet is the weight it can
sustain when attached to its armature. It is pro-
86 ARITHMETIC OF ELECTRICITY.
portional to the square of the number of lines of
force passing through the area of contact.
Role 65. The traction of a magnet In pounds Is equal
to the square of the number of lines of force per square
Inch, multiplied by tne area of contact and divided by
72,134,000. In centimeter measurement tne traction
in pounds Is equal to tne square of tne number of lines
of force per square centimeter multiplied by the area of
contact and divided by 11,183,000. The traction In
grams Is equal to the latter dividend divided by 24,655
(8 » x 081); for dynes of traction the divisor Is 25.132
<8»)
Examples.
A bar of iron is magnetized to 12,900 lines per
square inch; its cross-section is 3 square inches.
What weight can it sustain, assuming the armature
not to change the intensity of magnetization?
Solution: 12,900* X 3 = 499,230,000. This di-
vided by 72,134,000 gives 6.914 lbs. traction.
A table calculated by this rule is given. A
diminished area of contact sometimes increases trac-
tion, and a non-uniform distribution of lines may
occasion departures from it. The above rule and
the table alluded to are practically only accurate
for uniform conditions. The reciprocal of the rule
is applied in determining the lines of force of a
magnet experimentally.
Rule 66* The lines of force -which can pass through a
magnet core -with economy are determined by the tables,
keeping in mind that it is not advisable to let the per-
meability fall below 200—300. From them a number
Is taken (40,000 lines per square Inch for cast Iron or
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 87
100,000 lines per square Inch for wrought Iron are good
general averages) and is multiplied by the cross-sec-
tional area of the magnet core.
Rale 67. To calculate the magnetizing force in am-
pere turns required to force a given number of magnetic
lines through a given permeance, multiply the desired
lines of force by the reluctance determined as below.
Rule 68* a. The reluctance of a core or of any portion
thereof for inch measurements Is equal to the product
of the length of the core or of the portion thereof by
0.31 32 divided by the product of its cross-sectional area
and permeability.
b. The reluctance for centimeter measurements is
equal to the length of the core divided by the product of
1.2566, by the cross-sectional area and the permea-
bility.
Examples.
440,000 lines are to be forced through a bar of
wrought iron 10 inches long and 4 square inches in
area; calculate its reluctance and the magnetizing
force in ampere turns required to effect this mag-
netization.
Solution: The reluctance (a) = 10 X .3132 -*- (4 X
permeability). 440,000 lines through 4 square inches
area is equal to 110,000 lines through 1 square
inch; for this intensity and for wrought iron the
permeability = 166. 166 X 4 = 664. The reluc-
tance therefore = 3.132 + 664 = .0047. The mag-
netizing force in ampere turns = 440,000 X .0047
= 2068.
The same number of lines are to be forced
through a bar 25.80 square centimeters area and
88 ARITHMETIC OF ELECTRICITY.
25.40 centimeters long. Calculate the ampere
turns.
Solution: 440,000 lines through 25.80 sq. cent.
= ~^jp = 17,054 through 1 sq. cent., for which the
permeability = 161. The reluctance therefore, (b)
= 25.40 ■+■ (1.2566 X 25.80 X 161) = .0048. The
ampere turns = 440,000 X .0048 = 2112.
Magnetic Cibcuit Calculations.
Practically useful calculations include always the
attributes of a full magnetic circuit, because mag-
netization can no more exist without a circuit than
can an electric current. In practice an electro-
magnetic circuit consists of four parts: 1, The
magnet cores; 2 and 3, the gaps between armature
and magnet ends; 4, the armature core. To cal-
culate the relations of magnetizing force to magne-
tization the sum of the reluctances of these four
parts has to be found. A further complication
is introduced by leakage. The permeability of well
magnetized iron being so low, not exceeding 150 to
300 times that of air, a quantity of lines leak across
through the air from magnet limb to magnet limb.
Leakage is included in the sum of the reluctances
by multiplying the reluctance of the magnet core by
the coefficient of leakage, which is calculated for
each case by more or less complicated methods. For
parallel cylindrical limb magnets the calculation is
ELECTRO-MAGNETS, BYNAMOS AND MOTORS. 89
exceedingly simple. The calculation in all cases is
simplified by the fact already stated, that in air
permeability is always equal to unity, whatever the
degree of magnetization. For copper and other
non-magnetizable metals the variation from unity is
so slight that it may, for practical calculations, be
treated as unity.
Leakage op Lines op Fobce.
Leakage is the magnetic flux through air from
surfaces at unequal magnetic potential, such as
north and south poles of magnets. It is measured
by lines of force and is proportional to the relative
permeance of its path.
The coefficient of leakage of a magnetic circuit is
the quotient obtained by dividing the total magnetic
flux by the flux through the armature. The total
magnetic flux is the maximum flux through the
magnet core.
Rule 69. To obtain the coefficient of leakage divide
the permeance of the armature core and of the two saps
pins one-half the permeance of air between magnet
limbs by the permeance of the armature core and of the
two gaps*
Example.
The total flux through an armature core is found
to be at the rate of 70,000 lines per square inch, and
the armature core is 3 inches diameter and 10 inches
long. The average length of travel of the magnetic
90 ARITHMETIC OF ELECTRICITY.
lines through it is 11 inches. The air gaps are 10 X
3 inches area and i inch thick. The permeance be-
tween the limbs of the magnet is 500. Calculate
the coefficient of leakage.
Solution: 70,000 lines per square inch gives a
permeability of 1,921. By Rule 68 the reluctance
of the armature core is ^ £ im X .3132 = .000008.
The reluctance of a single air gap is i X .3132 =
.0052. Thus the armature reluctance is so small
that it may be neglected. The permeance of the
two air gaps is given by 0058 1 x2 = 100 (about). The
coefficient of leakage = 100 f ^° = 3.5.
As the coefficient of leakage is the factor used in
these calculations, the permeance of the leakage paths
is the desired factor for its determination. In the
case of cylindrical magnet cores parallel to each
other, they are obtained from Table XIII. given in
its place later. It is thus calculated and used.
The least distance separating the cores (b) is di-
vided by the circumference of a core (p) giving the
ratio (~) of least distance apart to perimeter of a
core. The number corresponding in columns 3 or 5
is multiplied by the length of a core. The product
is the permeance. Columns 2 and 4 give the re-
luctance. To reduce to average difference of mag'
netic potential divide by 2.
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 91
Example.
Calculate the permeance between the legs of a
magnet^ 3 inches in diameter and 12 inches high
and 5 inches apart.
Solution: The perimeter = 3 X 3.14 = 9.42. ^ =
i or .5 nearly. Prom the table of permeability
we find 6.278. Multiplying this by 12 we have
6.278 X 12 = 75.336, the permeance. Dividing by
2 we have ^^ = 37.668, the permeance for use in
the calculation of leakage coefficient.
It will be observed that this calculation is based
entirely on the ratio stated, and that absolute di-
mensions have no effect on it.
For flat surfaces, parallel and facing each other,
the following method precisely comparable to the
rule for specific resistance is used:
Rule TO. The permeance of the air space between flat
parallel surfaces is equal to their average area multi-
plied by 3*193 and divided by their distance apart, all
in inch measurements.
Example.
Determine the permeance between the two facing
sides of a square cored magnet 15 inches long, 3
inches wide and 8 inches apart.
Solution: 3 X 15 = 45 (the average area); 45 X
3.193 ■*■ 8 = 17.96. For use in calculations it
should be divided by 2 giving 8.98. This division
by 2 is to reduce it to the average difference of mag-
netic potential between the two magnet legs.
92 ARITHMETIC OF ELECTRICITY.
Calculations fob Magnetic Cibcuits.
A magnetic circuit is treated like an electric one*
The permeance (analogue of conductance) or reluc-
tance (analogue of resistance) is calculated for its
four parts, magnet core, two air gaps, and armature
core. The leakage coefficient is determined and ap-
plied. The requisite magnetizing force is calcu-
lated in the form of ampere turns (the analogue of
volts of E. M. P.). The preceding leakage rules
cover the case of parallel leg magnets. For others
a slight change is requisite in the leakage calcula-
tions, but in practice an average can generally be
estimated.
Example.
Assume the magnet and armature of a dynamo.
The magnet is of cast iron, each leg is cylindrical in
shape, 4 inches in diameter and 20 inches high.
Prom center to center of leg the distance is 9 inches.
The armature core of soft wrought iron is 4 inches
in diameter and 8 inches long, the pole pieces curv-
ing around it are 4 inches, measured on the curve
inside, by 8 inches long. The air gap is \ inch
thick. Calculate the reluctance of the circuit and
the ampere turns for 500,000 lines of force.
Solution: The pole pieces approach within 2*
inches of each other. This leaves 1| inches of the
diameter of the armature core embedded or included
within or embraced by them. One-half of this
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 9?
amount may be taken and added to 2i giving 3J as
the average depth of core for an area 4 X 8 = 32
square inches. The lines per square inch of arma-
ture core are ^ =15,625 lines per square inch.
By the table of permeability, 4650 is given for per-
meability for 30,000 lines in soft iron. For 15,625
lines per square inch 9,000 can safely be taken for
permeability. Its relative reluctance is therefore
^ x 8 ^ 000 = .000011 relative armature core reluctance.
(i)
The relative reluctance of one air gap (permeabil-
ity = 1) is i + 32 = .0078 and .0078 X 2 = .0156 =5
air gaps reluctance (2).
The ratio jj of the table for determining the leak*
age between cylindrical magnet legs is 4 x 5 8 14 - = .4.
5 is the distance between the legs. Permeance cor-
responding thereto is 6.897, which multiplied by 20,
the length of the legs, and divided by 2 for average
magnetic potential difference gives 68.97 for rela-
tive effective permeance (3).
The relative reluctance of the air gaps and arma-
ture core is .015611; the reciprocal or permeance is
64.06 (4).
For coefficient of leakage we have (64.06 ~f- 68. 97)
+ 64.06 = 2.08 (5).
To find the relative reluctance of the magnet core
whose yoke may be taken as of mean length inohes
94 ARITHMETIC OF ELECTRICITY.
and of area equal to that of the core (3.14 X 2 2 =
12.56) we have to first determine the permeability.
g %ffi = 40,000 lines per square inch, corresponding
to a permeability of 258. For the effective reluc-
tance of the magnet core introducing the factor of
leakage (2.08) we have the expression * + J£*£* M
- .0314.
To get ampere turns, we add the reluctances of
circuit, multiply by .3132 and by the required lines,
(.000011 + .0156 + .0314) X .3132 X 500,000 =
7362 ampere turns required.
In the above calculations, the multiplication by
.3132 was omitted to save trouble, relative reluc-
tances only being calculated, until the end when one
multiplication by .3132 brought out the ampere
turns. The leakage appears excessive partly be-
cause of the high reluctance of the two air gaps.
These should be increased in area and reduced in
depth if possible. The leakage is also high on
account of the legs of the magnet being close to-
gether. Were these separated, a larger armature
core might be used, justifying a lower speed or rota-
tion of armature, reducing reluctance of air gaps by
increasing their area, and reducing leakage between
magnet legs by increasing their distance. The
magnet legs might also be made shorter, thus reduc-
ing leakage.
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 95
Thus assume the magnet core of the same cross-
sectional area, bat only 10 inches long and with a
distance apart of legs of 7 inches, giving a 7 X 10
inch armature core and pole piece areas (air gap
areas) of 7 X 10 -■ 70 sq. inches.
For leakage ratio we have (£) ■■ j^g — .56 giv-
ing from the proper table 6.000 (about), ^ 10
= 30 relative permeance of air space between legs.
For air gaps reluctance £ -*- 70 = .00357 which
for the two gaps gives .00714 relative reluctance.
Treating the armature core as a prism 7 X 10 «= 70
sq. inches area and 5 inches altitude, we have for
lines per sq. inch 500,000 + 70 = 7000 giving it
about 9000 and reluctance as 5 + (70 X 9,000) =»
.000008 reluctance.
Air gaps and armature core reluctance ~ .007148
and permeance = -^^ = 139.
Coefficient of leakage — ^±*> = 1.21.
If the depth of the air gaps was reduced to J inch
the coefficient of leakage would then be about 1.11.
Every surface in a magnet leaks to other surfaces
and the leakage from leg to leg is sometimes but one
third of the total leakage. In practice the total
leakage often runs as high as 50#, giving a coefficient
of 2.00 and in other cases as low as 25£, giving a co-
efficient of 1.33.
96 ARITHMETIC OF ELECTRICITY.
Dynamo Abmatures.
An armature of a dynamo generally comprises two
parts — the core and the winding. The core is of
soft iron. Its object is to direct and concentrate
the lines of force, so that as many as possible of
them shall be cut by the revolving turns or convolu-
tions of wire. The winding is usually of wire. It
is sometimes, however, made of ribbon or bars of
copper. Iron winding has also been tried, but has
never obtained in practice. The object of the wind-
ing is to cut the lines of force, thereby generating
electro motive force. The number of the lines of
force thus cut in each revolution of the armature
is determined from the intensity of the field per
unit area, and from the position, area and shape of
the armature, coils and pole pieces. The number
thus determined, multiplied by the number of times
a wire cuts them in a second, and by the effective
number of such wires, gives the basis for determin-
ing the voltage of the armature.
Bale 7 1 • One -volt E. RE* F. is generated by the cutting
of 10 8 (100,000,000) lines of force in one second*
Examples.
A single convolution of wire is bent into the form
of a rectangle 7 X 14 inches. It revolves 25 times a
second in a field of 20,000 lines per square inch.
What E. M. P. will it develop at its terminals?
Solution: The area of the rectangle is 7 X 14 = 98
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 97
square inches. Multiplying this by the lines of force
in a square inch, we have 98 X 20,000 = 1,960,000.
Each side of the rectangle cuts these lines twice in
a revolution, and makes 25 revolutions in a second.
This gives 25 X 2 X 1,960,000 = 98,000,000 lines
cut per second, corresponding to 98 X 10* X 10 -8 =
98 X 10" 2 = Ak volts E. M. F. generated, or 1W0W0V0
= Tfifr VOltS.
The field of the earth in the line of the magnetic
dip — .5 line per square centimeter. Calculate a
size, number of layers, and speed of rotation for a
one volt earth coil.
Solution : We deduce from the rule the following :
Area of coil X revolutions per sec. X convolutions
of wire X .5 X 10 -8 — .5. We may start with revolu-
tions per second, taking them at 20. Next we may
take 50,000 convolutions. 20 X 50,000 X .5 —
500,000. This must be multiplied by 200 to give
10 8 ; in other words, the average area within the
wire coils must be 100 square centimeters, or 10 X
10 centimeters. 2 X 100 X 2000 X 500 X .5 — 10 8 ,
and 10 8 X 10- 8 — 1 volt
Rule T2. The capacity of an armature for current Is
determined by tbe cross-section of its conductors. This
should be such as to allow 520 square mils per ampere
= 1923 amperes per square inch area.
Example.
A drum armature coil is of 4 inches diameter,
and is wound with wire th of the periphery of the
98 ARITHMETIC OF ELECTRICITY.
drum in diameter; the wire is 100 feet long. Its
E. M. F. is 90 volts. What is the lowest admissible
external resistance?
Solution: The circumference of the drum is 3.14
X 4 =» 12.56 inches. The diameter of the wire is
^ — .0418 in. or 42 mils. The area of the wire is
21' X 3.14 = 1387 square mils. By the rule the
allowable current in amperes for a single lead of
such wire is -W = 2.66 amperes. But on a drum
armature the wire lies with two leads in parallel.
Hence it has double the above capacity or 2.66 X 2
=» 5.32 amperes. The resistance of such wire may
be taken at .137 ohm. By Ohm's law the total re-
sistance for the current named must be ^ or 17
ohms. The external resistance is given by 17 ~
.137 = 16.863 ohms.
These two rules enable us to calculate the ca-
pacity of any given armature. Certain constants
depending on the type of armature have to be intro-
duced in many cases.
Drum Type Closed Circuit Armatures.
For these armatures the following rules of varia-
tion hold, when they do not differ too much in size,
and are of identical proportions.
Bale 78* a. The B. RE. F. varies directly with the
square of the else of core and with the number of turns
of wire.
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 99
b. The current capacity varies with the sixth root
of the size of core for identical E. M. F.
c. The resistance varies directly with the cube of the
number of turns and inversely with the size of core*
d. The amperage on short circuit varies directly with
the cube of the size and Inversely with the square of
the number of turns*
In these rules the proportions of the drum are
supposed to remain unchanged. Size may be re-
ferred to any fixed factor such as diameter, as lineal
size is referred to.
These rules enable us to calculate an armature
for any capacity and voltage. As a starting point a
given intensity of field, speed of rotation, and num-
ber of turns of wire and size of wire has to be taken.
The wire is selected to completely fill the periphery
of the drum. Then a trial armature is calcu-
lated of the required voltage and its amperage is
calculated. With this as a basis, by applying Rule
73, sections a and b, the size of an armature for the
desired current capacity is calculated, the E. M. F.
being kept identical. As a standard for medium
sized machines 20£ of the turns of wire may be con*
sidered inactive.
Example.
Calculate a 100 volt, 20 ampere armature, whose
length shall be twice its diameter, to work at a
speed of 15 revolutions per second.
Solution: Take as intensity' of field 20,000 lines
per square inch. Allow 80# of active turns of
100 ARITHMETIC OF ELECTRICITY.
wire. Start with a core 8 X 16 = 128 square inches,
including 128 X 20,000 = 256 X 10* lines of force.
The given speed is 15 rotations per second. For
the number of active turns of wire per volt we
have to divide 10 8 or 100,000,000 by one half the
lines of force cut by one wire per second. This
number is 256 X 10 4 X 15, or 38,400,000; and
^;^ =2.6 turns. For 100 volts, therefore, 260
active turns are needed. If one half the lines were
not taken the result would be one half as great, be-
cause each line cuts each line of force twice in a
revolution, and in the computation a single cutting
per revolution only is allowed for.
The reason for thus taking one half the lines cut
by a single wire as a base is because in the drum arma-
ture the wires work in two parallel series, giving one
half the possible voltage. The actual turns are
260 -*- .80 = 325, say 324 turns. Assume it to be
laid in two layers giving 162 turns to the layer.
The space occupied by a wire is equal to the peri-
meter divided by the number of wires or t% = .154
in. Allowing 25# for thickness of insulation, lost
space, etc., we have .115 in. or 115 mils as the
diameter of the wire. In the drum armature as just
stated the wire is parallel, so that the area of one
lead of wire has to be doubled, giving 10,573 X
2 square mils as the. area of the two parallel leads.
This is enough for 40 amperes or double the amper-
age required. This capacity is reached by taking
ELECTRO-MAGNETS, DYliAmkJ^MOTbS& 101
520 square mils per ampere as the proper cross-sec-
tional area of the wire. (Rule 72.)
"We must therefore reduce the size to give }4 the
ampere capacity; this reduction (by Rule 73 b) is in
the ratio 1 : j£* = 1 : .89018 ; the size therefore is
8 X .89018 diameter by 16 X .89018 length = 7.12
X 14.24 inches.
Applying a for voltage we have for the same num-
ber of turns on the new armature a voltage in the
ratio of 1 : .89018 2 or about A of that required. We
must therefore divide the number of turns in the
trial armature by .89018 2 , giving for the number of
turns ^ = 409, say 410 turns.
To prove the operation we first determine the
voltage of the new armature. Its area is 7.12 X
14.24 = 101.4 square inches including 2,028,000 lines
of force. The active wires are 410 X .8 = 328.
We have for the voltage = , ^^xi64xi5 = 99>?8
VOlt8.
The relative capacity of the wire is deduced from
the square of its diameter. The circumference of
the new armature is 7.12 X 3.14 = 22.3568. There
are 205 turns in a layer giving as diameter of wire
~jjp==.1091 mils. This must be squared, giving
.0119, and compared with the square of the corre-
sponding number for the original armature. This
number was 25 -*- .162 = .154 inch. .154 a = .02371
102 • £a£fxkMT$cr or electricity.
and .01190 + .02371 = i (nearly), showing that
the new armature has one half the ampere capac-
ity of the old, or 40 X J = 20 amperes as re-
quired.
The gauge of the wire is reached by making the
same allowance for insulation and lost space, viz.,
25*. .1091 X .75 = .0818 in. or 81.8 mils diameter,
for size of wire. Of course there is nothing absolute
about 25* as a loss coefficient; it will vary with
style of insulation and even to some extent with the
gauge of wire. But as Rule 73 is based upon the
assumption that this loss is a constant proportion of
the diameter of the wire, too great a variation of
sizes should not be allowed in its application. In
other words the trial armature should be as near as
possible in size to the final one.
Suppose on the other hand that an armature for
100 amperes was required. This is for 2% times
40 amperes (the capacity of the first calculated or
trial armature).
Applying b we extract the 6th root of 2j£. (2J^)$
= 1.1653 (by logarithms or by a table of 6th
roots). The size of the new armature is therefore
8 X 1.1653 by 16 X 1.1653 or 9.3224 X 18.6448
inches.
Applying a for voltage we have for the same num-
ber of turns of the new armature a voltage in the
ratio of 1.1653 s : 1 or 1.358 times too great. We
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 103
must therefore multiply the original turns by the
reciprocal of 1.358, giving ^ = 239 turns.
To prove the voltage, we multiply 239 by .8 for
the active turns of wire, giving 191.2 turns. The
area of the armature is 9.32 X 18.64 = 172.7 square
inches. For voltage this gives 172.7 X 191.2 X 20,-
000 X 15 X 10- 8 = 98.6 volts (about).
To prove the capacity we must divide the circum-
ference of the new armature, 9.32 X 3.14 = 29.26
inches, by the turns of wire in one layer, 4* = 120
turns (about). This gives a diameter of 244 mils
(nearly). The ratio of capacities of the original and
this wire is .244* -*- .154 2 inches = .059536 -*- .02371
=2.51 corresponding to 40 X 2.51 = 100 amperes.
These results, owing to omissions of decimals, do
not come out exactly right and it is quite unneces-
sary that they should. The accuracy is ample for
all practical purposes. For armature dimensions it
would be quite unnecessary to work out to the sec-
ond decimal place. It would answer to take as ar-
mature sizes in the two cases given 7 X 14j£ inches
and 9# X 18# inches.
It is also to be noted that a very low rate of rota-
tion was taken. 25 to 30 turns per second would
not have been too much. The latter would give
double "the voltage and the same amperage.
Field Magnets of Dynamos.
The calculation for a magnetic circuit given on
104 ARITHMETIC OF ELECTRICITY.
pages 92 et seq., is intended to supply an example of
the calculation of the circuit formed by a field mag-
net and its armature, such as required for dynamos.
The leakage of lines of force is and can only be so
incompletely calculated that it is probably the best
and most practical plan to assume a fair leakage
ratio and to make the magnet cores larger than re-
quired by the lines of force of the armature in this
ratio., A low multiplier to adopt is 1.25, which is
lower than obtains in most cases; 1.50 is probably a
good average.
Rale 74* The cross-sectional area of the field-magnet
cores Is equal to the lines of force In the field divided by
the magnetic flax (column B) for the material selected
and corresponding: to the chosen permeability (m) 9 mul-
tiplied bjr the leakage coenlcient.
A good range for permeability is from 200 to 400
giving for wrought iron from 100,000 to 110,000
lines of force per square inch and for cast iron from
35,000 to 45,000 lines per square inch; for the field
from 15,000 to 20,000 lines per square inch may be
taken.
The permeability table gives data for different
qualities of iron.
Example.
Taking the 100 volt 100 ampere armature last cal-
culated, determine the size of field-magnet core3 to
go with it, and the ampere turns and other data.
Solution: Assume 20,000 lines of force per square
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 105
inch in the field, 45,000 in cast iron and 110,000 in
wrought iron core and a leakage coefficient of 1.25.
We have for total lines of force passing through
armature 172.7 X 20,000 = 3,454,000; cross-sectional
area for cast iron core ^^ X 1.25 = 96 square
inches; cross-sectional area for wrought iron core
8 ' 45 ^ff ) X 1.25 = 39 square inches.
As length of cores we may take 20 inches with
a distance between them of 10 inches. Assume
wrought iron to be selected. If cylindrical they
would be 7 inches in diameter to give the required
cross-sectional area. The yoke connecting them
would average in length 10 + 7 = 17 inches, giving
for magnet cores and yoke a length of 17 + 20 + 20
= 57 inches. The reluctance of cores and yoke
(Rule 68) = m^ (taking r = 200) which reduces
to .00132 (1).
The armature area is 172.7 inches. As average
length of the path of lines of force through it 5
inches may be taken. As it passes only 20,000 lines
of force per square inch of field its permeability is
high, say 9000. Its reluctance is given by 178x90 oo-
This is so low that it may be neglected.
The area of each air-gap may be taken as 173
square inches, and of depth of two windings plus
about A inch for clearance or windage giving
(.224 X 2) + .1 = about .6 inch for its depth. Its
106 ARITHMETIC OF ELECTRICITY.
reluctance is 6 *^ 182 *=* .00108. As there are two air
gaps we may at once add their reluctances giving
.00216 (2).
By Rule 67 the ampere turns are equal to the
product of the reluctances (1) and (2), by the lines
of force giving (.00121 + .00216) X (172.7 + 20,000)
= 11640 ampere turns.
The proper size of wire for series winding may be
determined by Sir William Thompson's rule that in
series wound dynamos the resistance of the field mag-
net windings should be f that of the armature. The
length of the wire in the armature is equal approxi-
mately, to the circumference 9.32 X 3.14 = 29.26
multiplied by the number of turns (240) giving
29.26 X 240 = 7022 inches.
The wire turns on the field magnets are found by
dividing the ampere turns by the amperes giving
H¥* = 232 turns. The circumference of the mag-
net leg is 7.0 X 3.14 = 22 inches. The total length
of wire is therefore, approximately, 232 X 22 = 5104
inches.
To compare the resistances we must use **5P for
the length of the armature wire, because it is in
parallel, and therefore is i the length and i the re-
sistance of the full wire in one length. Dividing by
4 introduces this factor.
As the resistances of the wires are to be in the ratio
of 2 : 3, we have by Rule 13 (calling the thickness
of armature Wire 244 X .75 = 183 mils to allow for
ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 107
insulation, etc.), 2:3:: 183* X 5104 : a 2 X *¥*, and
solving we find a? = 73026 . \ x = 270 mils.
For shunt winding Sir William Thompson's rule
is that the product of armature and field resistance
should equal the square of the external resistance.
The latter may be taken (Ohm's law) as equal to
louamperes = * °^ m * P ro P er ty the armature resist-
ance should be allowed for, but it is so small that it
need not be included. We have therefore, arma-
ture resistance X field resistance = l 2 X 1,
The armature resistance is .0419 ohms. There-
fore the field resistance is -~- 9 = 24 ohms. The cur-
rent through this is equal to W = 4 amperes
(nearly). Therefore ^4** = 2910 turns of wire are
needed. The length of such wire will be * ^g 2910
= 5335 feet. The resistance is about 4.4 ohms per
1000 feet corresponding to about .48 mils diameter.
The Kapp Line.
Mr. Gisbert Kapp, C. E. who has given much in-
vestigation to the problems of the magnetic circuit
and especially to dynamo construction, is the orig-
inator of this unit. He considered the regular C.
6. S. line of force to be inconveniently small. He
adopted as a line of force the equivalent of 6000 C.
G. S. lines and as the unit of area one square inch.
Therefore to reduce Kapp lines to regular lines of
force they must be multiplied by 6000, and ordi-
108 ARITHMETIC OF ELECTRICITY.
nary lines of force must be divided by 6000 to obtain
Kapp lines. These lines are often used by English
engineers. The regular system is preferable and by
notation by powers of ten can be easily used in all
cases.
CHAPTER X.
electric railways.
Sizes of Feeders.
To calculate the sizes of feeders for a trolley line
Rules 23, 24, and 25 in Chapter V. will be found use-
ful in conjunction with the following ones:
A. For load at end of feeders :
Rule 75. The cross-section of the feeder in cir-
cular mils is equal to the product of 10.79 times the
current in amperes times the length of the con-
ductor in feet, divided by the allowable drop in
-volts. #
Example.
What should be the cross section of a feeder 3,000
feet long carrying 90 amperes with 35 volts drop ?
Solution: 3,000 X 90 X 10.79 = 2,913,300. Di-
viding this by 35 gives 83,237 circular mils. This
would correspond to a No. 1 wire, which has 83,694
cir. mils area.
All computations of this kind should be checked
by table XVI. of current capacity on page 153 in
order to be sure that the wire will not become heated
above the allowable limit.
110 ARITHMETIC OF ELECTRICITY.
Referring the above example to this table, it is
seen that a No. 1 wire will carry 80 amperes with a
rise in temperature of 18° F., and 110 amperes with
a rise of 36° F. Hence the current of 90 amperes
will cause a rise of 24° F. This is calculated by
simple proportion; subtracting 80 from 90 and from
110 gives 10 and 30 as the respective differences and
shows 90 to lie at just one-third the distance from
80 to 110. Hence the resulting temperature rise
will be at one-third the distance between 18° and
36°. The difference between these last two figures
is 18°, one-third of which is 6°. Add this 6° to
18° gives us 24° F. as the answer.
It is often desirable to compute the drop on a
feeder carrying a given current; this is done by the
following:
Rule 76. The drop in volts on any conductor ia
found by multiplying- together 10.79, the current in
ampere* and it« length in feet, then divide this
product by it* area in circular mils.
Example.
What is the drop on a feeder 2,800 feet long, of
105,592 cir. mils area and carrying a current of 125
amperes ?
Solution: 10.79 X 125 X 2,800 = 3,776,500.
Dividing this product by 105,592 gives 35.76
volts drop.
B. For a uniformly distributed load:
ELECTRIC RAILWAYS. Ill'
The effect of a uniform distribution of load along
a main or trolley wire is the same as that of half
the total current passing the full length of the wire;
hence we require but half the cross-section needed to
deliver the current at the extremity of the wire.
This is readily done by substituting the constant
5.4 in place of 10.79 in the foregoing rules.
In designing electric railway circuits where the
track forms the path for the return current the rails
should be of ample area and well bonded, with an
extra bare wire connected to the bonds and materially;
reducing the drop in the track circuit.
Power to Move Cars.
At ordinary speeds on a level track in average con-
dition it is safe to assume that the force necessary
to move a car is 30 pounds per ton of weight of car.
Rale 77. To find the force required to pull or
push a cor on a level track in average condition,
multiply the weight of the ear in tons by 80.
Example.
Find the force required to drag a car weighing 7
tons on a level track.
Solution: 7 X 30 = 210 lbs. Ans.
Should it be required to find the force needed to
start a car on a level, or to propel it when round-
ing a curve, substitute the constant 70 in place of
30 in the foregoing rule.
112 ARITHMETIC OF ELECTRICITY.
Example.
What force is needed to start an 8 ton car on a
level track?
Solution: 8 X 70 = 560 pounds. Ans.
As the above does not take into account the speed
of the car we shall have to add this factor in order
to find the horse power needed to move it; we will
also allow for the efficiency of the motors.
Rule 78. To find the hone power required to move
a car along a level track multiply together the dis-
tance In feet traveled per minute and the force In
pounds necessary to move the car (as found by Rule
77), and divide the result by 33,000 times the ef-
ficiency of the motors.
Example. "
What horse power is needed to propel a loaded car
weighing 9 tons along a level track at the rate of
800 feet per minute, with motors of 70 per cent,
efficiency ?
Solution: Force to move car is 9X30 = 270
pounds. The product of 800 X 270 == 216,000 foot
pounds per minute. Dividing this by 33,000 gives
6.54 H. P. required to propel the car. Dividing by
the efficiency .70 gives 9.34 H. P. to be delivered to
the motors.
It will be noted in this solution that the quantity
216,000 should, according to the rule, have been di-
vided by the product of 33,000 times .70; it was,
ELECTRIC RAILWAYS. 113
however, divided by these two factors successively in
order to show the difference between the power actu-
ally moving the car and that supplied to the motors.
In computing the power taken by a car ascending
a grade the equivalent perpendicular rise of the car
together with its weight in pounds have to be consid-
ered in addition to the factors involved in the rule
just preceding.
Rule 79. To And the horse power required to pro*
pel a car up a grade, take the product of the
perpendicular distance In feet ascended by the car
In one minute multiplied by Its weight In pounds;
to this add the product of the horizontal distance
In feet traveled In one minute multiplied by the
force In pounds required to propel the car; divide
this sum by 33,000 times the efficiency of the mo-
tors.
Note. — The grade of a road or track is generally
stated as being so many per cent. This means that
for any given horizontal travel of a car its change of
altitude when referred to a fixed horizontal plane is
expressed as a certain percentage of the horizontal
travel. For illustration; if a car while traveling
horizontally 100 feet has a total vertical rise (or
fall) of 7 feet, the incline on which it moves is
termed a 7 per cent grade.
Example.
Find the electrical horse power taken by the motors
of an 8 ton car to propel it up a 5 per cent grade at
114 ARITHMETIC OF ELECTRICITY.
a speed of 1,000 feet per minute, the motors having
70 per cent efficiency.
Solution: Perpendicular rise of car is 1,000 feet
X .05 = 50 feet. Weight of car in pounds is 8 X
2,000 = 16,000 pounds. Product of lift and weight
is 50 X 16,000 = 800,000 foot pounds. Force re-
quired to propel car is 30 X 8 = 240 pounds. Pro-
duct of force and distance is 240 X 1,000 =240,000
foot pounds. Sum of the two products is 800,000 +
240,000 = 1,040,000 total foot pounds.
Product of 33,000 by efficiency is 33,000 X .70 =
23,100. Electrical H. P. is the quotient of 1,040,000
-*- 23,100 = 45.021 H. P. Ans.
CHAPTER XL
ALTERNATING CURRENTS.
By far the greater part of calculations in the do-
main of alternating currents lie in the relalm of
trigonometry and the intricacies of the calculus.
On this account it is hoped that the following pres-
entation of some of the simpler formulae may prove
welcome to the craft.
A current flowing alternately in opposite directions
may be considered as increasing from zero to a
certain amount flowing in, say, the positive direc-
tion, then diminishing to zero and increasing to an
equal amount flowing in the negative direction and
again decreasing to a zero value. This action is
repeated indefinitely. The sequence of a positive
and negative current as just described is called a
cycle.
The frequency of an alternating current is the
number of cycles passed through in one second. An
alternation is half a cycle. That is to say, an alter-
nation may be taken as either the positive or the
negative wave of the current.
116 ARITHMETIC OF ELECTRICITY.
The frequency may be expressed not only in cycles
per second but in alternations per minute.
Since one cycle equals two alternations we can
interchange these expressions as follows :
Rale 80. A. Having: g-lven the cycles per second,
to find the alternations per minute multiply the
cycles per second by 120. B. Having: flriven the al-
ternations per minute, to find the cycles per second
divide the alternations per minute by 120*
Examples.
If a current has 60 cycles per second, how many
alternations are there per minute?
Solution: 60X120 = 7,200 alternations.
A current has 15,000 alternations per minute;
how many cycles per second are there?
Solution: 15,000-1-120 = 125 cycles per second.
A bipolar dynamo having an armature with but
a single coil wound upon it (like an ordinary mag-
neto generator) gives one complete cycle of current
for every revolution of the armature. That is to
say, its frequency equals the number of revolutions
per second. A four-pole generator will have a fre-
quency equal to twice the revolutions per second, etc.
Rule 81. To find the frequency of any alternator,
diTide the revolutions per minute by 60 and multiply
the quotient by the number of pairs of poles in the
Held.
Example.
Find the frequency of a 16-pole alternator run-
ning at 937.5 revolutions per minute.
ALTERNATING CURRENTS. 117
Solution: 937.5 -f- 60 = 15.625 rev. per second.
15.625 X 8 = frequency of 125 cycles per second.
Electrical measuring instruments used on alternat-
ing currents do not indicate the maximum volts or
amperes of such circuits, but the effective values are
what they show. These effective values are the same
as those of a continuous current performing the
same work.
Rale 82. The maximum volts or amperes of an
alternating: current may be found by ' multiplying:
the average volts or amperes by 1.11. Reciprocally,
the average values can be found by taking; «707
times the maximum values.
Note that these llgrures are strictly true only for
an exactly sinusoidal current*
Example.
Find the maximum pressure of an alternating
current of 55 volts.
Solution: 55 X 1.11 = 61.05 volts. Ans.
Self-Induction.
In an alternating current circuit the flow of a
current under a given voltage is determined not
only by the resistance of the conductor in ohms but
also by the self-induction of the circuit. Suppose a
current to start at zero and increase to 10 amperes in
a coil of 1,000 turns of wire. This magnetizing
force, growing from zero to 10,000 ampere-turns,
surrounds the coil with lines of force whose action
118 ARITHMETIC OF ELECTRICITY.
upon the current in the coil is such as to resist its
increase. Conversely, when the current is decreas-
ing from 10 amperes to zero, the lines of force
change their direction and tend to prolong the flow
of current. This opposing effect which acts on a
varying or an alternating current is caller the counter
E. M. F. or E. M. F. of self-induction and is meas-
ured in volts.
Rale 88. The E. M. P. of self-induction of a tfiven
coil Is found by multiplying: together 12.5664, the
total number of tarns in the coil, the number of
tarns per centimeter lensrth of the coil, the see-
tional area of the core of the coil in square centi-
meters, the permeability of the magnetic circuit
and the current ; divide the resulting: product by
1,000,000,000, multiplied by the time taken by the
current to reach its maximum value*
Example.
Find the volts of counter E. M. P. in a coil of 300
turns wound uniformly on a ring made of soft iron
wire, the ring having a mean circumference of 60
centimeters and an effective sectional area of 25
square centimeters; its permeability to be taken as
200, and a current of 5 amperes in the coil requires
.02 second to reach its maximum value.
Solution: Product of 12.5664 X 300 X 5 X 25
X 200 X 5 is 471,240,000. Product of 10 9 X .02 is
20,000,000. Dividing the former product by the
latter gives 23.562 volts, Answer.
The coefficient of self-induction or, as it is more
ALTERNATING CURRENTS. 119
frequently termed, the inductance of a coil, is meas-
ured by the number of volts of counter E. M. F.
when the current changes at the rate of one ampere
per second. (Atkinson.)
The unit of inductance is the henry.
Rule 84. The Inductance of a coil la found by
multiplying together 12.56649 the total number of
turns in the coil, the number of turns per centimeter
length, the sectional area of the core of the coil in
square centimeters, and the permeability of the
magnetic circuit; divide the resulting product by
1,000,000,000.
Example.
Find the inductance of the coil specified in the
preceding example.
Solution: The factors are the same as before,
omitting the current and time. Product of 12.5664 X
300 X 5 X 25 X 200 is 94,248,000. Dividing this
by 1,000,000,000 gives the inductance .094248
henrys.
The E. M. P. of self-induction may be computed
when the inductance, the current and the time taken
for the current to reach its maximum are known.
Rule 85. To find the E. M. F. of self-induction
divide the product of the Inductance and current by
the time of current rise*
Example.
Using again the data of the foregoing examples,
find the counter E. M. P., the inductance being
120 ARITHMETIC OF ELECTRICITY.
.094248 henrys, the current 5 amperes, and the time
.02 second.
Solution: 5 X .094248 = .47124; dividing this
by .02 gives 23.562 volts, as before.
Rale 86. The resistance dne to ■elf-induction
equals 6.2832 times the product of the frequency
and the Inductance.
Example.
Find the inductive resistance of a circuit whos9
frequency is 60 cycles per second and the inductance
is .05 henry.
Solution: 6.2832 X 60 X .05 = 18.8496 ohms.
Ans.
The time constant of an inductive circuit is a
measure of the growth or increase of the current.
It is the time required by the current to rise from
zero to its average value. The average value of an
alternating current is .634 times its maximum value.
It must not be confused with its effective value,
which is .707 times the maximum.
The average value may be obtained by multiplying
the effective value, as shown by instruments, by .897.
Rule 87* To find the time constant of a coll or
circuity divide its inductance by its resistance*
Example.
What is the time constant of a coil whose induct
ance is 3.62 henrys and resistance is 20 ohms.
Solution: 3.62 -r- 20 = .181 second. Ans.
CHAPTER XII.
CONDENSERS.
A condenser, though it will allow no current to
pass through it, yet it will accumulate or store up
a quantity of electricity depending on various factors
which the following rules will show:
Rale 88. The quantity stored equal* the product
of the E. M. P. applied and the capacity of the
condenser*
Q, = ESC*
Rule 89. The capacity of a condenser equals the
quantity stored divided by the applied S3. M. P.
C = -.
B
Rule 90. The E. M. F. applied to a condenser equals
the quantity stored divided by its capacity*
11 = —.
C
The quantity stored in a condenser is measured
in coulombs (i. e., ampere-seconds) ; the E. M. P.
in volts, and the capacity in farads. Condensers in
practical use have, however, so small a capacity that
122 ARITHMETIC OF ELECTRICITY.
it is usually stated in microfarads and the quantity
in microcoulombs.
Examples.
A battery of 30 volts E. M. F. is connected to a
condenser whose capacity is one half microfarad.
What quantity of electricity will be stored?
Solution: 30 volts X .0000005 farads = .000015
coulombs. This solution could also be given direct-
ly in micro-quantities, thus: 30 volts X Y2 micro-
farad = 15 microcoulombs.
A condenser is charged with 7.5 microcoulombs
under an E. M. F. of 15 volts. What is its capacity?
Solution: 7.5 microcoulombs ~- 15 volts = .5
microfarad. Ans.
What E. M. F. is required to charge a condenser
whose capacity is .1 microfarad with 21 microcoul-
ombs of electricity?
Solution: 21 microcoulombs -5- .1 microfarad =
210 volts. Ans.
By connecting condensers in parallel the resulting
capacity is the sum of their individual capacities.
When they are connected in series the resulting capa-
city equals 1 divided by the sum of the reciprocals
of their individual capacities. It will be noticed that
these laws of condenser connections are the inverse
of those for the parallel and series connection of re-
sistances.
CONDENSERS. 123
When applying a direct current to a condenser,
as in the above examples, it flows until the increasing
charge opposes an E. M. F. equal to that of the
charging current.
With an alternating current a charge would be
surging in and out of the condenser, so that a real
current will be flowing on the charging wires in
spite of the fact that the actual resistance of a con-
denser, in ohms, is practically infinite.
Rule 01. The alternating current in a circuit hav-
ing capacity equals the product of 6.2832, the fre-
quency, the capacity, and the applied voltage.
Example.
Find the current produced by an E. M. F. of 50
volts and a frequency of 60 cycles per second in a
circuit whose capacity is 125 microfarads.
Solution: The capacity 125 microfarads equals
.000125 farads.
6.2832 X 60 X .000125 X 50 = 2.3562 amperes.
Ans.
Rule 92. The alternating H. M. F. required to he
impressed upon a circuit of a given capacity in
order to produce a certain current is equal to the
current divided by 6.2832 times the product of the
capacity and the frequency*
Example.
Find the E. M. F. necessary to produce an alter-
nating current of 50 amperes at 50 cycles per sec-
ond in a circuit of 80 microfarads capacity.
Solution: 6.2832 X .000080 X 50 = .0251328
124 ARITHMETIC OF ELECTRICITY.
Dividing the current 50 amp. by .0251328 gives 2,000
volts, nearly.
Since, in a condenser circuit, a real current flows
under a given E. M. F., the circuit may be treated as
though it was of a resistance such as would allow
the given current to flow.
Rale 03. The resistance due to capacity equals 1
divided by the product of 6.2832, the frequency and
the capacity.
Example.
Find the capacity resistance of a circuit having a
frequency of 60 cycles per second and a capacity of
50 microfarads.
Solution: 6.2832 X 60 X .000050 = .01885.
1 -f- .01885 = 53 ohms, very nearly.
By comparing this Eule 93 for capacity resistance
with Eule 86 on page 120, which is for inductive re-
sistance, it will he seen that they are mutually recip-
rocal and hence the effect of capacity is directly
opposite to that of self-induction and vice versa.
It follows from this that it is possible, by the
proper proportioning of the inductance and the
capacity, to have their effects neutralized, and when
this adjustment is effected the current will be con-
trolled by the volts and ohmic resistance the same
as if it were a direct current circuit.
Tfte impedance is the apparent resistance of an
alternating current circuit
CONDENSERS. 125
Rale 94. To And the impedance of a circuit whose
?hmic resistance can be neglected and which has
an inductance and a capacity in series, calculate
both the Inductive resistance and the capacity re*
sistance; their difference will be the impedance.
Example.
Find the current produced by an alternating E."
M. F. of 40 volts on a circuit of slight ohmic resist-
ance whose capacity is 100 microfarads, the frequency
being 60 cycles per second, and having in series an
inductance of .02 henry.
Solution: Inductive resistance is 6.2832 X 60
X .02 = 7.42 ; capacity resistance is 1 -r- 6.2832 X
60 X .000100 = 1 -r- .0377 = 26.52.
Impedance = 26.52 — 7.42 = 19.1 ohms.
Current, by Ohm's Law, = 40 -r- 19.1 = 2.08 am-
peres. Ans.
CHAPTER XIII.
DBMOKSTBATIOlSr OF BULES.
In the following chapter we give the demonstra-
tion of some of the rules. As this is not within the
more practical portion of the work, algebra is used
in some of the calculations. It is believed that rules
not included in this chapter, if not based on experi-
ment, are such as to require no demonstration here.
Rule 1 to 6, pages 13 and 14. Ohm's law was
determined experimentally, and all the six forms
given are derived by algebraic transposition from
the first form which is the one most generally ex-
Raie 8, page 1 9. This is simply the expression of
Ohm's law as given in Rule 1, because in the case of
divided circuits branching from and uniting again
at common points, it is obvious that the difference
of potential is the same for all. Hence the ratio as
stated must hold.
Rule 9, page 20. This rule is deduced from Rule
8. It first expresses by fractions the relations of
the current. Next these fractions are reduced to a
common denominator, so as to stand to each other in
128 ARITHMETIC OF ELECTRICITY.
the ratio of their numerators. By applying the new
common denominator made up of the sum of
the numerators the ratio of the numerators
is unchanged, and the ratio of the new fractions is
the same as that of their numerators, while by this
operation the sum of the new fractions is made
equal to nnity. Thus by multiplying the total cur-
rent by the respective fractions it is divided in the
ratio of their numerators, which are in the inverse
ratio of the resistances of the branches of the circuit
and as the sum of the fractions is unity, the sum of
the fractions of the current thus deduced is equal to
the original current.
Rale io f page 22. Resistance is the reciprocal of
conductance. By expressing the sum of the recip-
rocals of the resistances of parallel circuits we ex-
press the conductance of all together. The recipro-
cal of this conductance gives the united resistance.
Rule ii f page 22. This is a form of Rule 10.
Call the two resistances x and y. The sum of their
reciprocals is l+\ which is the conductance of the
two parallel circuits or parts of circuits. Reducing
them to a common denominator we have: iL -f «_
which equals ^^, whose reciprocal is ^-.
Rale it, page 31. Taking the diameter of a wire
as d, its cross sectional area is ^. The resistance
is inversely proportional to this or varies directly
with ^j^^r* As the resistance of a conductor
DEMONSTRATION OF RULES. 129
raries also with its length and specific resistance we
have as the expression for resistance:
Sp. Bea. x 1.2737 X 1
d 9
Bole 18, page 32. Assume two wires whose
lengths are I and Z 1 , their cross sectional areas a and
#!, their specific resistances 8 and s l9 and their resis-
tances r and r v From preceding rules we have for
each wire: r = s i (1) and r x = 8 t J 1 (2).
Dividing (1) by (2) we have:
If we take the reciprocal of either member of this
equation and multiply the other member thereby it
will reduce it to unity, or:
For convenience this is put into a shape adapted
for cancellation.
Rule 20, page 38. This is merely the expression
of Ohm's Law, Eule 3.
Rule 22, page 40. Gall the drop e, the combined
resistance of the lamps R, and the resistance of the
leads x. Then as the whole resistance is expressed
as 300 (because the work is by percentage) the differ-
ence of potential for the lamps is 100 — e. By
Ohm's law we have the proportion: 100-0 : e :: R :
re or
eR
Rule 25, page 44. From Rule 22 we have:
130 ARITHMETIC OF ELECTRICITY.
Call the resistance of a single lamp r, then we
have by Eule 12:
R =£ (2)
Substituting this value of B in equation (1) we
have:
_ er
~ n X (100— €) (3)
Prom Rule 24 we have, calling the cross-section a:
„ l X 10.79
Substituting for z its value from equation (3) we
have:
„ l X 10.79 X n X (100-e)
° " (5)
er
But as I expresses the length of a pair of leads,
not the total length of lead but only one-half the
total, the area should be twice as great. This is
effected by using the constant 10,79 X 2 = 21.58 in
the equation giving:
„ l X 21.58 X n X (100-g)
a = ■
er
Rule 28, page 48. Assuming the converter to work
with 100# efficiency (which is never the case), the
watts in the primary and secondary must be equal
to each other or:
C a E = (V E t , and E = %?.B lf
or the resistances of primary and secondary are in
the ratio of the squares of the currents. The direct
ratio is expressed by the ratio of conversion, when
squared it gives the ratio of the squares as required.
DEMONSTRATION OF BULE8. 131
Rule 87, page 59. Let d = diameter of the wire in
centimeters. The resistance of one centimeter of
such a wire in ohms = Sp. Resist. X 10" 6 X ^. The
specific resistance is here assumed to be taken in
microhms. The quantity of heat in joules devel-
oped in such a wire in one second is equal to the
square of .the current in 0. G. S. units, multiplied
by the resistance in 0. G. S. units and divided by
4.16 X 10 7 , the latter division efEecting the reduc-
tion to joules. 1 ohm = 10 9 C. G. S. units of re-
sistance. Multiplying the expression for ohmic
resistance by 10 9 we have: Sp. Eesist. X 10 8 X ^
1 ampere = 10" 1 0. G. S. unit. If we express the
current in amperes we must multiply it by 10" x , in
other words take one-tenth of it. Our expression
then becomes for heat developed in one second
, Sp. Resist. X 10* X 4
<*)•
X n d* X 4.16 X W
The area of one centimeter of the wire is nd
square centimeters. The heat developed per square
centimeter is found by dividing the above expression
by ic d giving:
ay- 1
Sp. Resist. X 10* X 4
w* d« X 4.16 X 10*
The heat developed is opposed by the heat lost
which we take as equal to dW per square centimeter
per degree Cent, of excess above surrounding me-
dium. Therefore taking t° as the given tempera*
132 ARITHMETIC OF ELECTRICITY.
ture cent, we may equate the loss with the gain
thus:
t° _ /c \« x Sp Resist. X 10* X 4
4000 "~ U0>? «■ d« X 4.16 X 10*
c» X Sp. Resist. X 10* X 4 X 40000
- »« X 4.16 X 10 7 Xt° *
c* X Sp. Resist. X .00089
t°
m
Rule 511, page 69. Call the external resistance r;
number of cells n; resistance of one cell R; E. M.
F. of one cell E; E. M. F. of outer circuit e.
Then from Ohm's law we have:
n nE
c= ^r+f (i)
which reduces to:
Cr
w ~ e-cr (2)
but C r = e.
•'* n " E-CiJ (3)
Rule 54, page 72. This rule is deduced from the
following considerations. The current being con-
stant the work expended in the battery and external
circuit respectively will be in proportion to their
differences of potential or E. M. F's. But these
are proportional to the resistances. Therefore the
resistance of the external circuit r should be to the
resistance of the battery R as efficiency: 1 — effi-
ciency or r : R :: efficiency : 1 — efficiency or
Jg = (1 ~^ncy >Xr ' The rest of the rule is deduced
from Ohm's law.
DEMONSTRATION OF RULES. 133
Rule 57, page 74. This rule gives the nearest ap-
proximation attainable without irregular arrange-
ment of cells. By placing some cells in single series
and others two or more in parallel, an almost exact
arrangement for any desired efficiency can be ob-
tained. Such arrangement are so unusual that it is
not worth while to deduce any special rule for
them. Thus taking the example given on page
74 the impossible arrangement of 1.4 cells in
parallel and 63 in series would give the desired
current and efficiency. The same result can
be obtained by taking 72 cells in 36 pairs with a re-
sistance of 36 x A = 3 ohms, and adding to them
27 cells in series with a resistance of 27 x£ = 4£
ohms, a total of 7* ohms. The E. M. P. is equal to
(36 + 27) X 2 = 126 volts. The total cells are
72 + 27 = 99.
Rule 58, page 76. Ono coulomb of electricity lib-
erates from an electrolyte .000010384 gram of
hydrogen. This has been determined experimen-
tally. Let H be the heat liberated by the chemical
combining weight of any body combining with
another. H is taken in kilogram calories. Hence it
follows that for a quantity of the substance equal to
.000010384 gram X chemical combining weight, the
heat liberated will be equal to H X .000010384,
which corresponds to a number of kilogram meters
of work expressed by .000010384 X H X 424. The
work done by a current in kilogram-meters =
134 ARITHMETIC OF ELECTRICITY.
volte X^coulomb. or f()r one coulomb = «tfta Thig ex .
presses the work done by one coulomb. Let the
volts = E, and equate these two expressions:
i = •000010384 X H X 424,
which reduces to
E - H X .043.
Auieei, page 78. For the work (in kilogram-
meters) done by a current (volt-coulombs) we have
the general expression:
_ Yolte x coulombs A11 EQ
Wa3 m — or ra (i)
Making W ■■ 1 (i. e. one kilogram-meter) and
transforming, we have, as the coulombs correspond-
ing to 1 kilogram-meter:
«- X (2)
One coulomb of electricity liberates a weight (in
grams) of an element equal to the product of the
following: .000010384 X equivalent of element in
question X number of equivalents «+■ valency of the
element. Therefore, the coulombs corresponding
to one kilogram-meter, liberates this weight multi-
plied by ~r or, indicating weight by G,
Qsx .000010884 X equiv. x number equlv. y 0.81
valency E (3)
but .000010384 X 9.81 = .000101867.
Q _ equlT. X n X .000101867
•"• Ex valency (4)
Bale 73, page 98-99. The voltage of an armature of
DEMONSTRATION OF RULES. 135
a definite number of turns of wire and a fixed speed,
varies with the lines included within its longitu-
dinal area, as such lines are cut jin every revolution.
These lines vary with its area, and the latter varies
with the square of its linear dimensions.
To maintain a constant voltage if the size is
changed, the number of turns must be varied in-
versely as the square of the linear dimensions.
This ensures the cutting of the same number of
lines of force per revolution.
If, therefore, its size is reduced from x to ^ the
turns of wire must be changed from x to &. The
relative diameters of the two sizes of wire is found
by dividing a similar linear dimension by the rela-
tive size of the wire. But i -«- a? = ^. = diameter of
X X 9
the wire for maintenance of a constant voltage with
change of size.
The capacity of a wire varies with the square of
its diameter and (|y) f — |;.
Therefore the amperage, if a constant voltage is
maintained, will vary inversely as the sixth power of
the linear dimensions of an armature.
CHAPTER XIV.
NOTATION IN POWEK8 OF TEN.
This adjunct to calculations has become almost
indispensable in working with units of the C. G. S.
system. It consists in using some power of 10 as a
multiplier which may be called the factor. The
number multiplied may be called the characteristic.
The following are the general principles.
The power of 10 is shown by an exponent which
indicates the number of ciphers in the multiplier.
Thus 10 a indicates 100; 10 8 indicates 1000 and so
on.
The exponent, if positive, denotes an integral
number, as shown in the preceding paragraph. The
exponent, if negative, denotes the reciprocal of the
indicated power of 10. Thus 10"* indicates t4tf; 10 -8
indicates nftnr and so on.
The compound numbers based on these are re-
duced by multiplication or division to simple expres-
sions. Thus: 3.14 X 10 7 = 3.14 X 10,000,000 =
31,400,000. S-UXlO^^^j^orj^S^. Re-
gard must be paid to the decimal point as is done
here.
NOTATION IN P0WEB8 OF TEN 137
To add two or more expressions in this notation
if the exponents of the factors are alike in all re-
spects, add the characteristics and preserve the same
factor. Thus:
(51 X 10*) + (54 X 108 )= 105 X W.
(9.1 X 10-*) + (8.7 X 10" 9 ) = 17.8 X 10^.
To subtract one such expression from another,
subtract the characteristics and preserve the same
factor. Thus:
(54 X 10*) - (51 X 10*) = 3 X 10«.
If the factors. have different exponents of the
same sign the factor or factors of larger exponent
must be reduced to the smaller exponent, by factor-
ing. The characteristic of the expression thus
treated is multiplied by the odd factor. This gives
a new expression whose characteristic is added
to the other, and the factor of smaller exponent is
preserved for both.
Thus:
(5 X 10 7 ) + (5 X 10 9 ) = (5 X 10 7 ) + (5 X 100 X
10 7 ) = 505 X 10 7 .
The same applies to subtraction. Thus:
(5 X 10») - (5 X 10 7 ) = (5 X 100 X IOt) - (5 X
10*)= 495 X 10 7 .
If the factors differ in sign, it is generally best to
leave the addition or subtraction to be simply ex-
138 ARITHMETIC OF ELECTRICITY.
pressed. However by following the above rule it
can be done. Thus:
Add 5 X 10* and 5 X 10 8 .
5 X 10* = 5 X 10* X 10"* : (5 X 10» X 10-^ + (5 X
10-t) = 500005 X 10-*. This may be reduced to a
fraction ^ = 5000.05.
To multiply add the exponents of the factors, for
the new factor, and multiply the characteristics for
a new characteristic. The exponents must be added
algebraically: that is, if of different signs the numer-
ically smaller one is subtracted from the other one,
its sign is given the new exponent.
Thus:
(25 X 10*) X (9 x 10 8 ) = 225 X 10 1 *.
(29 X 10-*) X (11 X 10 7 ) = 319 X 10*.
(9 X 10 8 ) X (98 X 10-«) = 882 X 10 8 .
To divide, subtract (algebraically) the exponent
of the divisor from that of the dividend for the ex-
ponent of the new factor, and divide the character-
istics one by the other for the new characteristic.
Algebraic subtraction is effected by changing the
sign of the subtrahend, subtracting the numer-
ically smaller number from the larger, and giving
the result the sign of the larger number. (Thus to
subtract 7 from 5 proceed thus: 5 — 7 = —2.)
Thus:
(25 X 10 6 ) •*• (5 X 10 8 ) = 5 X 10-*
(28 X 10" 8 ) •*• (5 X 10 8 ) = 5.6 X 10-*.
if
i
TABLES
1
.8
S
I
f
I
i
t
¥
t
§
§
I
140
IL— EQUIVALENTS OF UNITS OF ABEA.
Square
Millimeter
Square
Gentlmet'r
Circular
MIL
Square
MIL
Square
Square
Foot
Square Millimeter
1
0.01
1978.6
1560.1
.00155
.0000108
Square Centimeter
100
1
197,861
155,007
.155007
.001076
QrcularMfl.
.00060T
.0000051
1
.78540
8X10-*
Square MIL
.000645
.0000065
1.2788
1
.000001
Square Inch
645.182
6.451
1,278,288
1,000,000
1
.006944
Square Foot
«*,898.9
928.989
144
1
HL-
EQUIVALENTS OF UNITS OF VOLUME.
Cublo
Inch
Fluid
Ounce
Gallon
Cubio
Foot
Cubic
Yard
Cu. Cen-
timeter
liter
Cubio
Meter
Cubic Inch
1
.554112
.004329
.000578
16.8862
.016886
Fluid Oz.
1.80469
1
.007812
.001044
29.5720
.029572
Gallon
281
128
1
.188681
.00495
8785.21
8.78521
.008785
Cubic Ft
1728
957.506
7.48052
1
.087037
28315.8
28.8158
.028815
Cubic Yd.
46,656
25,852.6
201.974
27
1
764,505
764.505
.764505
Cu. Cent!
.061027
.088816
.000264
.000085
1
.001
.000001
liter
61.027
88.8160
.264189
.085817
1000
1
.001
Cu. Meter
61027
88816
264.189
85.8169
1.8080
1000
1
141
IV.— EQUIVALENTS OF UNITS OP WEIGHT.
Grain.
Troy
Ounce.
Pound
Ays.
Ton.
Milli-
gram.
Gram.
Kilo,
gram.
Metric
Ton.
Grain
1
.020888
.000148
64.799
.064799
.000065
TroyOunce
480
1
.068641
81,108.5
81.1085
.081104
PoundAvs.
7,000
14.6888
1
.000447
458.598
.458598
.000454
Ton
82,666.6
2240
1
.001016
1.01605
Milligram
.015482
.000082
.000002
1
.001
.000001
Gram
15.4828
.082151
.002205
1000
1
.001
Kilogram
15,482.8
82.1507
2.20462
.000984
1,000,000
1000
1
.001
Hetrio Ton
82,150.7
2204.62
.98421
1,000,000
1000
1
142
V.— EQUIVALENTS OF UNITS
Erg.
Meg-
Grain-de-
gree C.
Kilogram-
degree O.
Pound-
degree O.
Pound-
degree F.
Erf.
1
.000001
Meg.-erg.
1,000,000
1
.024068
.000024
.000068
.000096
Gram-degree 0.
41.6487
1
.001
.002206
.008968
Ktiogram-degreeC.
41,648.7
1000
1
2.2046
8.9688
Pound-degree C.
18,846.6
468.69
.40869
1
1.8
Pound-degree P.
10,470.1
261.996
.261996
.666666
*\
Watt-Seeond.
10'
10
.24063
.000241
.000681
.000966
Gram-oentlmeter.
981
.000981
.0000286
Kilogram-meter.
98.1X10*
98.1
2.86108
.002861
.005206
.009870
Foot-Pound.
18.6626
.826426
.000626
.000720
.001296
Hone-Power-Sec.
English.
7469.48
179.486
.179486
.8967
.71248
Hone-Power-8ee.
Metric
7867.6
177.076
177.076
.890876
.70276
143
OF ENERGY AND WORK.
Watt-
Second.
Gram-
Centim'tr.
Kilogram-
meter.
Foot-
Ponnd.
Horse-
power-
second
English.
Horse-
power-
second
Metric.
10-'
.001019
Erg.
.1
1019.87
.010194
.078784
.000184
.000186
Meg-erg.
4.16487
42,858.6
.428686
8.06866
.00667
.006647
Gram-degree O.
4154.87
428.686
8068.66
5.67
5.64708
Kflogram-degTeeC.
1884.66
192.114
1889.6
2.52668
2.66149
Pound-degree O.
1047.08
106.780
772
1.40864
1.42806
pound-degree F.
1
10,198.7
.101987
.787887
.0018406
.0018602
Watt-Second.
.000098
1
.00001
.000072
Gram-Centimeter.
9.81
100,000
1
7.28828
.018152
.018884
Kilogram-meter.
1.85626
18,825.8
.188268
1
.0018182
.001848
Foot-Pound.
746.948
76.0692
660
1
1.01888
Horse-Power-Sec
JEngHafr,
786.76
75
642.496
.986856
1
Horse-Power-Sec.
Metric.
144
VL- TABLE OF SPECIFIC RESISTANCES IK MICBOHMS AND
OF COEFFICIENTS OF SPECIFIC RESISTANCES OF METAIA.
Spedflo
Besist-
anoe.Mi-
crohms.
Ooeffl-
dents of
Sp. Bes.
Spedflo
Resist-
ance, Mi-
crohms.
Coeffi-
cients of
Sp. Bee.
Annealed Silver. ..,
Hard Silver..
Annealed Copper. . ,
Hard Copper ,
Annealed Gold
Hard Gold
Annealed Aluminum
Compressed 23no...
Annealed Platinum
" Iron....,
1.621
1.4KB
1.616
1.662
2.081
2.118
2.945
6.689
9.168
9.826
.9412
1.0228
1.0000
1.0228
1.2877
1.8107
1.8224
8.6204
6.6671
6.0796
Annealed Nickel.
Oompres'dTtn...
* Lead.
" Antimony
" Bismuth
liquid Mercury
2 Silver, 1 Platinum.
German Silver .. .,
2 Gold, 1 Silver..
12.60
18.86
19.86
86.90
182.70
99.74
24.66
21.17
10.99
7.7970
8.2678
12.2884
22.2158
82.1170
61.7206
16.2699
18.1002
6.8008
SPECIFIC RESISTANCE OF SOLUTIONS AND LIQUIDS.
XAnHOBSSBf AKD OTHD8.
Names of Solutions.
Temper-
ature
Centi-
grade.
Temper-
ature
Fahren-
heit
Spedfle
Resistance.
Ohms.
Copper Sulphate, concentrated
9«
48.2°
29.82
" with an equal volume of water
««
«♦
46.64
•• with three volumes of water
<«
«i
77.68
Common Salt, concentrated
18°
66.40
6.98
'* with an equal volume of water . .
a
it
6.00
" with two volumes of water. . . .
t«
it
9.24
" with three volumes of water. .
tt
<•
11.89
Zinc Sulphate, concentrated
14"
67.2°
28.00
44 with an equal volume of water . .
t«
«<
22.76
•* with two volumes of water
<«
(«
29.75
Sulphuric Add, concentrated
14.8°
14.6*
67.8°
68.1°
6.82
50.5*, Spedfle Gravity 1.898. . .
1.086
44 29.6*, Spedfle Gravity 1.216. . .
12.8»
64.6°
.88
" 12* Spedflo Gravity 1.080...
12.8°
66.0°
1.868
Nitric Add, Spedflo Gravity 1.86 (Blavier) ....
14°
67.2°
1.46
U M «4 M
84°
76.2°
L2S
Distilled Water, (Temp'tura unknown) (Poufllet)
988.
14i>
m— RELATIVE RESISTANCE AND CONDUCTANCE OF PX7RE
COPPER AT DIFFERENT TEMPERATURES.
II
ji
Relative
Resistance.
Relative
Conductance
11
n
fi
Relative
Reslstanoe.
Relative
Conductance
0*
82*
1.
t
16°
60.8*
1.06168
.9419
1
88.8
1.00881
.99620
17
62.6
1.06568
.98841
2
86.6
1.00766
.9925
18
64.4
1.06969
.98494
8
87.4
1.01186
.98878
19
66.2
1.07866
.98148
4
89.2
1.01515
.98608
20
68.
1.07754
.92804
5
41
1.01896
.98189
21
69.8
1.08169
.92462
6
42.8
1.0228
.97771
22
71.6
1.08568
.92120
7
44.6
1.02668
.97406
28
78.4
1.08964
.91782
8
46.4
1.08048
.97042
24
75.2
1.09866
.91445
9
48.2
1.08485
.96679
25
77.
1.09769
.9111
10
60
1.08822
.96819
26
78.8
1.10162
.90776
11
61.8
1.04210
.95960
27
80.6
1.10667
.90448
12
68.6
1.04699
.95608
28
82.4
1.10972
.90118
18
65.4
1.0499
.96247
29
84.2
1.11882
.89784'
14
57.2
1.06881
.94898
88
86.
1.11785
.89457
15
69
1.06774
.94541
146
VIIL-AMKEIOAN WIRE GATJGB TABLE.
Properties of Copper Wire : Spedflo Grayity, 8.878 ; Specific CondacttTftj, 1.766 at T5°. F.
i
SlDL
WeiOUT A.SD Ltssni,
B*MM4irC*\.
M
1
&
B
B
o
Plnm-
«ter In
Mill,
Square of
Diameter
M 'v
Qrtifirt
Foot.
Po'nfo
pet
10O0
Fu#t
Fcot
per
Pound.
Ohm-'
per 1000
Feet
par
Ofcua
Ohm a p*r
Pound.
1*1-
u 4i
0000
460.000
211000.0
447T.2
689 60
1.561
.001
19920,7
/» T85
430
(KM}
4i^.f4<i
167W4 9
8550.5
607. -22
1.971
.068
15304,0
000125 ,
262
00
364.800
138079.0
2>l.'i -
402.25
£.486
.080
12084.2
,000198 1 208
«>
85*4.900
liiWlrt r,
2286 2
819,17
3.183
,101
9945,8
,000815
165
1
289.800
Ni^4.4P
1770.9
252 98
8.052
.127
7882,9
.000501
lft<
*J
2.17 aso
MAT* X
1404.4
200,68
4,904
.100
6251.4
.0O0T99
108
B
^>.4^»
02638.68
lUM
150.09
6.2S6
.202
4907.8
.001268
31
4
204.810
41742,67
683.3
126.17
7.025
.254
8931.6
.00*016
60
D
181.940
88102.16
700.4
100.06
8.005
.321
8117J
.008206
52
6
tag .aw
26250 48
000.4
70.84
12,604
,404
2472,4
,005095
41
T
m.'^t
20H16.72
44W.4
62.02
15.898
.AO0
1060.6
,008106
82
e
138,490
16300.64
848.8
49.90
20,040
.643
1660.0
.01269
26
•
114.430
13094.22
8T7.1
82.58
25,265
.811
1238,8
,02043
20
10
101.880
10881.07
218,7
81.88
81.867
1.023
977.9
moe
10
13
90.742
8234 11
174.3
24.80
44u;ij
1.2*9
775,5
.061B1
13
12
BO SOS
fa-lit w
i*M
19.74
00.650
l r 62P
015.02
.03287
10,9
Vi
n.96i
0178,89
108.6
1 :>«;:>
68.898
2 r 048
4*VJ5
.18087
8,1
14
64.034
4106.70
86, ST
12-41
80.080
3.585
8St,flQ
.20830
6.4
Lfi
07,088
8256 76
68.83
8.84
101,626
8,177
806.74
.88133
0.1
IS
00,820
2082.67
54. OT
7.81
128.041
4.683
WM
JMtt
4,0
IT
45.267
2048.19
■l;\ B8
6.18
161 + 501
0.183
192.91
,88744
M
]8
40,308
1624 88
84.87
4.01
908.868
li.Wrt
15299
1.8818
8,0
19
85.880
1202,45
26.00
3.780
204 136
3.477
117,96
2.2393
1.96
90
fti.wi
1021.51
21.00
8.086
824.045
10.394
90,21
8 8488
1.60
21
28.462
B10, 08
17,14
2.448
40r497
IK. 106
70,80
5.3580
1.2B
n
25.84T
642.47
18.08
1.942
014.033
u; ftifl
60,51
s.&ow
1.08
a\
bjri
0O9 40
10,77
1.080
640.778
20 842
41 ■>■
13 384
.80
JS
20 100
404 ol
8.05
1,221
819.001
26.284
88.05
21.024
.03
ar»
n.800
820 41
6,77
.067
fi^.iifi
88.185
80.18
84 298
.00
so
16,940
254 0*
5.38
.768
1302,083
41,739
an tr;(
54.410
.40
ST
14.105
201 49
4 26
.608
1544.73T
0-2, 6*7
1>M
86.057
,81
2B
12.641
159.79
8 39
,484
2066 116
60.440
15 05
187,288
.25
29
u aw
126.72
2.69
->-"r
260* J 67
83.752
11.04,
213. J04
.30
Hi'
10.020
100.50
2.11
.802
8811.208
100.641
9.466
840.805
.16
HI
8.92b
79. Tl
X.«7
/m
4W LiMi
133.191
7. DOS
507,286
.IB
82
T.90D
03,20
1.83
.100
5268.158
f.v- nn
.v;^>
8S4.267
.098
ftg
7 ii- 1
BO, IS
1.00
.151
6623.51 T
iu«an
4.721
1403,78
.078
14
6,304
88.14
.847
.121
h2r}4.4&-t
267,160
8.748
224)7 .03
,00$
85
0,614
81.62
.638
.094
10688.80
886,61
2,809
3088,12
,049
B6
5.000
25,00
.025
.075
18883.88
424.60
2.800
5661.71
.000
m
4,458
19. ea
.ISO
.060
16666.66
535,83
1,ftfl8
3922.20
.081
»s
8. B65
15,72
.315
,045
22222,22
675,23
1.481
16000.0
,046
a©
8,08!
13.47
.200
.m%
26*16,79
Bl,1M
1.174
22410.5
.0341
40
8.144
0.88
.210
.030
38333 83
1074.11
.981
80303,8
.015
147
EL— CHEMICAL AND THEBMO-CHEMICAL EQUIVALENTS.
FORMATION Or OlLDBS.
1
Name of Compound.
Formula.
Valency.
Chemical
Equiv-
alents.
Combin-
ing
Weights.
Thermo-
Chemical
Equiv-
alents.
Water
Iron Protoxide
HSO
FeO
Fe»03
ZnO
CuO
HgO
II
II
III
II
II
II
18
72
160
81
79.4
216
9
86
68.8
40.6
89.7
108
84.6
84.6
Iron Sesquioxide
81.9x8
PfncPxM*
48.2
Copper Oxide
Mercury Oxide
19.2
16.6
Formation or Salts.
Name of
Base.
Va-
lency.
Nitrates
Sul-
phates
Chlo-
rides
Cya-
nides.
Iron
n
Formula.
Chemical Equivalents
Combining Weights
Thermo-Chemical Equlv'lts
Fe sr
90
18.9
FeSO*
186
68
12.6
FeClt
127
68.6
00
66
8.2
Zinc
n
Formula
Chemical Equivalents
Combining Weights
Thermo-Chemical Equlv'lts
ZnfN03)»
94.6
9.8
ZnSO*
161
80.6
11.7
ZnCls
186
68
66.4
ZnCyt
117
68.6
7.8
Copper
n
Formula.
Chemical Equivalents ....
Combining weights
Thermo-Chemical Equlv'lts
Cu(NOS)8
187.4
98.7
7.6
CuSO*
169.4
79.7
9.2
CuClt
184.4
67.2
81.8
CuCy*
126.4
62.7
7.8
Mercury
II
Formula.
Chemical Equivalents
Combining weights
Thermo-Chemical Equlv'lts
163
7.6
HgSO*
280
140
9.2
186.6
9.46
V
126
16.6
148
X.— CHEMICAL AND KLECTEO-CHEMICAL EQUIVALENTS.
Name
Hydrogen
Gold
SflYar...
Copper (Caprie) ,
Mercury (Mercuric).
" (Mercuroui)
Iron (ferric) ,
** (ferrous)
Nickel ,
Zinc
Lead ,
Oxygen .... ,
Chlorine
Symbols
H
An
Ag
On..
Hg„
Hg.
Fe...
7e„
Ni
Zn
Pb
O
CI
Valen-
I
III
I
II
n
i
m
ii
u
h
ii
n
i
Chemical
Equivalent*
1
196.6
108
68
200
200
66
66
60
66
207
16
86.6
Combining
Weights
1
65.6
108
81.5
100
200
18.7
28
29.6
82.6
108.5
8
86.5
Electro-
Chemical
Equivalent*
.0106
.6877
1.184
.8807
1.06
2.10
.1964
.294
.8098
.8418
1.0868
.064
.8728
XL— MAGNETIZATION AND MAGNETIC TRACTION.
B
B.
Dynes
Grammes
Kilogrs.
Founds
Lines per
lines per
per
per
per
per
eq. cm.
sq. in.
sq. oentlm.
sq. oentlm.
sq. oentlm.
sq. Inch*
1,000
6,450
89,790
40.66
.0466
.677
2,000
12,900
159,200
162.8
.1628
2.808
8,000
19,860
868,100
866.1
.8661
5.190
4,000
25,800
686,600
648.9
.6489
9.228
5,000
82,250
994,700
1,014
1.014
14.89
6,000
88,700
1,482,000
1,460
1.460
20.75
7,000
46,150
1,950,000
1,987
1.987
28.26
8,000
51,600
2,547,000
2,696
2.696
86.95
9,000
58,060
8,228,000
8,286
8.286
46.72
10,000
64,500
8,979,000
4,056
4.066
57.68
11,000
70,950
4,816,000
4,907
4.907
69.77
12,000
77,400
5,780,000
5,841
5.841
88.07
18,000
88,850
6,725,000
6,855
6.856
97.47
14,000
90,800
7,800,000
7,550
7.650
118.1
15,000
96,750
8,958,000
9,124
9.124
129.7
16,000
108,200
10,170,000
10,890
10.89
147.7
17,000
109,660
11,500,000
11,720
11.72
166.6
18,000
116,100
12,890,000
18,140
18.14
186.8
19,000
122,550
14,680,000
14,680
14.68
208.1
20,000
129,000
15,920,000
16,280
16.28
280*
149
2 JL— PERMEABILITY OP WROUGHT AND CAST IRON.
8QUAR1 CKCTIMBTEB WtAflURMCENT.
Annealed Wrought Iron.
Gray Cast Iron.
B
M
H
B
M
H
5,000
9,000
10,000
11,000
12,000
18,000
14,000
15,000
16,000
8,000
2,250
2,000
1,692
1,412
1,088
828
526
820
161
90
54
80
1.66
4
5
6.5
8.5
12
17
28.5
60
105
200
850
666
4,000
5,000
6,000
7,000
8,000
9,000
10,000
11,000
800
500
279
188
100
71
68
87
5
10
21.5
42
80
127
18S
292
17,000
18,000
19,000
20,000
BQUAKB INCH MmABTTBXMXST.
Annealed Wrought Iron.
Gray Cast Iron.
B.
/<•
H.
B,
/<•
H,
80,000
40,000
50,000
£0,000
70,000
80,000
90,000
4,660
8,877
8,081
2,159
1,921
1,409
907
408
166
76
85
27
6.5
10.8
16.5
27.8
86.4
56.3
99.2
245
664
1,661
8,714
5,165
25,000
80,000
40,000
60,000
60,000
70,000
768
756
258
114
74
40
82.7
89.7
155
489
807
1,480
100,000
110,000
120,000
180,000
140,000
160
PEBMJUBUJTY 07 SOFT OHABOOAL WBOUGHT IROH
(tHKiVOSD BXDWBLL.)
SQUAB! <
B
M
H
7,890
1899.1
8.9
11,660
1121.4
10.8
16,460
888.4
40
17,880
160.7
116
18,470
88.8
208
19,880
46.8
427
19,820
88.9
666
0QVARS IXOH MMAltr
B.
M-
H.
47,414
1897
26.0
ft 104
1122
M.l
99,191
888
268
111,189
160
788
118,604
88.8
1886
124,021
46.8
2740
127,106
88.9
8768
(-Both in lines of foroe.
B — Magnetic Flux. )
H — Magnetlxing Fort*, f
fl — g the Permeability or multiplying power of tbo oora.
151
XIIL-MAGNETIO BELUCTANCE OF ATE BETWEEN TWO PARALLEL
OYLINDEBS OF IBON.
b
p
_
Ihoh TJhitb.
Batio of least
distance apart
to circumference.
0.1
.1964
5.1055
0.0771
12.968
0.2
.270T
8.6917
0.1066
9.877
0.8
.8251
8.0768
0.1280
7.815
0.4
.8688
2.7158
0.1450
6.897
0.5
.4046
2.4716
0.1598
6.278
0.6
.4861
2.2988
0.1717
5.825
0.8
.4887
2.0465
0.1924
5.198
1.0
.5816
1.8807
0.2098
4.777
1.8
.5684
1.7996
0.2288
4.571
1.4
.6007
1.6645
0.2865
4.228
1.6
.6289
1.5902
0.2476
4.089
1.8
.6541
1.5287
0.2575
8.888
2.0
.6774
1.4764
0.2667
8.750
4.0
.8857
1.1968
0.8290
8.040
6.0
.9819
1.0782
0.8669
2.726
8.0
1.0047
.9958
0.8955
2.628
10.0
1.0544
.9484
0.4151
2.409
In this table in columns 2 and 8 the Unit length of a cylinder is taken as 1 centi-
meter ; in columns 4 and 5 as 1 inch, p — circumference of cylinder b — shortest
distance apart.
XIV.-TABLE OF 6TH BOOTS.
Num-
ber
Sixth
Boot
Number
Sixth
Boot
Num-
ber
Sixth
Boot
Number
Sixth
Boot
.69855
1
.95820
1*
1.0177
If
1.0978
.70717
I
.96850
H
1
0192
1*
1.1019
.72806
1
.97006
n
1
0226
1*
1.1068
.74185
♦
.97468
u
1
0260
If
1.1087
.76478
f
.97798
n
1
0608
If
1.1107
.79870
I
.98055
H
1
0879
If
1.1119
.88268
ft
.98258
n
1
0491
1ft
1.1129
.89090
n
1
0199
2
1.1287
.98462
n
1
0888
152
XV.-8TANDARD AND BIRMINGHAM WIBB GAUGES.
Btavdabd.
BnUfXNGHAM.
Number of
Gauge.
Diameter
in Mils.
Square of
Diameter or
CtroTrMils.
Number of
Gauge.
Diameter
in Mils.
Square of
Diameter or
CtroTr Mils.
0000000
500
250000
0000
454
806116
000000
464
215296
000
425
180625
00000
489
186824
00
880
144400
0000
400
160000
840
115600
000
879
188884
1
800
90000
00
848
121104
2
294
80656
824
104976
8
269
67081
1
800
90000
4
288
56644
2
276
76176
5
220
48400
t
258
68504
6
808
41209
4
289
58824
7
180
82400
5
818
44944
8
165
27225
e
198
86864
9
148
21904
T
176
80976
10
184
17956
8
160
25600
11
120
14400
9
144
20786
19
109
11881
10
128
16884
18
095
9025
11
116
18456
14
088
6889
12
104
10816
15
072
5184
18
099
8464
16
065
4226
14
080
6400
17
068
8864
15
079
5184
18
049
2401
16
064
4096
19
049
1764
17
056
8186
20
085
1225
18
048
2804
21
089
1024
19
040
1600
22
028
784
SO
086
1296
28
025
625
81
069
1024
84
022
484
22
028
784
25
020
400
98
024
576
26
018
824
24
022
484
95
020
400
96
018
824
I ■
©
8 a
S°
^ g
o g
S3 8
S a
is
I
III
•"10
»5
me
E
w
153
e i b a
§ s a 8
||S§S3§sgss
g g 8 §
S S
sasssessss
§ § I
I 1
g © »-l « CO •*
& IIIS
2 I H I i i 5 I I
I I
I I § i I I §
IlllllHiS!
gigfifgggs'
oo k- t-
S co ss
'S
I
1
154
XVH.-WATTS AND HORSE POWER TABLES FOB VARIOUS
PRESSURES AND CURRENTS.
These tables will be found very convenient for quickly finding the
watts and electrical horse power on lighting and power circuits.
To find the watts or h. p. for any current up to 1,000 amperes at a
standard voltage add the watts or h. p. corresponding to the units,
tens and hundreds digits of the current.
Example: Find the electrical h. p. of 496 amperes at 101 volts.
Solution: The h. p. for 400 amp. is 66.3, for 80 amp. it is 4.82 and for
6 amp., .845. Adding these quantities gives 61.865 h. p., which we will
call #1.4 h. p., as the tabular values are computed to three figures
only, which are sufficient for engineering purposes.
To find values for voltages higher or lower than in the tables,
select a voltage 1-10 or ten times that required, and multiply the re-
sult by 10 or 1-10. Thus : to find h. p. at 7 amp., 56 volts, take 7 amp.
at 560 volte«5.16 h. p.; multiply by 1-10, which gives .516 h. p. To
find h. p. at 9 amp., 1,300 volts, take 9 amp. at 120 volts=1.45 ; multi-
ply by 10=14 6 h. p.
To read in kilowatts place a decimal point before the watts when
less than 1,000 in value, or substitute it for the comma in the larger
values,
HORSE POWER AT VARIOUS PRESSURES AND CURRENTS.
100 volts.
105 volts.
110 volts.
Amperes.
Watts.
100
":i6
Watts.
105
h.p.
.fil
Watts.
no
h.p.
.147
2
200
.268
210
.282
220 ,
.295
8
300
.402
315
.422 '
330 '
.442
4
400
.686
420
.563
440
.690
6
600
.670
525
.704
560
.737
6
600
.804
630
.845
660
.885
7
700
.938
735
.985
770
1.03
8
800
1.07
840
1.13
880
1.18
9
900
1.21
945
1.27
990
1.33
10
1,000
1.34
1,050
1.41
1,100
1.47
20
2,000
2.68
2,100
2.82
2,2u0
2.95
30
3,000
4.02
3,150
4.22
3.300
4.42
40
4,000
5.36
4,200
563
4,400
6.90
60
5000
6.70
5,250
7.04
5,600
7.37
60
6,000
8.04
6,300
8.45
6,600
8.85
70
7,000
9.38
7,350
9.85
7,700
10.3
80
8,000
10.7
8,400
11.3
8,800
11.8
90
9,000
12.1 i
9,450
12.7
9,900
13.3
100
10,000
13.4
10,500
14.1
11,000
14.7
. 200
20.000
28.8
21,000
31,500
28.2
22,000
29.5
300
30,000
40.2
42.2
33 000
44.2
400
40,000
63.6
42,000
56.3
44,000
59.0
600
50,000
67.0
52,500
70.4
55,000
73.7
600
60,000
70,000
80.4
63,000
84.5
66,000
88.5
700
93.8
73,500
98.5
77,000
108
800
80,000
90,000
107
84,000
113
88,000
118
900
121
94,500
127
99,000
133
1,000
100,000
134
105,000
141
110,000
147
155
HORSE POWER AT VARIOUS PRESSURES AND CURBENTS.
(Continued.)
115 volts.
120 volts.
125 volts.
Amp.
ms
h :&
Watts.
120
h -.?61
Watts.
135
h.p.
.168
2
HO
.808
240
.322
m
.335
3
345
.462
36^
.483
375
.505
4
AW
.617
480
.644
610
.670
5
575
.770
60
.805
G25
.840
6
M0
.925
720
.966
750
1.01
7
905
1.08
840
1.13
875
1.18
8
wo
1.23
Ofti
1.29
UNO
1.34
9
1,035
139
l,<£fl
145
U35
1.51
10
1,150
1.54
uuo
161
ym
168
20
a>aoo
3.08
2,*w
8.22
£.500
3.35
80
3,450
4.62
Moo
483
3,750
5.15
40
4,600
6.17
4,#>0
6.44
5,HD
6.70
60
&.T50
7.70
(1,1 mo
8.04
ti.250
8.40
60
6,900
9.25
7,300
966
7. , ii
101
70
B.O50
10.8
M<JU
11.3
8,750
11.8
80
9,200
12.3
0,000
12.9
lOiitf
13.4
90
IO&jQ
13.9
iu y m
14 5
]],£5G
15.1
100
11,500
15 4
]2a 0&
16.1
13,500
16.8
800
12U0O
30.8
S4,< 00
82.2
85,000
335
800
H4.5O0
462
ieutto
483
:;t,:^)
505
400
4*s,nno
617
ft&OQQ
64.4
fAi.Hi)
67.0
500
57,500
77.0
enw*
80.4
UL*,-VHJ
840
600
OftJ-00
92.5
72, mQ
96.6
75,000'
1)1
700
mi r*fl
108
&1.0G
113
K.b
118
800
88,000
123
8fi.ro *
129
KXMOO
134
900
i*i;i, on
139
lOh.H^I
145
Ufc.fiOn
351
1,000
115,000
154
120,000
161
l£5,u0
168
156
HORSE POWER AT VARIOUS PRESSURES AND CURRENTS.
(Continued.)
200 volts.
210 volts.
220 volts.
Amp.
Watts.
200
.$»
Watts.
210
.£2
Watts.
220
h .&
2
400
.586
420
.663
440
.690
8
800
.804
630
.845
660
.885
4
800
1.07
840
1.13
88')
1.18
6
1,000
1.34
7, aw
1.41
1,100
1.47
A
1,200
1.61
l*S80
1.69
1,320
1.77
7
1,400
1.88
3,470
1.97
1,640
2.06
8
1,600
2.14
i.i>M)
2.25
1.760
2.36
9
1,800
£41
KW0
263
1,960
2.65
10
2,000
£.iW
2,lo0
2.82
2,200
2.95
20
4,000
&.*i
4^1(0
5.68
4,400
5.90
80
6,000
BjCM
MOD
845
6,600
8.85
40
8,000
io.t
8.4<i0
11.3
8,800
11.8
60
lo,000
l'j.4
10,600
14. 1
11,000
14.7
60
12,000
16 + l
12.600
16.9
13,200
17 7
70
14 MO
IBJ
J 4, 7(0
19.7
15,400
20.6
80
hi.iJOO
2L4
1(3 ^W
22.5
17,600
23.6
90
18.U0O
£4 1
18,9(0
25.3
19,800
26.5
100
KkOOO
26.3
0,000
28.2
22,000
295
200
^OJKIO
63.6
42,000
66.3
44,000
59.0
800
"MHO
80.4
(S^OtiO
84.6
66,000
88.5
400
8O.C0O
107
84.0(10
113
88,000
118
600
lnyjco
134
] 05,0(0
141
110,(K0
147
800
lflMoo
161
1W *0
169
132,000
177
700
u\m
188
l47,rlJ0
197
154,000
206
800
mjm
214
168,000
225
176,000
236
900
I'vtOO
241
189,000
253
198,000
265
1,000
fen J, 000
268
2lU,WJ0
282
220,000
295
157
HORSE POWER AT VARIOUS PRESSURES AND CURRENTS.
(Continued.)
230 volts.
240 volts.
260 volts.
Amp.
Watts.
230
h.p.
.808
Watts.
MO
.fc
Watts.
300
h :&
2
|80
.617
480
.644
m
.670
8
090
.926
720
.966
750
1.01
4
920
1.28
MO
1.29
1,000
1.34
5
1,150
1,54
1.200
1.61
i, aw
1.68
6
1,330
LSfi
U40
1.98
LBOO
2.01
7
1,010
%M
L680
2.26
h7*0
2&*
8
1.840
8,47
1,930
2.68
3,1)00
2.68
9
2,'*7G
2.77
2, IftO
2.90
3,2*0
8.02
10
2,300
3.08
2J00
3.22
: T ',:Vi0
3a^
20
4,000
6 17
4.800
6.44
5.000
6.70
80
CkflOO
M6
7,2*0
9.66
7..M-)
10.1
.40
0,200
12.3
^000
12.9
10,000
13.4
60
11,600
15.4
la^oo
16.1
IB, W0
16 8
60
13 + Jto0
18 5
H iOl
19.3
14,000
20.1
70
lti,100
si .a
J'. SQ0
22.6
17,600
23.6
80
38,400
24,7
M.i.:,'(J0
268
solooo
26.8
90
20,7111
27.7
2^*10
29.0
23,500
80.2
100
2&(HJU
30. B
84,000
82.2
*sr».UXJ
33.8
XX)
uloqq
nt.7
4HOU0
64.4
60,000
67.0
300
O'.i ' ■ xi
92.5
72,1100
96.6
76.000
101
400
02,000
123
fl(t,( IX)
129
100,000
134
600
115,000
164
120,000
161
1-5.000
168
600
138,000
18A
144,000
193
160.000
201
700
161,000
216
IfW.lOO
226
175,0
236
800
184,000
247
l!--;.^K)
258
2T0,000
268
900
207,000
277
819,000
290
226.000
802
1,000
2W*X)
808
240,000
332
260,000
336
153
HORSE POWER AT VARIOUS PRESSURES AND CURRENTS.
(Continued.)
aoo volte.
550\
olta.
GOOv
Gltfi.
Watts
Wflita
Watt*
Amp.
and k. w.
BOO
h -£
and k. w.
5 II
fcutt.
,74
and Li . w.
eoo
b jSi
2
j,IH«0
l£i
1,110
1,47
1,300
l.ttl
8
I..VII
2 01
1,660
a.st
l.MO
S.41
4
3 HI in
2. m
2,3i]
2.95
2,400
&gg
fi
BJ500
3.35
a 710
::."'!■
:mr.
402
e
liJUNI
4 + <J2
3,40
4 43
3.6U0
4M
7
3,500
4. OS
3,860
6,1*
4 300
503
8
4JXN»
5 t ;w
4jm»
590
4600
643
ft
4,500
6.44
4,ftri0
£.04
5,400
7 24
li)
ftk.w.
0.7
fc.5 k,w.
7-4
6 k. w.
8,04
an
10
13 4
IUD
14.7
12
16,1
30
15
201
I&fl
231
10
n.i
40
30
sas
330
30.3
*4
SS.&
50
3ft
IW .ft
27,*
30 .ft
;<ii
40 t 2
60
30
40,2
33.0
44.3
30
4S + B
70
95
4flB
38.5
51.0
42
5<n
so
40
63 fi
44
fift.O
43
04.3
ftO
45
txu
43.5
flft.4
M
72.4
10Q
fiO
A7
55
74.0
00
80.4
200
100
134
no
147
120
101
300
Lflfl
201
Iflft
221
iao
HI
4UJ
sou
3(58
390
3»5
2411
333
600
200
aat
375
3«S
300
402
600
»«0
*03
330
441
300
483
700
■{5*1
4*9
MB5
MO
430
ftfl3
suo
403
53fl
440
590
480
*W3
qui
4^
fclitt
405
6H4
540
724
i,ouo
50LI
O
550
740
600
804
INDEX.
Alternating current sys-
tem 47-49
Alternating currents .... 115
Amperage of armature ... 99
Ampere 11-12
Ampere-second 57
Ampere-turns 87
Ampere-turns, rules for
calculating 87
Armature amperage of, on
short circuit 99
Armature calculations. .99-103
Armature, capacity of . . . . 99
Armature, general fea-
tures of 96
Armature, resistance of . . 98
Armatures, rules for cal-
culating 98-99
Armature, voltage of.... 99
Armature winding iron or
copper 96
Bbidge, Wheatstone, prin-
ciple of 25
Batteries, illustrations of
arrangements 66
Batteries or generators in
opposition 15
Batteries, storage, resist-
ance of. 65
Battery arrangement and
size for given efficiency. 73
Battery, arrangement of
cells in 67
Battery calculations, dis-
crepancies in 72
Battery, chemicals con-
sumed in 78
Battery constants 65
Battery, current of 68
BatterV, effective rate of
work of .,,,.., 7,7-78
Battery, efficiency, to cal-
culate 72
Battery, electromotive
force of 67
Battery, how rated 65
Battery, resistance of . . . . 67
Battery, to calculate cur-
rent and arrangement. 69-72
Battery, to calculate its
voltage 76-77
Battery t work of. 77-78
Calobib, gram and kilo-
gram, defined 54
Calorie, relation to watts
and ergs 54-55
Capacity of armature
winding 97
Cars, power to move .... Ill
Cell constants 65
Chemicals consumed in a
battery 78
Circuits, divided, branched
or shunt 19-25
Circuits, portions of 17-19
Circuits, single conductor,
closed 14-15
Circular mils calcula-
tions 44-45
Circular mils 43-45
Condensers 121
Conductance 36-37
Conductors of same mate-
rial 26
Conductors, resistance of
different 26
Contact, area of in elec-
tro-magnets and arma-
tures 86
Convection and radiation,
loss of heat by 58
Conversion, ratio of . , . . t 47
100
INDEX.
Converters, rale for wind-
ing 49
Cores of field-magnet, to
calculate 104
Counter-electromotive force
of plating-bath 80
Current, battery required
for a given 60-72
Current, distribution In
parallel circuits 20-21
Current, its heating effect
and energy 50
Currents passed by
branches of a circuit. • 10
Current, work of 00
Current yielded by a bat-
tery 65
Definitions, general 9
Demonstrations of rules. 127
Derived units 10
Dimensions, units of 10
Drop of potential In
leads 88-41
Drum type closed circuit
armatures 98-99
Duty and commercial ef-
ficiency 63-64
Duty of generators 63
Efpictivd B. M. F 16
Efficiency, its relation to
resistances 64
Efficiency of generators,
commercial 64
Efficiency of generators,
electrical 63
Efficiency, to calculate
battery for a given.. 73-74
Electrical railways 109
Electro-magnets and dy-
namos 82
Electro-magnets, general
rules for 85
Electro-magnets, to mag-
netize 86-87
Electro-magnets, traction
of 85-86
Electromotive force of
battery 67
Electro - plating calcula-
tions 79
Energy defined 9
Energy in circuit, rules
for determining 50-53
Energy of current 50-53
B. 11 F., its meaning... 10
E. M. F., primary and sec-
ondary 47
Erg 54
Fobcb defined 9
Force, magnetic 84
Fundamental units 10
Field-magnet cores, to cal-
culate 104
Field-magnet for dynamo
or motor 104-107
Field, unit Intensity of
magnetic 82-83
Genbbatom, efficiency of,
63-64
Generators or batteries In
opposition 15
Hbat, absolute quantity in
circuit 54
Heating effect of current,
50-53
Heat, specific 57
Horse-power, electrical . . 61
Horse - power, reduction
from kilogram-meters . . 60
Horse-power, to calculate
for lamps 62
Joulb, his law of heating,
effect of currents 50
Joule or gram-calorie.... 54
Kapp, Gisbert, his line of
force 107-108
Kilogram-meters 60
Lsads, tapering in size. . 41-42
Leakage between cylindri-
cal magnet leg, to cal-
culate 91
Leakage between flat mag-
net surfaces, to calcu-
late 91
Leakage of lines In a mag-
netic circuit 95
Leakage of lines of force. 89
Line of force, Kapp's. . 107-108
Lines of force 82
Lines of force cut per sec-
ond for one volt 96
Lines of force, leakage of. 89
Lines of force, to diminish
leakage of 94-95
INDEX.
161
Magnetic circuit 84
Magnetic circuit calcula-
tions 88-89
Magnetic circuit, the law
of 85
Magnetic circuits, four
parts of 88
Magnetic circuits, general
calculations for 92
Magnetic field 82-83
Magnetic flux 82-83
Magnetic force .-. 84
Magnetic potential, aver-
age difference of 90
Magnetism, no insulator of 84
Magnetizing force 87
Magnet legs, long and
short 94
Mass defined 9
Metal 8, deposition of, by
battery 79
Mho, unit of conductance 36
Mil, circular, calculations
based on 44-45
Mils, circular 43
Mils, circular, applied to
alternating current ... 48
Multiple arc connections,
to calculate 38-42
Nbutbal wire in three
wire system 46
Notation in powers of ten 136
Ohm 11-12
Ohm's law 13-25
Ohm's law, its universal
application 14
Ohnrs law, six expres-
sions of 13-14
Parallel connections of
equal resistance 23
Parallel leads, resistance
of 22
Permeance 83-84
Permeability 85
Permeability, average
range of 104
Potential, diagram for cal-
culating fall of 41
Potential difference .... 38-42
Potential difference, drop,
or fall of 17
Potential, drop of in leads,
S8-42
Power to move cars Ill
Powers of ten, notation in 136
Primary E. M. F 47
Proving armature calcula-
tions 103
Radiation and convection,
loss of heat by 58
Railways, electric 109
Rate of heat-energy, units
of 63
Ratio of conversion 47
Reluctance 83-84
Reluctance, calculation of. 87
Resistance 26-35
Resistance and efficiency,
how related 64
Resistance defined 9
Resistance of battery 67
Resistance of circuit, note
relative thereto 14
Resistance of parallel
leads 22-23
Resistance referred to
weight of conductor . . . 35-36
Resistance specific 29
Resistances, two in bat-
tery circuits 65
Resistance, universal rule
for 31-32
Rules, demonstrations of. 127
Safety-catches for fuses,
how calculated 59
Secondary E. M. F 47
Self- Induction 117
Series winding for dyna-
mos 106
Shunt circuits 19
Shunt circuits, resistance
of 22
Shunt winding for dyna-
mos 107
Sizes of feeders 109
Space defined 9
Specific heat 57
Specific resistance 30-31
System, alternating cur-
rent "47-49
Systems, special 46-49
System, three wire 46-47
Tables —
American wire gauge
table 146
INDEX.
Tables — Continued.
Chemical and electro-
chemical equivalents . . 148
Chemical and thermo-
chemlcal equivalents . . 147
Current capacity of bare
or insulated overhead
wires 153
Equivalents of units of
area 140
Equivalents of units of
energy and work . . . 142-143
Equivalents of units of
length 139
Equivalents of units of
volume 140
Equivalents of units of
weight 141
Magnetic reluctance of air
between two parallel
cylinders of iron 151
Magnetization and mag-
netic traction 148
Permeability of soft char-
coal wrought Iron .... 150
Permeability of wrought
and cast iron 149
Relative resistance and
conductance of pure
copper at different tem-
peratures 145
Specific resistance of solu-
tions and liquids 144
Standard and Birmingham
wire gauges 152
Table of specific resist-
ances in microhms and
of coefficients of specific
resistances of metals. . 144
Table of the sixth roots. . 151
Three wire system 48
Three wire system, saving
in size of wires 46
Three wire system, the
neutral wire 46
Time defined 9
Units, concrete statement
of 12
Units, fundamental 10
Units, original and de-
rived 10
Volt 10-12
Voltage of battery, to cal-
culate 76-77
Volt amperes.' 56-60
Watt 56-60
Watts and horse power
tables for various pres-
sures and currents 154
Weight defined 9
Weight of conductor, re-
sistance referred to . . . 35-36
Wheats tone bridge, prin-
ciple of 25
Winding, series and
shunt 106-107
Wire, rule for heating of,
by a current 59
Work defined 9
Work of current 60
Scientific and Practical Books
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ASKZK80N. Perfumes and Their Preparation,
Containing complete directions for making handker*
chief perfumes, smelling salts, sachets, fumigating
pastils; preparations for the care of the skin, the
mouth, the hair; cosmetics, hair dyes, and other toilet
articles. 300 pages. 82 illustrations. $3.00.
BABB. Catechism on the Cumbustion of Coal and
.the Prevention of Smoke.
A practical treatise for all interested in fuel econ-
omy and the suppression of smoke from stationary
steam-boiler furnaces and from locomotives. 85 illus-
trations. 349 pages. $1.50.
BARROWS. Practical Pattern Making.
A fully illustrated book which teaches you just what
you should know about pattern making. It contains a
detailed description of the materials used by pattern
makers, also the tools, both those for hand use, and
the more interesting machine tools; having complete
chapters on the Band Saw, the Buzz Saw and the
Lathe. Individual patterns of many different kinds
are fully illustrated and described, and the mounting
of metal patterns on plates for molding machines is
included. Price, $2.00.
Battleship Chart.
An engraving which shows the details of a battle-
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ana you could see all the interior. The finest piece
of work that has ever been done. So accurate that it
is used at Annapolis for instruction purposes. Shows
all details and gives correct name of every part. .28 x
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BAUBB. Marine Engine* and Boiltrt: Their De-
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A large practical work of 722 pages, 650 illustra-
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engineers and naval constructors. $9.00 net.
BAXTSB, JB. Switchboards.
The only book dealing with this important part of
electrical engineering. Takes up all sixes ana kinds
from the single dynamo in the engine room to the
largest power plant work. Includes divert and alter-
nating currents; oil switches for high tension; arc and
incandescent lighting; railway work, and all the rest,
except telephone work. $1.50.
BAXTBB, JB, Commutator Construction.
The business end of a dynamo or motor is the com-
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and what to do to put 'em right again. Price, 26
cents.
BBHJAimr. Modern Mechanism.
A practical treatise on machines, motors, and the
transmission of power, being a complete work and a
supplementary volume to Appleton s Cyclopedia ox
Applied Mechanics. Bound in half Morocco. Price,
$5.00.
BLACSAXXt Air-Brake Catechism.
This book is a complete study of the air-brake
equipment, including the £. T. Equipment, as well as
all the latest devices. All parts of the air brake, their
troubles and peculiarities, and a practical way to find
and remedy them, are explained. This book contains
over 1,500 questions with their answers. 812 pages.
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BLACSAXX. Hew York Air-Brake Catechism.
This is a complete treatise on the New York Air-
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and the methods of locating and remedying the same.
200 pages, fully illustrated. $1.00.
BOOTH AND KEBSHAW. Smoke Prevention and
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As the title indicates, this book of 107 pages and
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$2.00.
BUCHETTL Engine Tests and Boiler Efficiencies.
This work fully describes and illustrates the meth-
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evaporative power of fuels. Combustion of fuel and
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BYRON. Physios and Chemistry of Mining.
For the use of all preparing for examinations in
Mining or qualifying for Colliery Managers' Certifi-
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Car Charts.
Shows and names all the parts of three types of
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COCKIN. Practical Coal Mining.
An important work, containing 428 pages and 213
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who are interested in or connected with the industry.
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COLYDf. American Compound Locomotives.
The latest and most complete book on compounds.
Shows all types, including the balanced compound
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COLVItf AND CHENEY. Engineer's Arithmetic
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OOLVXY. Link Xotioae and Yalta Setting .
A handy little book for the engineer or machinist
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why. Piston and slide valves of different types are
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COLVIK AHD CHEVEY, Xachine 8hop Arithmetic
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COLVIK. The Railroad Pockotbook.
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00LYDT-8TABEL. Thread! and Thread Cutting.
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a lot of useful hints and several tables. Price, 25
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COLVIK. Turning and Boring Tapers,
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CBAVE. American Stationary Engineering.
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in print on this important subject. Bound in fine sea' 1
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ENGSTROM. Bevel Gear Tables.
No one who has to do with bevel gears in any wfcy
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FOWLER. Boiler Boom Chart.
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FOWLER. Locomotive Breakdowns and Their
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GODDARD. Eminent Engineers.
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GRIMSHAW. Saw Filing and Management of Saws.
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GRIMSHAW. "Shop Kinks."
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GBIHSHAW. Steam Engine Catechism.
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GBXX8HAW. Locomotive Catechism.
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HABBISOtf. Electrio Wiring, Diagrams and
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HISCOX. Gas, Gasoline, and Oil Engines.
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HISCOX. Compressed Air in All Its Applications.
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HISCOX. Mechanical Movements, Powers, and
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HISCOX. Mechanical Appliances, Mechanical Move-
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896 pages, 1,000 specially made illustrations. Price,
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HISCOX. Modern Steam Engineering in Theory and
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HOBABT. Brazing and Soldering.
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Horse Power Chart.
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Inventor's Manual; How to Make a Patent Pay.
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HLEDTHAHS. Boiler Construction.
The only book showing how locomotive boilers are
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KBATT88. Linear Perspective Self -Taught.
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LE VAN. Safety Valves; Their History, Invention,
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MEINHA&DT. Practical Lettering and Spacing.
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PAB8ELL & WEED, Gas Engine Construction.
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FEBBIGO. Change Gear Devices.
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PERRIGO. Modern American Lathe Practice.
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PBATT. Wiring a Home.
Shows every step in the wiring of a modern house
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BEAGAH, JR. Electrical Engineers' and Students'
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BICHABD8 AHD COLVI2J. Practical Perspective.
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Makes everything plain so that any mechanic can un-
derstand a sketch or drawing in this way. Saves tune
in the drawing room and mistakes in the shops. Con-
tains practical examples of various classes of work.
60 cents.
BOTJUXION. Drafting of Cams.
The laying out of cams is a serious problem unless
you know how to go at it right This puts you on
the right road for practically any kind of cam you
are likely to run up against. Price, 25 cents.
BOTJUXION. Economics of Manual Training.
The only book that gives just the information need-
ed by all interested in manual training, regarding
buildings, equipment and supplies. Shows exactly
what is needed for all grades of the work from the
Kindergarten to the High and Normal School. Gives
itemized lists of everything needed and tells just
what it ought to cost. Also shows where to buy sup-
plies. $2.00.
SAUNIEB. Watchmaker's Hand-Book.
Just issued, 7th edition. Contains 498 pages and is
a workshop companion for those engaged in watch-
making ana allied mechanical arts. 250 engravings and
14 plates. $3.00.
8L0ANZ. Electricity Simplified.
The object of "Electricity Simplified" is to make
the subject as plain as possible and to show what the
modern conception of cectricity is. 168 pages. $1.00.
GOOD, PRACTICAL BOOKS
SLOANS* How U Become a Snooeiafnl Sleotrlelan.
It is the ambition of thousands of young and old
to become electrical engineers. Not every one is pre-
pared to spend sereral thousand dollars upon a col-
lege course, even if the three or four years requisite
are at their disposal. It is possible to become an elec-
trical engineer without this sacrifice, and this work is
designed to tell "How to Become a Successful Elec-
trician" without the outlay usually spent in acquiring
the profession. 189 pages. $1.00.
SLOANS. Arithmetic of Electricity.
A practical treatise on electrical calculations of all
kinds reduced to a series of rules, all of the simplest
forms, and involving only ordinary arithmetic; each
rule illustrated by one or more practical problems,
with detailed solution of each one. 138 pages. $1.00.
SLOANS. Electrician's Handy Book.
An up-to-date work covering the subject of practical
electricity in all its branches, being intended for the
every-day working electrician. The latest and best
authority on all branches of applied electricity. Pocket-
book size. Handsomely bound in leather, with title
and edges in gold. 800 pages. 600 illustrations. Price,
$3.60.
SLOANS. Electric Toy Making, Dynamo Building,
and Electric Motor Construction,
This work treats of the making at home of electrical
toys, electrical apparatus, motors, dynamos, and in-
struments in general, and is designed to bring within
the reach of young and old the manufacture of genu-
ine and useful electrical appliances. 140 pages. $1.00.
SLOANS. Bubber Hand Stampa and the Manipula-
tion of India Bubber.
A practical treatise on the manufacture of all kinds
of rubber articles. 146 pages. $1.00.
SLOANS. Liquid Air and the Liquefaction of
Oases.
Containing the full theory of the subject and giving
the entire history of liquefaction of gases from the
earliest times to the present. 865 pages, with many
illustrations. $2.00.
SLOAXE. Standard Electrical Dictionary.
A practical handbook of reference, containing defi-
nitions of about 6,000 distinct words, terms, and
phrases. 682 pages. 398 illustrations. $8.00.
GOOD, PRACTICAL BOOKS
•TARBUCK. Modern Plumbing Illustrated.
A comprehensive and up-to-date work illustrating
and describing the Drainage and Ventilation of dwell-
ings, apartments, and public buildings, etc Adopted
by the United States Government in its sanitary work
in Cuba. Porto Rico, and the Philippines, and by the
principal boards of health of the United States and
Canada. The standard book for master plumbers,
architects, builders, plumbing inspectors, boards of
health, boards of plumbing examiners, and for the
property owner, as well as for the workman and his
apprentice. 800 pages. 66 full-page illustrations.
$4.00.
Tonnage Chart.
Built on the same lines as the Tractive Power
Chart; it shows the tonnage any tractive power will
haul under varying conditions of road. No calcula-
tions are required. Knowing the drawbar pull and
grades and curves you find tonnage that can be
hauled. 60 cents.
Tractive Power Chart
A chart whereby you can find the tractive power
or drawbar pull of any locomotive, without making a
figure. Shows what cylinders are equal, how driving
wheels and steam pressure affect the power. What
sized engine you need to exert a given drawbar pull
or anything you desire in this line. Printed on tough
jute paper to stand rolling or folding. 60 cents.
T/sheb. The Modern Machinist.
A practical treatise embracing the most approved
methods of modern machine-shop practice, and the ap-
plications of recent improved appliances, tools, and
devices for facilitating, duplicating, and expediting
the construction of machines and th*ir parts. 857
engravings. 822 pages. $2.60
VAH DEBVOORT. Modern Machine Shop Tools;
Their Construction, Operation, and Manipula-
tion.
An entirely new and fully illustrated work of 656
pages and 678 illustrations, describing in every de-
tail the construction, operation, and manipulation of
both Hand and Machine Tools. Includes chapters on
filing, fitting, and scraping surfaces; on drills, ream-
ers, taps, and dies; the lathe and its tools; planers,
shapers, and their tools; milling machines and cutters:
Sear cutters and gear cutting; drilling machines and
nil work; grinding machines and their work; hard-
ening and tempering; gearing, belting, and transmis-
sion machinery; useful data and tables. $4,00.
GOOD, PRACTICAL BOOKS
WALLIS-TAYLOR. Pocket Book of Refrigeration
and Ice-Making.
This is one of the latest and most comprehensive
reference books published on the subject of refriger-
ation and cold storage. $1.50.
"WOOD. Walschaert Locomotive Valve Gear.
The only work issued treating of this subject of
valve motion. 150 pages. $1.50.
WOODWOETH. American Tool Making and Inter-
changeable Manufacturing.
A practical treatise of 560 pages, containing 600
illustrations on the designing, constructing, use, and
installation of tools, jigs, fixtures, devices, special
appliances, sheet-metal working processes, automatic
mechanisms, and labor-saving contrivances; together
with their use in the lathe, milling machine, turret
lathe, screw machine, boring mill, power press, drill,
subpress, drop hammer, etc., for the working of met-
als, the production of interchangeable machine parts,
and the manufacture of repetition articles of metal.
$4.00.
WOODWOETH. Dies, Their Construction and Use
for the Modern Working of Sheet Metals.
A new book by a practical man, for those who wish
to know the latest practice in the working of sheet
metals. It shows how dies are designed, made and
used, and those who are engaged in this line of wor*
can secure many valuable suggestions. Thoroughly
modern. 384 pages. 505 illustrations. $3.00.
WOODWOETH. Hardening, Tempering, Annealing,
and Forging of Steel.
A new book containing special directions for the
successful hardening and tempering of all steel tools.
Milling cutters, taps, thread dies, reamers, both solid
and shell, hollow mills, punches and dies, and all kinds
of sheet-metal working tools, shear blades, saws, fine
cutlery and metal-cutting tools of all descriptions, as
well as for all implements of steel, both large and
small, the simplest and most satisfactory hardening
and tempering processes are presented. The uses to
which the leading brands of steel may be adapted are
concisely presented, and their treatment for working
under different conditions explained, as are also the
special methods for the hardening and tempering of
special brands, 320 pages. 250 illustrations, $2.50,
JUST PUBLISHED
HYDRAULIC ENGINEERING
By Gardner D. Hiscox, a practical work on
hydraulics and hydrostatics in principle and prac-
tice, including chapters on :
The measurement of water flow for power and
other uses; siphons, their use and capacity; hydrau-
lic rams; dams and barrages ; reservoirs and their
construction ; city, town and domestic water sup-
ply; wells and subterranean water flow; artesian
wells and the principles of their flow ; geological
conditions ; irrigation water supply, resources and
distribution in arid districts; the great projects for
irrigation.
Water power in theory and practice, water wheels
and turbines; pumps and pumping devices, cen-
trifugal, rotary and reciprocating; air-lift and air-
pressure devices for water supply— hydraulic power
and high-pressure transmission; hydraulic mining;
marine hydraulics, buoyancy displacement, tonnage,
resistance of vessels and skin friction. Relative
velocity of waves and boats ; tidal and wave pow-
er, with over 300 illustrations and 36 tables of
hydraulic effect. A most valuable work for study
and reference. About 400 octavo pages. Price,
$4.00.
THE TELEPHONE HAND-BOOK
By H. C. Cushing, Jr., E.E., and W. H. Radcliffe,
E.E. A practical reference book and guide for tele-
phone wiremen and contractors. Every phase of
telephone wiring and installation commonly used
to-day is treated in a practical, graphic and concise
manner. Fully illustrated by half-tones and line-
cut diagrams, showing the latest methods of install-
ing and maintaining telephone systems from the
simple two-instrument line, intercommunication
systems for factories, party lines, etc 175 pages,
100 illustrations, cloth, pocket size. $1.00.
THIS BOOS IS DUE ON THE LAST DATE
STAMPED BELOW
AN INITIAL FINE OF 25 CENTS
WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO SO CENTS ON THE FOURTH
DAY AND TO $1.00 ON THE SEVENTH DAY
OVERDUE.
DEC 16 1932
SEP 15 1937
AUG * **
o*r> "
* 'AP™'
LD 21-50m-8,-32
7B 09539
3S3491
UNIVERSITY OF CALIFORNIA LIBRARY