ELECTRICAL ENGINEERING
ADVANCED COURSE
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ELECTRICAL
ENGINEERING
ADVANCED COURSE
BY
ERNST JULIUS BERG Sc. D.
i
PROFESSOR OP ELECTRICAL ENGINEERING UNION
COLLEGE, SCHENECTADY, N. Y.
AUTHOR OF "ELECTRICAL ENGINEERING," FIRST COURSE
FIRST EDITION
McGRAW-HILL BOOK COMPANY, INC.
239 WEST 39TH STREET. NEW YORK
LONDON: HILL PUBLISHING CO., LTD.
6 & 8 BOUVEBIE ST., E. C.
1916
13
COPYRIGHT, 1916, BY THE
McGKAW-HiLL BOOK COMPANY, INC.
TUB MAIT-IC FKBSS T O H K
PREFACE
This volume contains abstracts of a series of lectures given to
graduate students in electrical engineering at Union College. It
is primarily intended to prepare the student to understand and
to deal mathematically with phenomena which are incidental to
abnormal or transient conditions in electric circuits.
The first part is practically a reprint of a series of articles
published by the author some years ago in the General Electric
Review. These cover the simple transients in circuits containing
concentrated inductance, capacity, and resistance, which have
been treated by many authors, notably by BEDELL AND CREHORE
in their " Alternating Currents," published 1893.
The second part deals with the somewhat more difficult prob-
lems of transients in circuits of distributed inductance, capacity
and resistance. These were treated mathematically very fully
almost thirty years ago by Heaviside in a series of papers on
" Electromagnetic Theory/ 7 later published in book form. In
1909 Steinmetz's " Transient Phenomena" appeared. This
book covered in a broad sense very much the same ground as that
of the authors given above, but covered it in an essentially differ-
ent way; introducing for the first time asfa-r as the author knows
a really advanced book on practical electrical engineering
problems.
The third part of the book deals with problems in electro-
statics. These again have been very fully treated almost fifty
years ago by Maxwell in his famous books on "Electricity and
Magnetism. " Since that time a large number of papers and books
have appeared on the subject, notably by Heaviside, Kelvin,
Gray, Jeans and Webster, and quite recently by Coffin in his
interesting little book on "Vector Analysis."
While the literature on this phase of engineering is thus very
extensive, it has, for all purposes, been closed to the practical
engineer because of his lack of sufficient mathematical knowledge.
Dr. W. S. Franklin has, however, recently published a number of
papers, which in a beautifully simple way have demonstrated
that these advanced problems can be solved with simple
mathematics.
vi PREFACE
The last part of the book gives an outline of the theory oi
electric radiation. The mathematical theory was again given
almost fifty years ago by Maxwell. Hertz's verification of
Maxwell's theoretical work given twenty years later and pub-
lished in his "Electric Waves" is today almost the last word in
the theory of wireless transmission of energy. Yet it would be
out of place to omit a reference to the recent excellent papers and
books by Marconi, Lodge, Flemming, Pierce, Zenneck, Cohen,
Austin and a score of others.
It is evident then that the field covered in this volume is not
new. Nevertheless, the book seems justified because it endeavors
to give the theory in a way comprehensible to students who have
had only the ordinary undergraduate course in electrical engi-
neering. It is hoped that the volume will also serve a useful
purpose in bringing to the attention of students a field of mathe-
matics of extreme practical importance that is hardly known
to them.
The author is greatly indebted to one of his graduate stu-
dents, MR. M. K. TSEN, who not only examined the manuscript
in detail, but checked and elaborated upon the theoretical work.
He is also indebted to DR. A. S. MCALLISTER, who kindly criticized
the manuscript prior to its publication and offered valuable
suggestions.
CONTENTS
CHAPTER PAGE
INTRODUCTION 1
PART I
TRANSIENT PHENOMENA
I. CIRCUITS CONTAINING CONCENTRATED INDUCTANCE AND
RESISTANCE 3
II. PROBLEMS INVOLVING MUTUAL INDUCTANCE 33
III. CIRCUITS OF RESISTANCE AND VARIABLE INDUCTANCE ... 56
IV. CHARACTERISTICS OF CONDENSERS 68
V. A CIRCUIT CONTAINING DISTRIBUTED RESISTANCE AND IN-
DUCTANCE 106
VI. CIRCUIT CONTAINING DISTRIBUTED LEAKAGE CONDUCTANCE
AND CAPACITY 110
VII. CIRCUIT CONTAINING DISTRIBUTED RESISTANCE AND
CAPACITY 113
VIII. DISTRIBUTED INDUCTANCE AND CAPACITY 127
IX. DISTRIBUTED RESISTANCE INDUCTANCE LEAKAGE CON-
DUCTANCE AND CAPACITY 143
X. PERMANENT CONDITIONS WHEN ONE OF THE FOUR CONSTANTS,
R, L, G, AND C is NEGLIGIBLE 148
XL DISTRIBUTION OF FLUX OR CURRENT IN A CYLINDRICAL OR
FLAT CONDUCTOR 150
PART II
PROBLEMS IN ELECTRO-STATICS
XII. FUNDAMENTAL LAWS 157
XIII. METHODS OF IMAGES, APPLIED TO THE PROBLEM OF POINT
CHARGES + 10 AND 5, SEPARATED 5 CM 168
XIV. APPLICATION OF THE POTENTIAL FORMULA V = 2.*
TO SOME MAGNETIC PROBLEMS 180
XV. DIVERGENCE OF A VECTOR, POISSON AND LAPLACE EQUATIONS. 186
XVI. LEGENDRE'S FUNCTION 189
XVII. DISTRIBUTION OF CHARGE ON AN ELLIPSOID 199
XVIII. CONCENTRIC SPHERES 209
XIX. CYLINDRICAL CONDUCTORS 218
XX. MUTUAL AND SELF-INDUCTION OF ELECTRO-STATIC CHARGES
OR FLUXES MAXWELL'S COEFFICIENTS 232
XXI. TWO-CONDUCTOR CABLE 237
vii
Vlll
CONTENTS
XXII. THE ELECTRO-STATIC EFFECT OF A THREE-PHASE LINE ON AN
ADJACENT WIRE OR WIRES 249
XXIII. THE CURL OF A VECTOR 257
XXIV. THE EQUATION OF THE ELECTROMOTIVE FORCE 260
XXV. SOLUTION OF ALTERNATING CURRENT IN CYLINDRICAL CON-
DUCTOR SKIN EFFECT '. 271
XXVI. ELECTROMAGNETIC RADIATION. . 278
APPENDIX I: PARTIAL DIFFERENTIATION . .
APPENDIX II: ELEMENTS OF VECTOR ANALYSIS.
319
327
INDEX
331
ELECTRCAL ENIN-EERING
ADVANCED COURSE
PART I. TRANSIENT PHENOMENA
CHAPTER I
CIRCUITS CONTAINING CONCENTRATED INDUCT-
ANCE AND RESISTANCE
The study of transients in circuits of concentrated inductance
and resistance involves as a rule a knowledge of the solution of
linear differential equations of the first order.
One example of such a differential equation is:
2+/i(*)=/i(*) (1)
where fi(x) and fz(x) may be functions of x or constants, but
must not be functions of y.
For the sake of convenience fi(x) will be denoted by P and
fz(x) by Q. P and Q in the most general case are then functions
of x but not of y. Thus, equation (1) becomes
% + Pdv = Q (2)
A solution of this equation can be obtained, in several ways,
all of which, however, involve "educated guesses."
Let, for instance,
y = uv (3)
where u and v are unknown functions of x, which will be deter-
mined in the most advantageous way.
Since dy dv du
y = uv, -7- = u -; \~ v-j- (4)
1 dx dx dx
Substituting (3) and (4) in equation (2),
dv du
or
Since u is entirely arbitrary, this expression can be greatly
4 ELECTRICAL ENGINEERING
simplified, by selecting such a value of u as to make the coefficient
of v or the parenthesis zero. Therefore let:
,
dx u
.'. logu = - fPdx + C.
Since the simplest possible function is sought, let that particu-
lar one be chosen, which makes C = zero. Thus:
log u - - fPdx,
and u = e-SPdx ( 6 )
Substituting now this value in (5), there is obtained,
.'. v = fef pd * Qdx + C.
and since y = uv,
y = t-fFd* [ffP** Qdx + C] (7)
Special cases:
First. Let P be constant, a; and Q be a function of x
and y = e~ ax [fe ax Qdx + C] (8)
Second. Let P be a function of x, but Q be a constant, b.
and y = e -^ Fdx [bfef pd * dx + C] (9)
Third. Let both P and Q be constants, a and 6 respectively,
dy
'
or > _ _ i n -ax
Fourth. Let P be a function of x and Q be zero.
and, y = <
INDUCTANCE AND RESISTANCE 5
If P is a constant a, then y = Ce~ ax .
Fifth. Let P be zero and Q be a constant, 6,
. dy^ b
and, y = bx + C. (12)
Two useful integrals that can, of course, easily be solved but
will frequently appear are given below for the sake of convenience.
e ai cos ut dt = - -5 [o> sin ut -\- a COS co/1.
a" + or
rf sin at dt = - ~^\a sin co^ co cos co/1.
a 2 + a; 2
/
A study will now be made of the equation of the current flowing
in such circuit when the impressed e.m.f. is steady and also when
it varies with time. Referring to Fig. 1, it is evident that the
following e.m.fs. exist:
FIG. 1.
First, the impressed e.m.f., E\
Second, the e.m.f. consumed by the resistance = ir;
Third the e.m.f. consumed by the self-inductance = j^ --yr or
di
L di>
where E is the impressed e.m.f. in volts,
r the resistance in ohms,
N the number of turns of the coil,
L the inductance in henrys (assumed constant),
-77 the rate of change of flux at a particular instant, t, and
i the current in amperes at any particular instant.
6 ELECTRICAL ENGINEERING
The e.m.f. consumed by self-inductance can be expressed as
J* or L -jT- because the inductance by definition is :
_AT0
= 10 8 i
thus Nd<t> _ di^
10*dt " dt'
The equation connecting these e.m.fs. is obviously:
' *-* + L (14)
That is, at any instant the impressed e.m.f. E is numerically
equal to the e.m.f. consumed by the resistance and the e.m.f.
consumed by the inductance. Note that e.m.fs. consumed by
but not e.m.fs. of resistance and self-induction are considered.
The latter are:
T di
ir and L rr
dt
Equation (14) can be written:
l+z'-f f
Compare this equation with (2) and note that P = T an d Q =
LJ
E
-j- are constant when the impressed e.m.f. is constant and not
function of t. Thus the solution is found in equation (10) and
* /~v 1 ' 1 "^ / -f t~*\
i = Ce L + - (16)
The integration constant C is determined from the fact that
time is required to impart energy, that is, in this case to produce
or alter a magnetic field.
Before the switch is closed, there is obviously no field sur-
rounding the turns. Shortly after, however, there is a current
and thus a field which appears simultaneously with the current.
Thus since a magnetic field can not be produced instantaneously,
no current can pass at the very first instant. Thus for t = 0,
i = 0. Therefore
= C.--L- +?,
INDUCTANCE AND RESISTANCE
but c = 1,
therefore = C + ^, and C = - ^;
and . Ef
I = 11
r \
This equation shows, that as t increases, the current increases,
and finally reaches a value,
Assume now that after the current has reached this value, the
circuit is disconnected from the generator, and at the same instant
short-circuited. What can be expected to happen?
The Dying Away of a Current in an Inductive Circuit. Re-
ferring to Fig. 2, since the coil is surrounded by a magnetic field,
and the field can not be destroyed
instantaneously, and since the mag-
netic field can not exist without a
current, it is' evident that the cur-
rent can not disappear instantan-
eously, but must die away gradu- FIG. 2.
ally.
Referring to equation (15) which is the general equation of the
current and remembering that the impressed e.m.f. E is zero,
we have:
the solution of which has been shown to be:
i = Ce'i*'
To determine the integration constant, it is remembered that
at the very first instant when t = O, there was a definite current /
in the circuit.
Thus, i = I when t = 0,
which substitued above gives:
C = I,
and the equation of the decaying current becomes:
= - (is)
8 ELECTRICAL ENGINEERING
If dW is the energy delivered during a short interval dt, then
the rate of energy supply, or power is:
_.
dt
The practical unit of power is the watt, which is work done at
the rate of 1 joule per second. At any instant the power is the
product of the instantaneous values of e.m.f. and current.
Thus the power equation corresponding to equation (14) is:
Ei = i X ir + iXL~
= i*r + Li j t (19)
It is seen from this equation that when the instantaneous
value of the current is i, energy is being dissipated at the rate of
i 2 r joules per sec., or watts, in heat, and is being stored in the
magnetic field at the rate of Li y- watts. The energy that has
been supplied to the circuit t sec. after the switch is closed and
the current started is:
Eidt joules (20)
f
Jo
The energy dissipated in heat
= I Prdt (21)
and the energy stored in the magnetic field
, V
(22)
where 7 is the particular value of i when the time is t.
In almost all calculations of transient phenomena, the ex-
pression e~ ax is met with, e is the base of the natural logarithm.
It has the numerical value of approximately 2.718. To calculate
the numerical value of any particular expression, the ordinary
logarithms are used. Thus, for instance, to find the value of
y = c~- 2 , the method is as follows:
log y = - 0.2 log e = - 0.2 X 0.434 = - 0.0868
+ 0.9132 - 1,
therefore y = 0.819,
therefore
INDUCTANCE AND RESISTANCE
In Fig. 3 are shown the values of this function for a large number
of values of the exponents. Since this curve is plotted on
rectangular coordinate paper, it is rather unsatisfactory for
small values of the exponent, and the table below has therefore
been worked out.
FIG. 3.
X
er x
X
e~ x
X
e~ x
X
e~ x
0.00
1.0
0.25
0.78
0.80
0.449
1.8
0.165
0.02
0.98
0.30
0.741
0.85
0.427
2.0
0.135
0.04
0.96
0.35
0.705
0.90
0.407
2.5
0.084
0.06
0.942
0.40
0.67
0.95
0.387
3.0
0.05
0.08
0.923
0.45
0.638
1.0
0.368
4.0
0.018
0.10
0.905
0.50
0.607
1.1
0.333
5.0
0.0067
0.12
0.887
0.55
0.577
1.2
0.301
6.0
0.0025
0.14
0.870
0.60
0.549
1.3
0.273
7.0
0.0009
0.16
0.852
0.65
0.522
1.4
0.247
8.0
0.00034
0.18
0.835
0.70
0.497
1.5
0.202
9.0
0.00012
0.20
0.819
0.75
0.472
1.8
0.165
10:0
0.00004
Example No. 1. A coil having 1000 turns and 5 ohms resist-
ance is connected to a source of constant potential of 100 volts.
10
ELECTRICAL ENGINEERING
(a) Show at what rate energy is being delivered to the entire
circuit and to the resistance. Show at what rate it is being
stored in the magnetic field as the current is increasing after the
circuit is closed.
(b) What is the rate of change of the flux when the current is
10 amp.?
Referring to equation (13),
* T? (23)
therefore the rate of energy supply to the entire system is Ei
watts,
and
Ei
+
10 8 dt
(24)
The current will begin at zero value and finally reach a value of
E
i = I = = 20 amp.
2000
1800
1600
"1400
1200
c
fc 1000
400
200
Rate of Energy Supply to Inductive Circuit
Constants of Circuits
e = 100 Volts
r = 5 Ohms
N = 1000 Turns
8 10 12
Current in Amperes
FIG. 4.
14
1G
20
The rate at which energy is dissipated in heat is i*r and the
rate at which energy is stored in the magnetic field is:
3? - w - * v (25)
The three curves in Fig. 4 show these rates.
It is interesting to note that energy is being stored at the
greatest rate when the current is one-half of the the final value.
This can readily be proven by differentiation of equation (25)
and equating the result to zero, thus,
E - 2ir = 0,
INDUCTANCE AND RESISTANCE 11
therefore . E_ /
~ 2r ~ 2*
The rate of change of the flux as the current changes is obviously
d<j> _ E - ir
~dt ~ F X IF 8 '
Therefore when the current is 10 amp. the rate of change is
5,000,000 lines per sec. The rate of change is greatest at
first and becomes zero when the current reaches its final value.
The determination by calculation of the inductance L of a
circuit is usually very difficult, in fact almost impossible except
in the very simplest cases, such as parallel long circular con-
ductors. Approximations of one nature or another have almost
always to be resorted to. Usually the inductive circuit contains
iron, and in that case the reluctance (and hence the inductance)
is not constant but changes with the degree of magnetization.
Later in this volume the effect of the changing inductance in
iron circuits will be considered, but at present it shall be assumed
that L is a constant regardless of the value of the current.
The inductance of the field circuit of a dynamo can readily be
determined for any particular field current by experiment. All
that is needed is to run the machine at some speed and to read
the voltage and field current. These data in addition to those
of the field and armature windings suffice. By definition,
total flux X turns
current X 10 8
The total flux per pole is determined from the voltage, speed and
armature winding. Consider a 10-kw., two-pole, direct-current,
110-volt generator, having 2.5 megalines of flux per pole, and
1500 field-turns per pole. Assume that at normal voltage its
field current is 3 amp. and that the field spools are connected in
series. Thus
T 2.5 X 10 6 X 1500 X 2 or
L = 3 x 1Q8 = 25 henrys.
Example No. 2. Figs. 5 and 6 represent the direct-current
generator referred to above. M is the armature and F the field.
If a voltmeter of 11,000 ohms resistance is connected as shown
and switch S is opened without arc when the field current in
ammeter A is 3 amp., what will be the effect on the voltmeter and
will the ammeter and voltmeter read in the same direction as
12
ELECTRICAL ENGINEERING
before the switch was opened? Before the switch is opened the
current flow is as shown in Fig. 5. As the switch is opened
the field flux can not die away instantaneously. The field cur-
rent therefore can not die away instantaneously, but continues to
flow through the only available path, which is that of the volt-
meter. Since the resistance of the voltmeter is 11,000 ohms it is
evident that the voltage across the instrument becomes at the
very first instant very high.
FIG. 5.
x~i r
i t
FIG. 6.
It tends to become ir = 3 X 11,000 = 33,000 volts.
Thus the voltmeter will probably burn out as the needle
swings to the opposite side of the scale. The ammeter needle
will remain stationary for the first instant and gradually come
down to zero.
This problem gives an idea of the nature of the shock that is
experienced where the field current of a generator is carelessly
interrupted and permitted to pass through a person. Depending
upon the nature of the contact the resistance of a body may be
from 1000 to 10,000 ohms. If, therefore, a person touches both
sides of the field winding when the field circuit is interrupted,
he will experience a very severe shock. The energy stored is
usually quite considerable. In this case it is J^L/ 2 = J X 25 X
9 = 113 joules. Since 1 joule is 0.74 ft.-lb., the energy available
is 84 ft.-lb., i.e., that of a pound weight dropping 84 ft.
It may bs asked, what would happen if the voltmeter were not
connected across the field winding? Where would the initial
rush of current, of 3 amp. flow, when the switch was opened?
In reality it is impossible to open the field switch without an
arc; therefore the current can not be interrupted instantaneously.
INDUCTANCE AND RESISTANCE
13
Furthermore the circuit is more complex than assumed. The
field winding has considerable capacity and therefore acts as if it
were shunted by a condenser. A portion of the 3 amp. will
therefore flow as condenser current, but a large portion will
appear as secondary currents in the iron circuit of the poles.
This phenomenon will be understood later from the investigation
of circuits having mutual inductance.
The problem is instructive in that it explains frequent burnout
of voltmeters, and in that it teaches that the voltmeter should
always be disconnected before the switch is opened, or otherwise
be connected on the armature side of the field switch. It teaches
also that in opening the field switch a relatively low resistance
should be shunted across the field winding to prevent high vol-
tage, and finally that it is well to open the field switch slowly.
The importance of shunting the field circuit is best illustrated by
a numerical example.
j Example No. 3 (Fig. 7). Assume that the field circuit having a
resistance of 36.5 ohms is shunted by a resistance of 50 ohms, and
assume again, for the sake of simplicity, that the field current of
it \
r = 36.5
L= 25
FIG. 7.
3 amp. is Interrupted without arc and that L is constant at 25
henrys. The total resistance in the circuit is then 50 + 36.5
ohms or 86.5 ohms. Determine the current in the field winding
and the shunted resistance and the voltage across the field coils
which is the same as the voltage across the resistance after the
switch is opened.
Referring to equation (18)
= 3e
-3.46
For t
0.05
0.10
0.20
0.5
1.0
-3.45<
1
0.84
0.71
0.50
0.18
0.03
i
3
2.32
2.13
1.5
0.54
0.09
iR
150
116.0
107.0
75.0
27.0
3.0
14 ELECTRICAL ENGINEERING
It is seen that in this case the maximum voltage across the field
coils, which, of course, occurs at the moment of opening the
switch, is 150 volts, as compared with 33,000 when the voltmeter
shunted the field coils. The field current i dies away very rapidly.
In 1 sec. it has almost disappeared. The energy stored in the
field is spent in heating as an i z r loss.
Example No. 4. Prove that in discharging an inductive circuit
all energy stored is spent in heat.
The instantaneous value of the current was found to be:
i = Ie~l l ,
therefore the energy expended in heat from time zero to infinite
time is:
2r
J"<-- /*
i*rdt = Pr\
=o Jo
[T 2r -i oo T2 r
-*'*]. --^
It is of interest to study the rate at which the field flux, or what
is equivalent, the field current, can build up when closing the
field winding on a constant-potential busbar, and to see how much
more rapidly the field current can be made to build up when a
considerable resistance is inserted in series with the field coils.
It will be assumed that use is made of the winding described
in example 3, that is, one with a resistance of 36.5 ohms and
inductance of 25 henrys. This circuit is connected to a
direct-current busbar having a constant potential of 110 volts.
Referring to equation (17),
i = l ~^ =3[1 -6- 1 - 46 '].
The lower curve in Fig. 8 shows the result of this calculation.
If, instead of exciting the winding from a 110-volt main, it is
connected to a 220-volt circuit and sufficient resistance is inserted
in series to keep the permanent current at 3 amp., the rise in
current will be more rapid than in the first case, as shown in the
upper curve of Fig. 8.
There is an interesting mechanical analogy for the starting or
stopping of a current in an inductive circuit.
To bring a train up to speed a certain force is necessary; this
force must overcome the friction and provide the necessary
acceleration.
INDUCTANCE AND RESISTANCE
15
Let F be the total force necessary, and fv the force of friction
and wind resistance which, for simplicity's sake, is assumed to
be proportional to the velocity v, and the mass M.
Then
F = fv + mass X acceleration
or,
dv f F^
dt ~~ M v " M'
.2 .4
.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6
Time in Seconds
FlG. 8.
If the drawbar pull F as well as the coefficient of friction / be
assumed constant during acceleration,
F f t
-*
where C is the integration constant
16 ELECTRICAL ENGINEERING .
If the train start from rest, then for t = O, v = O.
F F
.'. O = j + C, or, C - -,
. -[,-.-
By comparing this with the equation for the starting of a current
Fr r ~~i
in an inductive circuit, which is, i = 1^1 e~ L * J,it is seen that
in electrical problems, the current corresponds to Velocity, the
e.m.f. to the mechanical force, the ohmic resistance to frictional
resistance and the inductance to the mass.
The analogy can be carried further. The energy stored in the
magnetic field, %LI 2 , corresponds to the kinetic energy of a
moving body, %Mv 2 . The electromagnetic momentum LI cor-
responds to the mechanical momentum Mv, etc.
A problem involving mechanical as well as electrical transients
will next be considered.
Find the equation of the dying away of the field current in a
direct-current self-excited shunt motor disconnected from the
circuit and permitted to decelerate to standstill.
Let the moment of inertia of the revolving part be /. Let the
full speed be N revolutions per second corresponding to an angu-
lar velocity of 0:0 radians per second. Let the power required to
run the motor at full speed but at no-load be P hp., and assume
that this power is represented by friction loss in the brushes and
bearings, which is a very close approximation, particularly after
a few seconds of deceleration, when the core loss becomes very
small; and neglect the i 2 r loss. Assume that the saturation curve
is a straight line, so that proportionality exists between the field
current and the flux.
Let the normal field current be J . Let the normal flux per
pole corresponding to this current be <. Let the armature e.m.f.
at full speed and flux be E, and the total field-circuit resistance
be r, and let the motor have p poles and each field spool have n
turns.
Mechanical Calculations. 1. Determine the angular velocity
o. It is, a = 2irN.
2. Determine the friction torque, or moment Q. We have
X lb.
INDUCTANCE AND RESISTANCE 17
3. Determine the stored energy. In general W = ^Mv 2 , in
the case of a revolving wheel; if p is the radius of gyration,
W = %M (2ir pN) 2 = %Ia 2 ,
where / = Mp 2 = moment of inertia.
Thus, with revolving masses, / takes the place of M, and a. of v.
During deceleration, no external force or torque is applied.
Thus,
= Q + I -JT = torque of friction + torque of deceleration.
(For sake of simplicity the small power given electrically is
neglected.)
da = -jdt,or,a = j t + C;
for t = 0, a. = a . .'. C = a -
Q.
. . a. = ao -j t.
If T denotes the time at which the rotor stops, then for t = T,
a = 0.
Ci ^ T ' T*
= a ~jT, ..7 - Q O .
And, o,_ / t\
\ l "" T)
a = Oi jfj t = o
Check whether all energy is spent at t = T, neglecting the sup-
ply of energy from the diminution of magnetic field and the con-
sumption of energy in heat. The stored energy is 3^/a 2 .
The energy consumed by friction during deceleration is:
I force X vel. dt = \ Q2* Ndt = \ Qadt = Q
substituting, ^ _ /
~ Q Q:O '
= Q L>| o - 7 Y Q2 a 2 = ~' Q ' E ' D '
Electrical Calculations. If the field current remained constant
during the deceleration (which it obviously does not), the arma-
18 ELECTRICAL ENGINEERING
ture voltage at speed a would be:ei = E, so that due to the loss
of speed alone, the armature voltage is reduced from E to E.
If the field current is reduced from / to i, the flux is reduced
from <l> to <p, and therefore the e.m.f . at constant speed is reduced
in the proportion y--
i
Thus, at field excitation i and speed a, the armature voltage is
Jo a
but E
Jo~ '
But the relation between the current and the e.m.f. is:
di
or,
- -t -
.-. i = C ^ Lr = Ce LT 2 ;
for < = 0, i = 7 . .*. C = J .
/. i = / oe "2TZ.
The motor stops, when t = T, and when the current is:
_rT
10 = J ^
Remembering the equation of the decaying current in an
_rT
ordinary inductive circuit, i = I ^~, it is evident that in the
case of a decelerating self-exciting machine the current does not
die as fast.
After the motor has stopped, the current obviously dies down
according to the law:
- ~I (t ~ T)
NOTE. For a more detailed discussion of this see STEINMETZ'S "Tran-
sient Phenomena."
INDUCTANCE AND RESISTANCE 19
Verify the curve (a) in Fig. 9 in the case of a four-pole, 7.5-hp.
motor having the following constants:
P = 4 7 = 2.75 amp.
N = 20 E = 110
7 = 0.25 r = 40 ohms, total
$ = 1.5 megohms per pole n = 1000 per pole
P = 0.72 hp.
In large machines, the windage loss is frequently greater than
the loss in the bearings.
The windage loss may be assumed to be proportional to the
square of the velocity. In other words the torqu.e necessary to
overcome the windage is proportional to the speed.
Assuming again that the electric power is small and that it can
be neglected then the equation connecting speed and time during
deceleration becomes:
0= Qi + Qia + JF^'
or, da Q<2 Qi
~
when . r Qi
a = oto, 1 = 0. . . O = o ~T 7T*
" a= ~ ft
when
i = T, the motor stops, a = 0,
-
> 2 + Qi
- y T, or, log - 1 -
I , ^2 + Q
r, a a .
e = Eo - j- = ir,
OLQ IQ Q!o
a . T di
. . ir = IT + L --1
20 ELECTRICAL ENGINEERING
s+f ('-=)-
ao
which, transformed, becomes:
where 4=14-
v _
and
I +(!-'>-.
but
when = 0, z = 7o;
rA
.". C =
If the problem given above is modified, so as to include a
windage loss at full speed of 0.15 hp. as well as the bearing loss
of 0.72 hp., the constants are:
r = 40, I = 0.25,
L = 2l, Q 2 - - 0.00525,
o = 125.8, Qi = 3.16.
T becomes 9.04 sec., and
A = 5.71,
^ = 10.60,
K = 0.021,
_. 7 (-10.60* +504.6 -504.6e--2 1
. . & i 06
In curve 6, Fig. 9, is given the relation between the current
and time in this case.
INDUCTANCE AND RESISTANCE
21
The problems considered up to this point have all involved
very simple integrations. Frequently, however, this is not the
case, and to solve the differential equations, it is necessary to
make algebraic transformations.
The most important of these transformations is to separate
fractions into partial fractions.
3.0
2.5
1,2.0
1.0
0.5
(b
5 6 7
Time, Seconds
FlG. 9.
10 11
Almost any algebra deals with this; nevertheless it may be
opportune to refer to it briefly here, although it is suggested
that the student's memory be refreshed by reading, for instance,
WILLSON'S " Advanced Algebra," from which the following is
largely abstracted.
If FJ-\ is a fraction, that is, the numerator is of lower degree
than the denominator.
It is known that F(x) can always be expressed as the product
of linear factors, which are not necessarily real.
If the factors are real, then F(x) can be expressed as the product
of real linear and quadratic factors. Two cases will be here
considered.
First. No factors are repeated.
Example. F(x) = (aiX + &i) (a& + & 2 ) (a$x 2 -f- & 3 # -f- C 3 ).
Then
A l
-f
F(x) aix + bi a 2 x -[- 6 2 a 3 x 2 + 6 3 a; +
where AI, A 2 , A 3 , and B 3 are constants, which can readily be
found, since if the expression:
22 ELECTRICAL ENGINEERING
aox n -f aix n ~ l + a 2 x n ~ 2 + - i = b x n + to"- 1 + b 2 x n ~ 2 +
holds for all values of x, then the coefficients of like powers of x
must be equal, thus a = &o, i = &i> etc -
Show that
* 2 l 1 2 5 4z + 5
Second. Some factors of the denominator are repeated.
F(x) = (aix + bi) 2 (a 2 x + 6 2 ) (3^ 2 + to + c 3 ) 2
Then
+
Prove that
2x* + b 2 7 15
(x - l) 2 (x - 3) 2 (x - I) 2 ' 2(x - 1) ' (x - 3) 2 2(x - 3)
The application of this transformation is found in any transient
phenomenon in which disproportionality between magneto-
motiveforce and resulting flux exists.
As an example the condition governing the self-excitation
of a direct-current shunt-wound generator will be considered.
(For a more detailed discussion see STEINMETZ'S " Transient
Phenomena.")
It will be assumed that the relation between the flux <p and the
field current i can be sufficiently closely represented by FROE-
LICH'S equation:
9 - ki (1)
1 +fcii
Let e c be the e.m.f. generated per megaline of flux at normal
speed, and e Q be the normal e.m.f. at normal flux <f> .
Then e c
The e.m.f. e corresponding to any other flux <p is:
e = e c <p
The e.m.f. consumed by the resistance is ir.
The e.m.f. consumed by the changing flux is 77^ -
1UU dt
if (p is expressed in megalines
and n is the total number of turns enclosing the flux.
INDUCTANCE AND RESISTANCE 23
Thus . n d<p
h 100 dt
from (1) . _ _JP__
K ~~~ K\(f>
<pr
. . e c <p = T 7
Separating the variables
W/c/ / j i v \Jj\D
n e<f>(K KHP) <pr
<p(e c k r e c knp)
To integrate this the fraction is broken up into partial fractions
thus:
100 dt [A B
T-
and A is found to be
and B is found to be
n [<f> ek r k^e c (p
k
e c k r
k\r
e c k r
Integrating each term we get after a slight transformation
~ - ~ '08 * -
If at the time of closing the field circuit the remanent flux
is <p r and the corresponding voltage = e r then for t = 0, <p = p r ,
e = e r .
When C is determined and the final expression becomes:
n FT i e i e c k r kie ~\
t = ln/ > / i x ke c log r log ; ,
100e c (e c k r) L & e r & e c k r kie r l
The voltage ultimately reached is e = e$ when t = .
Thus , e c k r n/ioo
loe; r i = oo thus
P If - -- 7* A % -|P
e c fc r
e c /c r K\en = or eo = ;
The greatest value of r which gives a positive value of e
r = e c k.
The condition of self-excitation is thus r ^ e c k.
24 ELECTRICAL ENGINEERING
Up to this point, the problems have involved inductive circuits,
on which a direct-current e.m.f. has been impressed. In case of
alternating current the impressed e.m.f. varies from instant to
instant and, while a harmonic e.m.f. is usually assumed, fre-
quently the variation represented by a wave is much more com-
plex. As long, however, as the e.m.f. is obtained from a dynamo
of symmetrical poles, no matter how shaped, the e.m.f. wave
can be expressed by a series of sine functions of odd frequencies.
In the study of transient phenomena in connection with alter-
nating current, the equations are derived for the fundamental
wave only, that is, the instantaneous values of the e.m.f. are
represented by e = E sin 6.
If it is desired to know the result with distorted waves, the
simplest method is to treat each harmonic independently and to
add the instantaneous values so obtained. If the effective
value is desired the square root of the sum of the squares of the
effective value of each wave should be taken.
As stated previously, the instantaneous value of the e.m.f.
is generally expressed in two ways, either e = E sin coi or e =
E sin 6, or the expression may be of more general form : e
E sin (at -f a) and e = E sin (6 + a). In these expressions,
e is the particular value of the e.m.f. at time t, or at phase angle
6, and E is the maximum value of the e.m.f. In the first case,
the angle ut is expressed in radians, not in degrees, w is the an-
gular velocity = 2 TT/, where / is the frequency. The relation
between radians and degrees is 360 = 2ir radians, thus 1 radian
is -77 = 57.3. To reduce equation e = E sin (cot + ) to
ZTT
degrees it should therefore be written e = E sin (57.3 ut + a),
where in all cases a is expressed in degrees, as is customary. To
reduce the expression to radians it should be written
Note in connection with this that in the expression, y = sin x,
x is expressed in radians, not in degrees. To bring it to degrees
the equation becomes y = sin 57.3 x.
In the development the value of the sine function
Sin x = x - I + I - ^ +
x is again expressed in radians.
INDUCTANCE AND RESISTANCE
25
It is important to have this clearly in mind. It is well worth
while to plot some curves of distorted waves from equations in-
volving phase angle as well as radians.
Example No. 5. Verify the e.m.f. wave in Fig. 10, e = EI
sin ut + Es sin (3 ut + a) for E l =-- 10, E 3 = 5, a = 30 and
the frequency 25 cycles.
1.2
1.0
8
G
4
2
2
4
G
8
1.0
1.2
/
t
f ~\
\
\
\
/
/
\
V
7
\
/
\
/
.4
.6
.*
1
JO 1
20 1
160 U80 200 220 2
Angle\in Degrees
10 2
2
50 3
30 3
20 3-
^3
\
/
\
/^
/
\
/
/
,
V^
/
\
t
\
^2
.01
.02
Time in Seconds
FIG. 10.
.04
Prove by integration that with a distorted wave :
e = EI sin (ut -f i) + E 3 sin (3 cot + 3 )
the effective value is e// = "\/6i 2 e// + ^3 2 e//
Thus in this instance, since the effective value of the funda-
~F 10
mental wave is ^ = 7= = 7.07, and that of the triple har-
monic is 7= = T= = 3.53, the effective value of the wave
V2
recorded by a voltmeter is e = \/7.07 2 + 3.53 2 = 7.9.
Referring to Fig. 11: prove that ammeter A when placed in a
circuit carrying 10 amp. direct current, 8 amp. 60-cycle current,
and 5 amp. 125-cycle current reads 13.7 amp.
26
ELECTRICAL ENGINEERING
Harmonic e.m.f. Impressed on a Circuit of Resistance and
Inductance in Series. Let time be counted from zero value of
the impressed e.m.f. and let the e.m.f. be rising.
Thus e = E sin ut where e is the instantaneous value of the
harmonic e.m.f. at time t. E is the maximum value, co = 2 TT/,
is the angular velocity, / the frequency, r the resistance and L
the inductance of the circuit.
10 Amperes, D.C.
8 Amperes, 60 Cycle
5 Amperes, 25 Cycle
FIG. 11.
If i is the instantaneous value of the current when the e.m.f.
is e then:
e E sin cot = ir + L -77
at
or
di
r . E
(27)
(28)
By comparing this equation with equation (2), it is seen that
T V
T = P and j- sin ut = Q.
P is not a function of the independent variable t, but Q depends
thereon, thus the solution is given in equation (8).
It is
(29)
i = e L l \ I e + L Y sm ut dt + C
The solution of this equation depends upon solving
C + r -tE , E r ,r t
I ^L jr sm wt dt = Y~ I L sm <*t dt.
E.
j- is a constant and can be left out of consideration at present.
INDUCTANCE AND RESISTANCE 27
It is also convenient to substitute a single letter for Y' Let then
L
The immediate problem then is to solve ft at sin co dt.
An integral involving exponentials or sine functions is usually
easy to solve, because the differential of the functions are similar
to the functions.
If y = ax then ^ = ae ax .
dx
Similarly if y = sin ux, then -5- = co cos coz,
or if y = cos ux, then - = co sin cox.
Thus i at , _ j. at
Ui
f
and / . , 1 ,
sin co* dt cos cot at.
J w
Fortunately for the engineer there are only very few methods of
integration that need to be known. One of these is " Integration
by Parts."
That is: fudv = uv - fvdu (30)
In integral ft* sin to* dt, let u = e at and dv = sin co* dt.
.'. du = ae at and v = cos co*
CO
/. yV* sin co* dt = - cos cot + \ - e at cos co* dt (31)
CO J CO
This equation is indeed more complicated than the original. It
is evident, however, that by again integrating the last term in 31,
an integral results which contains an exponential term e at and a
sine term instead of the cosine term. Thus the final expression
will contain integrals of the same trigonometrical and exponential
functions, which therefore can be solved directly. However, it
is somewhat more convenient to use another method.
Referring again to (30) let in this case:
u = sin co* and dv = e at dt
.'. du = co cos co* dt and v = - e at
a.
.'. ft at sin co* dt = sin co* - ( - e at cos co* dt (32)
a J a
28 ELECTRICAL ENGINEERING
By multiplying 31 by - and 32 by and adding the two equa-
OL Ct>
tions, it is readily seen that
./V" sin ( dt = ^-. (^ - ^-') (33)
co- + a- \ co a I
T
Substituting = r and remembering that x t the reactance cor-
responding to the inductance L is 2irfL = coL and that the
impedance z = \/r 2 + x 2 .
Then
I L * sin wt dt = e + L -^,[ r sin coi x cos coi]
(34)
Let the angle of lag of current be ft thus
tan ft = - and r = z cos ft (35)
x = z sin ft (36)
Substituting the values in 34:
-t + r t L
sin coi dt = e L - sin (cot ft) (37)
z
Referring to equation 29
E -L t
i sin (cot ft) + Ce L (38)
The integration constant C is determined from the particular
problem under consideration.
Assume that it is desired to find the value of the current at any
instant after the switch is closed and the alternating e.m.f. is
impressed upon the circuit, and that the switch is closed at time
t = i b when the instantaneous value of the e.m.f. is e = E
sin coii.
Since, as has previously been discussed, it is impossible to
establish a magnetic field instantaneously, the current can not
flow at the first instant. Thus for t = t\, i = 0. Substituting
these values in equation 38, then:
7? r
= - sin (coii ft) + Ce~L tl ,
E Z l
.'. C = e L sin (coii /3),
Substituting this in (38)
sin (coi ft) e * lfa) sin (uti ft) (39)
i =-- ~\s
INDUCTANCE AND RESISTANCE 29
It is often convenient, to eliminate t entirely from the expres-
sion and to use the phase angle 8 only and to express 6 in degrees.
That is, the e.m.f. is expressed as e = E sin 6. In that case
B = ^t = 27T/7.
The exponential term 6~ a ~' becomes ~^ (fl " 0l) = ~* ((? ~' l)
T_ (9-fr)
if 6 and 6\ are expressed in radians or & 57.3 if and 6\ are
expressed in degrees.
Thus when 6 and d\ represent degrees
Er - r <-_*> -]
i = -I sin (e - (3) - e * 57.3 sin (0i - /3) (40)
The equation is, however, always written
i = | [sin (0 - i?) - ~* "-' !) sin (0! - 0)] (41)
and it is understood that the exponential term should be ex-
pressed in radians.
Equation (41) can, of course, be derived directly by using the
phase angle instead of cat.
Thus T di
E sin ir = er + L -7-
cfo
may be written
E sin - ir + x ^, (42)
where x is the reactance.
Thus, x = 2-irffj = coL
and cot = 6.
.". d# = codt or dt = -
CO
Prove that equation 41 is the solution of
di
E sin 6 ir -\- X~TC,
The exponential term in equation (41), while of importance
during the first second or so, ceases to affect the result very
shortly after the switch is closed.
Thus the equation for the current after the system is stable is
i = E z sin (8 - 0) (43)
The current lags behind the e.m.f., E sin 0, by an angle 0, whose
x
tangent is -
30
ELECTRICAL ENGINEERING
The effective value of the e.m.f. is
E
= V5
and of the current j E
~~ V2Z
It is of interest to note that the transient term is a maximum
when sin (B l - /3) = 1, that is 0i - = 90, or 0i = 90 + /3.
This value of 6\ also gives the maximum value of the perma-
nent current.
1.4
1.2
1.0
.8
.6
.4
.2
-.2
-.4
-.6
-.8
-1.0
-1.2
-1.4
I
1
/
z
I
/'
6
L Impressed Voltage
' Current when Swi
) " "
' ft t 1
v
tch is Closed at
1
\
k
&
270
79 Giving tli
Value of
-
=
=
-
.',' ,',' ,'
e Permaaeu
Current
^
>
7T-
~r
s
y
S
J
Mill
/
y
\
1 LU=
x
A
-
s*
/
S
x
/
^s^
OS.
s^
~^^
\
/
//
/
^
s\%
/
\
k
/
/
y
/
/
s
i:
\\
\\
/
w/
\
\s
i
/
y
,
/
/
\
X
\\
^
/
//'
7*
/
\
5
/
^
/
/
/
/
\
\
\
\\
^
/
420^
M
7
\
\
j
W_
iKT
1201 1501 1
5Q
2
^
^
\\\
,300 330
1
3U
390 l
////
/^53
4i
5
5
w
5
6
An
glei
a Degrees
\
Y
^U |y
1 \
y////
\
\
V
V^k/
//'
^//
^
\
s
\
_^^:
-^//
^
>
X.
>s
^K
t
'//
^
X
\
'
^S
s\
"v
?
FIG. 12.
The exponential term is zero, that is, there is no transient
effect if #1 j8 = 6 or #1 f3 or, in other words, if the circuit is
closed at such a time as would give zero value of the permanent
current.
Fig. 12 shows a series of such transient currents. Each curve
corresponds to the closing of the switch at a particular value 0i
of the phase of the e.m.f.
Thus, for instance, curve D shows the starting current when
the e.m.f. wave has a phase angle of +60, that is, when 0i =
60. These curves are calculated with the following constants
E = 1 r = 0.196 x = 0.98.
Problem No. 6. Check some curve in Fig. 12.
It is of interest to study the rate at which energy is being sup-
INDUCTANCE AND RESISTANCE
31
plied at any instant. This is equal to the product of the e.m.f.
and the current:
p = d = Es'm B X |[sin (6 - 0) - e~x (0 ~ dl} sin fa - 0)] (44)
By simple transformations the equation becomes
- cos (20 - 0)
P :
-^(0-00 -I
- e * sin (0i 0) sin 0J (45)
A -Impressed Voltage
- Power Input
C- Permanent Power
FIG. 13.
The first term in equation 45 must represent the power at any
instant after the conditions have become stable. This power is
expressed by
E 2
P 1 = 22 [cos - cos (20 - ft] (46)
It consists of two terms, on e a constant term ^ cos /3, the
other a term which changes with double frequency; the net
result of which over a complete period is zero, since the positive
values are as large as the negative. Thus while the instantan-
eous values of the power vary from instant to instant and may
32 ELECTRICAL ENGINEERING
alternate from positive to negative values there is a definite
average power delivered, which is
E 2
P = 2z cos P
The exponential part of the power,
P 2 = - e ~ r x (e ~ 9l) sin (0! - 0) sin (47)
is gradually decreasing in magnitude as well as oscillating at
normal frequency.
In Fig. 13 are given three curves; the first, A, is the wave of
the impressed e t m.f.; the second, B, the power input; and the
third, (7, the power curve after conditions are stable. These
curves have been based upon the constants given in problem 6
and are well worth reproducing by calculation.
The curves show that during the transient period the instant
of maximum power is practically the same as that for permanent
condition. They also show that the first rush of power is greater
than that which corresponds to permanent condition, the reason
being that the change of flux during the first part of the cycle
is greater than during the corresponding time under stable
condition.
CHAPTER II
PROBLEMS INVOLVING MUTUAL INDUCTANCE
Up to this point the problems considered have dealt with cir-
cuits of inductance and resistance only. However, in many
circuits of commercial interest there are secondary circuits which
are more or less closely coupled with the primary, and which
influence the former materially. As instances of such circuits
may be given the secondary winding of a transformer, the eddy
currents in pole pieces of generators and motors, induced cur-
rents in telephone lines running parallel to transmission lines, etc.
Sometimes the secondary circuits carry currents by virtue of
impressed e.m.fs., but frequently the currents are the result of
the action of the primary currents. With a change of primary
current obviously there is a change of the flux produced by the
current and if this flux interlinks with the second circuit, e.m.fs.
are induced therein, the values of which become higher as the
interlinkage becomes more nearly perfect. While it is impossible
to arrange two circuits so that all flux interlinking one will also
interlink the other, the condition can be approached reasonably
close under the most favorable conditions.
The limiting case is, of course, perfect mutual induction, which
condition will therefore first be considered briefly.
Two Coils of Perfect Mutual Inductance. Assume then that
it is possible to place two coils so close together that there is no
leakage flux between them, that is, so that all flux that surrounds
one coil also surrounds the other. Let the first coil, the primary
coil, have NI turns and r\ ohms resistance, and the secondary
coil N% turns and r 2 ohms resistance. Determine the open-circuit
voltage of the second winding. When the first is connected to a
source of constant potential E, we have obviously:
F 'r
r
The rate of change of flux is thus
d4> E i
dt ~~ A
33
34 ELECTRICAL ENGINEERING
Therefore the voltage of the second coil e 2 is
N z d4> N 2
~WTt~ ~Nl (1
At the instant of starting, when ii is zero, the secondary voltage
is e 2 = - - -rp E, that is, it is proportional to the ratio of turns.
7^
When the primary current reaches its constant value J =
the secondary voltage e 2 is zero. If the secondary winding has
more turns than the primary, then at first the secondary voltage
is higher than the impressed voltage. It decreases rapidly,
however, and soon becomes zero.
Prove that the two voltages are equal numerically when
Assume that two coils, which, when considered alone, have
resistances and inductances of r\, r% and LI, L 2 , respectively, are
placed so close together that there is perfect mutual inductance
between them (which of course is in reality impossible). Find
the open-circuit voltage of the second coil if the first coil is
connected to a source of constant potential.
In the primary we have:
E = i fl + L, J-
The counter e.m.f. of self-induction of the primary coil is
- LI -j7 and thus the voltage of the second coil is
N 2 dii
Check the values of the primary current and secondary voltage
as given in full lines of Fig. 14.
for
E = 10 n - 0.10 L! = 2.5 NI = 10
r 2 = 0.50 L 2 = 10 N 2 = 20
In the case referred to above the primary current will rise from
zero to a final value of 100 amp., while the secondary voltage
decreases from 20 volts to zero.
If when the primary current has reached its final value the
coil is suddenly short-circuited, what will the primary current
and secondary voltage be?
PROBLEMS INVOLVING MUTUAL INDUCTANCE 35
The primary current will decrease according to equation:
Check numerically the two dotted curves in Fig. 13.
During the discharge of the primary the number of coulombs
are
| hdt = (
Jo Jo
= 100 =
r i
2500 coulombs.
1UU
90
80
70
S 60
r
< 40
30
20
10
-2
-4
-6
s - s
3-10
p>
-12
-14
-16
-18
20
\
I
E
-
\
_^,
-i
^
-^
s
^
~-
N
/
*
*
\
/
Two Coils of Perfect Mutual
Inductance
A-Primary Current
B- Secondary Voltage
\
^/
\
/
-
/
/
S N
X
/
\
^
/
>
/
a
/
1
ri
2
m
)
V
o
or
a
s
_4
50
\
e^
n
S
2C
!
\
J-
^
^
^
^Z
?
k^
^
/
/
/
\
/
/
/
~-~*
/
x
j
x^
/
N
/
*s
/
^
^
---,
^
/
FIG. 14.
Obviously, when connecting the primary to the source of supply,
the number of coulombs required up to the time when the current
becomes stationary is infinite, since it takes infinite time for the
current to reach this value.
Two coils of resistances and inductances of ri, r and LI L are
connected in series and placed so close together that it is assumed
that they have perfect mutual inductance. What will be the
resultant resistance and inductance (a), if the coils are wound
3
36
ELECTRICAL ENGINEERING
in the same direction; (6), if the coils are wound in opposite
directions?
The inductance of an air coil is subject to rigid mathematical
determination, but the complete solution is very cumbersome.
However, one of the best approximations, that of BROOKS and
TURNER, published as an Engineering Experiment Station
Bulletin by the University of Illinois, is:
L =
cm-
IQfr + 12c + 2R
10 9 (6 + c + R) ^ 106 + lOc + 1.472
X 0.51og 10 (lOO +
For coils which are not extremely thin or extremely long, this
equation becomes approximately:
L = 7
cm"
(2)
(b + c +
Where L is expressed in henrys
cm = centimeter length of wire
b and c are the height and thickness respectively of the coil
and R the outside radius, all in cm.
00000
OO(
00
f
ooooo
ooooo
00<
oo<
00
00
ft
ooooo
ooc
00
ooooo
oo<
oo
i
R -->
1* ' '
FIG. 15.
The maximum inductance is obtained when b = C and R
(see Fig. 15). Then
2C
L =
0.27Cm
C X 10 9
henrys.
It is seen that the inductance is proportional to the square of
the total length of wire, which is, of course, proportional to the
turns. Thus the inductance is proportional to the square of the
number of turns, or
L = KN\
(a) Coils in the same direction.
PROBLEMS INVOLVING MUTUAL INDUCTANCE 37
Let N be the number of turns in the first coil, or,
N =
and NI the number of the turns in the second coil, or,
JLl
r
The total number of turns in the two coils when considered as
one coil (which is permissible when perfect mutual inductance is
assumed) is
The combined inductance is then
Lo = KN 2 = (VL + VLi) 2 = L + Li + 2-v/LLL
The resistance is obviously r = r + r\.
(b) By similar reasoning it is found that if the turns are in
opposite directions
Lo = L + LI 2\/LLi and r = r + ^i-
From the above it is evident that the equation for the starting
current, for instance, is:
Two Coils of Perfect Mutual Inductance Connected Simulta-
neously to Sources of Constant e.m.f s. E and EI. Let r, r\ and L,
LI be the resistance and inductances respectively, and assume
that the circuits are closed at the same instant. Assume first
that the coils are connected in the same direction, that is, in such
a way that the permanent current in both coils will produce mag-
netic fields of the same polarity. It is evident that in this case the
impressed e.m.f. has to overcome not only the resistance and in-
ductance drop due to the current in the coil, but also the e.m.f.
which by transformer action is induced in one coil by the change
of current in the other.
Consider one coil alone, for instance the second coil: The
counter e.m.f. of this coil is i~5T' If it has N\ turns, the
voltage per turn is jj- rr' Since it has been assumed that
/V i u/t
38 ELECTRICAL ENGINEERING
there is no leakage field between the two coils, it is evident that
this same voltage per turn is induced in the first coil by the cur-
rent in the second. Thus the " transformer " e.m.f. in the first
coil having N turns is -^- LI -rr, and similarly the transformer
e.m.f. produced in the second coil by the current in the first is
Ni T di
~W L dt
But N IT.
therefore the e.m.f. in the first coil caused by the mutual flux is
L T dii _ ^/rr dil '
LI ^ ~ VLLl dt
Thus it is seen how, when the mutual inductance usually denoted
by M is perfect, M = \/LLi. In reality M is always smaller
than \/LLi. The general equation dealing with e.m.fs. consumed
by resistance, inductance and mutual inductance, are then
To solve for instance i the following transformation is conven-
ient, multiplying (3) by LI and (4) by M and add the equations
so obtained.
li is: LiE - ME, = Liir + LL l ~
at
- Mi.r - M 2 ^ (5)
Since with perfect mutual inductance
M 2 = LLi (6)
Liir - LiE + MEi
dii _ LJT <ti
' ' ~di ~ Mrl di'
Substituting this in equation (3) :
PROBLEMS INVOLVING MUTUAL INDUCTANCE 39
or
5j hi
dt r Lr l + Lir Lr l + LIT
Thus > -
(9)
To determine the integration constant C, it would be a mistake
to assume that the current i is zero when t = 0. All that is
known is that the combined coil can not be surrounded instan-
taneously by a flux it takes some time to produce or alter a
magnetic field, because a transfer of energy is involved. It is
possible that currents will flow the very first instant, currents
which produce m.m.fs. of equal magnitude but in opposite direc-
tion. One particular case of this would be where the currents
were zero, but this is not a general solution.
What is known, then, is that no flux will exist the first instant.
Thus the m.m.fs. must be equal and opposite, and since the cross-
section of the magnetic flux and the direction of the turns are
assumed the same in both coils, it follows that for
or N . . IL
* i = -Ni iu "Ste
Substituting this value in equation (7)
LiE +
(ID
Mr,
or L\EJ "
Lri +
for t = 0.
i E
' r ~
_ MEir
. . \j
r(Lri
ME
77
r(L
Similarly ii is found to be
. E MEjr + LErt , t , 10 ,
I = --- 77 = - r- 6 Lri+Lir (12)
r r(Lr l + Lir)
(13)
i r 1 (Lr 1 +L 1 r)
Problem No. 7. Prove by complete calculation that if the
terminals of the second coil are reversed the following are the
equations of the currents
40
ELECTRICAL ENGINEERING
E LEr l -
I =
r r(Lr l
r- 6 Ln+Lir
Lri+Lir
(14)
(15)
In the case that the two coils are excited from the same direct-
current busbars when E = E\ the equations become :
For coils wound in the same direction:
E
+
i+Lir
m
Ln+Lir
For coils wound in opposite direction:
Lri Mr m t
ll
1
Lri
e Ln+Li
100
CO
Starting Currents
Two Coils of Perfect Mutual Inductance
A - Coil No.l
B - Coil No.2
D - Coil No'2 ) Wound PP site Directions
Wound in same Direction
(16)
(17)
(18)
(19)
FIG. 16.
In Fig. 16 are given four curves showing the currents in two such
coils of perfect mutual inductance, having the following constants :
r = 0.10
ri = 0.50
L = 2.5
Li = 10.0
E = E l = 10 volts.
PROBLEMS INVOLVING MUTUAL INDUCTANCE 41
It is assumed that they are connected in parallel to the same
source of direct current at a constant potential of 10 volts. The
full-drawn curves correspond to the condition in which the turns
are in the same direction; the dotted curves to that in which the
turns are in opposite directions. It is well to verify these curves
by calculation. It is of interest to note from the full-drawn
curves that, while the two coils are
connected to the same source of con-
stant potential, during the first few
seconds the currents actually flow in
opposite direction. The second coil
having twice as many turns as the
first, and therefore a smaller final
value of current, has a current of
negative value at the first instant of
one-half the magnitude of the current in the first coil. Eventu-
ally the currents become positive and are proportional inversely
as the ohmic resistances.
It is of interest to deduce the equations of the currents in the
two coils when the first is connected to a source of constant poten-
tial, and the second is short-circuited upon itself, as shown dia-
grammatically in Fig. 17.
FIG. 17.
CO
40
20
Starting Current
Two Coils of Perfect Mutual Inductance
A - Coil No.l
B - Coil No.2
FIG. 18.
Prove that with the coils wound in the same direction:
11- - Lri
r
MEr
Ln +Lir
t
TTr* 1
Ln+Lir
(20)
(21)
42
ELECTRICAL ENGINEERING
In Fig. 18, which gives the values of the currents, it is of interest
to note that the current in the second coil, under this condition,
remains negative and approaches a value of zero. The initial
values of the currents are twice as great as before. Thus the
impedance is greatly reduced, as would be expected by the pres-
ence of the short-circuited winding.
As a further illustration consider:
Two similar coils having perfect mutual induction and calcu-
late the currents in the two coils when a sine wave of e.m.f. is
impressed upon one coil while the other is short-circuited.
Referring to Fig. 19
t , r, x
The equations evidently become:
di di\
di
FIG. 19.
E . di dii E
- sin e and ^- - ^ = cos
(22)
Ex
~
and
dii
E
dB 2x Zl = ~ 2~r COS
i = - e
~ cos d0 -f- Ce
- ~ - cos (6 - ?) + Ce L 9
/JQ I
(23)
where
Z =
2 + (2x) 2 .
The condition determining the integration constant is that
when the switch is closed no flux exists, thus i ii.
Let then the switch be closed when = 0^
Thus from (22)
and from (23)
E
- 2ii = sin 0!
-yr ~ cos (0! - ^) +
JQ I
(24)
(25)
PROBLEMS INVOLVING MUTUAL INDUCTANCE 43
From (24) and (25)
= ^ X ^ [f5 COS (Sl ~~ ^ ~jr sin
' >
""" -T 1 ]
The transient term disappears when
x sin 0i
y- COS (01 (p) =
Expanding and substituting it is readily seen that this occurs
when tan 0i = -, that is when 0i = <p.
The transient is a maximum when:
d [x . sin
-FT COS (0i to)
7*
that is when tan 0i = ^- or
0i - <? - 90.
Limiting Cases. (a) r small compared with 2x. :.<p = 90
and cos (0 <p) = sin and cos <p = 0.
The transient effect is greatest when
# . E
.*. ii = -- sin and % = ^- sin (27)
(6) r large compared with 2# when
E E
il == ~ir and Z< == ~ 2r (28)
When dealing with commercial problems involving mutual
inductance it is never possible except as a first approximation to
assume perfect mutual inductance M 2 is always smaller than LL.
In that case the solution demands differential equations of the
second or even higher order.
Fortunately, however, the solutions of these equations are as
a rule simple.
The majority come under the class of linear differential equa-
tions with constant coefficient or they are of the types given
below. 1
1 For a complete discussion see '*A Course in Mathematics,"
WOODS AND BAILEY, vol. II; ''Differential Equations," JOHNSON; "Differ-
ential Equations," MURRAY, or indeed almost any book on differential
equations.
44 ELECTRICAL ENGINEERING
d z y
First. -r^ 2 = f(x) = an expression that is a function of x.
Second. ~r~ 2 = fix, -j-j = an expression that is a function of x
and the first derivative of y with
respect to x.
Third. ^f 2 =
Fourth. ~T\ = f(y).
Fifth. Linear differential equations -r 2 + a -j- + by = X.
d^v C13&
First. -j\ = f(x) = a function of the independent variable
only. By integration we get:
<fy = |/(a:)da; + d
dx J
and y = J ' J*f(x)dx 2 + CiX + C 2 .
This is equally true for ~[ = f(x)
Let dy d*y rfp
da; P ' da; 2 " dx
.' .X -j- = f(x, p) which is of the first. order of p and x.
If we can find p from this equation then we can find y because
dy
dx = V'
d z y
"** S-/(*Z)
Let dy . d 2 y _ dp _ dp d/y _ dp
dx ~~ ^ ' dx 2 ~ dx ~ dy dx ~ ^dy
. ' . X p -7- = f(y, p) which equation is of the first order of p and y.
If we can find p from this equation we can find y } since -^ = p.
Fourth.
PROBLEMS INVOLVING MUTUAL INDUCTANCE 45
dii
multiply both sides by 2 -7- dx
Integrating + C
== =
^fWv)dy + C
+ c
i. Linear differential equations of the second order with
constant coefficients. If the coefficients are not constant, the
solution is quite complicated. It is therefore omitted here. As
a matter of fact in almost all problems the coefficients are con-
stants.
where a and b are constants and X is either a function of z or a
constant or zero.
For convenience the symbol D which represents the differential
operator -r- is introduced.
D means one operation of differentiation in respect to x t D 2 two
operations of differentiation in respect to x.
Thus equation (29) can be written
D 2 y + aDy + by = X or
(D 2 + aD + b)y = X (30)
from this follows that
= X
y ~ D 2 + aD + 6'
Obviously this does not mean an ordinary fraction but is simply
a symbol to express the solution of equation (29). The denom-
inator on the right-hand side is not a number but is an operator
which has a definite meaning, so for instance (D 2 + aD + b)e mx
46 ELECTRICAL ENGINEERING
really is the sum of three terms the first of which for instance is
obtained by differentiating t mx twice with respect to x, the second
once with respect to x, etc.,
The expression is:
raV* + ame mx + be mx
Equation (30) can, as will be shown, be written
(D - m l )(D - m^y = X (31)
(D m 2 )y means -j- - m z y.
Thus
d /dy \ (dy
(D - mi)(D - m 2 )y = ~
d z y dy dy
dx-*- m *dx- m *dx
(32)
(Incidentally it is seen that the same result would be obtained by
the simple multiplication of (31) treating D as an ordinary quan-
tity). Comparing equations (29) and (32) it is seen that
- (mi + mz) = a
and mim 2 = b.
From these equations mi and m 2 can be expressed in terms of
a and b which are known.
A slight consideration will show that mi and mz are also the
roots of the so-called auxiliary equation m 2 + am + 6 = 0, which
corresponds to the so-called complementary function
and the auxiliary equation corresponding thereto
m 2 + am + b = O
\ l~2
.*. m = h * \ - - h
2 - \4
and . m- , (33)
PROBLEMS INVOLVING MUTUAL INDUCTANCE 47
The rule then is to solve the so-called auxiliary equation
m 2 + am + 6 =
and find the values of mi and ra 2
Then write,
(D - m l )(D - m 2 )y = X.
To solve for y
Let u = (D - m 2 )y (35)
(D - mi}u = X
and du
dx
U = 1 xy c - 1 x Xdx _|_ ^m^
From (35)
An
Cm ~x XT*
l l -A i
y = ee-
Instance: d 2 y dy
m 2 - 3m + 2 =
m = 1.5 V2.25 - 2 = 1.5 + 0.5
m 2 = 1
(Z) - 1)(Z> - 2)y = cosx
(D \}u = cos x
du
-j -- u = cos x
dx
u = t x j*e~ x cos x dx + Cie x
but
e x j^e~ x cos x dx = % (sin x cos #) by simple integration
(D - 2)y = y^ (sin x - cos x) + CV = X l
dy (sin s - cos x)
dx ~ 2y ~ ~^T Cie = Xl
y = S*fe-**Xidx + C 2 e 2 *.
The three integrals involved are readily solved and the result
s
cos a; snx
- - 3 -
In many cases a simpler integration is obtained by the following
method which involves the breaking up of a fraction in partial
fractions.
48 ELECTRICAL ENGINEERING
It is known from algebra that if the equation given below
holds for all values of x then the coefficients of the like powers of
x are equal.
a Q x n + aix n ~ l + a n = b x n -f- bix n ~ l + b n
Thus a = b
0,2 = z
Equation -j 2 + a -5 h fa/ = ^ can be written
(D - mOCD - m,)?, = X or =
But it is known from algebra that the fraction
1 A B
(D - mi) (D - w 2 ) D - mi D - m 2
where A and 5 are to be determined.
_ 1_ = A(D -m 2 ) +B(D -mi)
(D - mi) (D - m 2 ) ~ (D - mi) (D - m 2 )
This equation shall hold for all values of D.
Rearranging the equation we get:
1 D(A + B) - (Am 2 +
(D - mi) (D - m 2 ) ~ (D - mi) (D - m 2 )
On account of the identity the coefficient for D must be zero and
the constant terms must be equal, thus
and Ani2 + Bm\ = 1
B = - - - and A =
m 2 mi m 2 mi
m 2 mi D mi D m 2
_JLJr_^__ i 1
mi m 2 LZ> mi D m 2 J
Let y = u -\- v
u = ^^ 1 D^ 1 (36)
and 1 X
~
PROBLEMS INVOLVING MUTUAL INDUCTANCE 49
Equation (36) written out is
du _ X
dx
u =
Similarly
v = - - -ft-** Xdx
The general solution is thus
y = Cie m i* + C 2 e m 2* + -
,m 2 x
" <-*** Xdx.
mi m z
When X is a constant = K
The solution evidently becomes:
^ = C^ mx + Cuc w * + y-
When X is zero the equation is called the complementary func
tion and its solution is found by making K = in the above.
The solution of the equation,
is y = Cie mlx + C 2 e w 2 a; .
Before leaving the subject it is necessary to discuss the values
of mi and m 2 which involve a square root in (33) and (34) which
might be real, imaginary or zero.
a 2
(a) The square root is real, that is -j > 6.
We have shown then that mi and mz depend upon the auxiliary
d^v du
equation -j 2 + a -r- + by = and that the solution of this
equation is,
OL'
(b) the square root is imaginary, that is b > -
50 ELECTRICAL ENGINEERING
"fVlOTl / & . -4, / O \ / d . >* /. Ct \
,, = rJ- + '\*-i> -4-c*(~2 -'\ 6 --4>
where a = A/6 - -j'
But e' ox = cos ax + j sin ax
. ' . CV * = Ci cos ax -|- Cij sin ax
. * . C 2 e~ ;oa: = C 2 cos ax Czj sin ax
y = ~ a z [(d + C 2 ) + cos ax (Ci - C 2 ) sin axj].
In order that y shall be real it is necessary that C\ + C 2 and
j (Ci C 2 ) shall be real, in other words, Ci C 2 must be im-
aginary.
Let A = Ci + C 2 and B = j(d - C 2 )
y = t~ 2 [A cos ax + B sin ax] (38)
= Cie~f sin (ax + C 2 ) (39)
where
and
tan C 2 = j} m
(c) the square root is zero, that is -j- = 6, or m\ = m 2
This is not a complete solution of the complementary function
because we have only one integration constant.
The equation can obviously be written.
(D - mi) 2 ?/ =
= (D - mi}(D - mi)y = O (40)
Let
(D - m,}y = v (41)
Then 40 becomes
(D - mi)v =
or
dv
-; m\v = O
ax
PROBLEMS INVOLVING MUTUAL INDUCTANCE 51
From (41)
(D - mi)y =
or
f x - mi y =
y = e m l a: [/V w l*C
or
i* [C lX + C 2 ] (42)
Two Coils Having Resistance, Self -inductance and Imperfect
Mutual Inductance Constant Impressed e.m.f. Let the constant
e.m.f. impressed on the first coil be E and that on the second
coil EI. Let their resistances and inductances be respectively
r, TI and L, LI and let M < LLj.
It follows that
E = ri + L -j \- M -r^ (43)
and dii , ,,di
E 1 = ^ 1 rl + LI -j h M -r- (44)
Differentiate (44).
From (43) is found
Substitute (46) and (47) in (45) and arrange the equation
with reference to the derivatives.
/72/ J 7 '
/. ^ (LLi - M 2 ) + -ft (Lir + Lri) + irr, = Er, (48)
Or dH vLir + Lri~\ di
dt 2 ' LLLi - M 2 J dt ' LLi - M 2
= zzr--ir 2 (49)
E7
Similarly ii = ~ + BI*** + Brf- (51)
4
52 ELECTRICAL ENGINEERING
where mi and m% are the roots of the equation.
Lir -f Lri rri
m + LL^W* m + LL^-W -
(L,r + Ln) + VW-LrO' + Wrr,
It is evident, from the factors under the square root sign, that
in this case the two roots are real.
Thus the solution is
L\r -f- Lri -
mi=
2(LLi -
,.
and Lir + Lri , ,
2
The integration constants AI, A 2 and J3i, 7? 2 are readily de-
termined, since in this case (where the mutual inductance is not
perfect) currents can not flow without producing some flux, and
thus, since the establishment of flux requires time, the currents
can not appear instantaneously.
Therefore at t = 0, i = ii = 0.
Referring to (50) and denoting the final current
(where / == ^) (56)
by = 7 + ^,4-^2, or A 2 = - (A l + 7)
we get i = I + Aiei' - (Ai + I)e m * (57)
and ^ = I I + B^" 1 * - (B l + 7i)e^ (58)
These equations still contain the two unknown quantities
A 2 and B 2 . To determine them, multiply (43) by LI and (44)
by - M.
L,E = L<ir + Lla + L,M (59)
- ML 1 - M 2
di
- ME! = Liir - Mi^ + (LL - M 2 ) ^
~ LlE + (LLl ~
The value of -r is found by differentiating (57) and the value of
PROBLEMS INVOLVING MUTUAL INDUCTANCE 53
i directly from equation (57). Substituting these values in (61)
and remembering that for t = 0, i\ = 0, the integration constant
AI is found to be:
Ltf - ME, + w,/(LLi - M 2 )
- - 2
.'. A 2 = -- (A! + /) (63)
Similarly n LE l - ME + m a /i(LLi - M 2 )
(mi - m 2 ) (LLi - M 2 )
B 2 = - (Bi + /O (65)
The equations of the currents are found by substituting these
constants in equations (57) and (58). They are so long and
cumbersome, however, that it seems unnecessary to insert them
in this text.
Assume that the two coils are identical and wound in the
same direction, and are connected across the same constant
potential busbars. What are the equations of the currents?
mi and m z are found from equations (54) and (55).
m > = - L^M (66)
m z = - L T _ M (67)
AI = #1 is found from equation (62) by substituting these values.
E _
thus A 2 = B 2 = 0.
Referring to equation (57) :
(68)
This shows that the mutual inductance acts as self-inductance.
It is also evident that if the two coils are wound in opposite
directions the circuit is almost non-inductive. It would be non-
inductive if M = L; that is, with perfect mutual inductance.
It is of particular interest to study the relations of the currents
in two such identical windings inductively related when one is
supplied with current from a source of constant potential and
the other is short-circuited.
54 ELECTRICAL ENGINEERING
It is well to deduce the equations from the two general ex-
pressions:
and . T dii . , f di
= t,r, + L, ^ + M %
However, it is evident that having once determined the
general equations (57), (58), (62), (63), (64) and (65), it is possible
to give the equations for the case in consideration by putting
#1 = 0;
that is,
T _ I _
= n =
i = I + Aie"* -. (Ai + 7)e m (69)
and n = Bi m i* - #ie w 2< = B! ( i< - *) (70)
Referring to equation (62) and substituting equations (66)
and (67)
and
(71)
Referring to equation (64) and making similar substitutions
we get
M' J
(72)
It is evident that these equations do not lend themselves to
the limiting condition M = L, on account of the assumption
made in determining the integration constants; that is, that leak-
age flux exists between the two coils. To get these values,
equations (20) and (21) should be used.
In Fig. 20 are given some very interesting curves which show
how the current in the short-circuited winding depends upon the
leakage flux between the windings. These curves represent the
conditions of two identical coils having a resistance of 0.10 ohm
PROBLEMS INVOLVING MUTUAL INDUCTANCE 55
and an inductance of 2.5 henrys, placed at various distances
apart so that the mutual inductance is M = L in curve a,
M = 0.9L in curve b, M = 0.7L in curve C, M = 0.5L in
curve d, and M = 0.1L in curve e.
-4
2
-1
10 20 30 40 50
70 80 90 100 110 120 130 140 150 160 170 180
Time in Seconds
FlG. 20.
One of the coils is connected to a source of constant potential,
e = 1 volt, while the other is short-circuited.
Prove that the time for the maximum value of the secondary
current is:
Z^-M 2 L + M
Iog L - M'
CHAPTER III
10000
8000
CIRCUITS OF RESISTANCE AND VARIABLE INDUCTANCE
In the discussions given so far it has been assumed that the
inductance L has been constant. In almost all cases of interest
to engineers this is, however, not the case because almost all
magnetic circuits contain iron, and the permeability of iron is not
constant but depends upon
the magnetization. In other
words the flux produced by a
given current is not pro-
portional to the current.
Fig. 21 gives the saturation
curve of an entirely closed
iron magnetic circuit, as
shown in Fig. 22. It is the
familiar hysteresis loop, which
shows how the magnetism
lags behind the m.m.f. pro-
ducing it.
This particular sample has
a remnant magnetism of 7600
lines per cm. 2 , so that this
density corresponds to an
exciting current of amp.
The maximum density is
10,000, which corresponds to
an exciting current of 4.5
FIG. 21. amp. If, after the maximum
density is reached, the current is gradually reduced the rela-
tion between existing current and density is found in curve a.
The flux does not disappear until the current is 2.6 amp. in op-
posite direction to the original 4.5 amp.
If, instead of being entirely of iron, the magnetic circuit con-
sisted partly of air circuit and partly iron (Fig. 23), the influence
of the air circuit would as a rule be so much greater than that of
56
-8000
-10000
CIRCUITS OF RESISTANCE AND INDUCTANCE 57
FIG. 22.
FIG. 23.
the iron that the shape of the saturation curve would become
materially modified. Thus the saturation curve of a dynamo,
having a magnetic circuit
largely of iron but also of at
least a small air gap, can be
represented by a set of curves
similar to those in Fig. 24.
If the air circuit is very small
the two curves corresponding
to a and b in Fig. 21 can be observed. If the gap is reasonably
large the two curves merge into one as shown in the dotted line.
FROLICH evolved an equation of such a saturation curve for a
magnetic circuit consisting partly of iron and partly of an air
gap; which, modified by KENNELLY, can be written thus
ki
: T+fo#
where <f> is the flux corresponding to an exciting current of i amp.
4 6
Current in Amperes
FIG. 24.
If the number of turns of the exciting winding is known then
the inductance for any particular value of current i can be
determined. It is
N(f>
-. where N = number of turns.
:.L =
A T /clO- 8
i + kj
The general equation thus becomes:
d , ^ di
r
NklQr
1 + kii
dt
iNkkMr* di
(1 - kii) 2 dt
(73)
58 ELECTRICAL ENGINEERING
To solve this equation the variables are separated
di
or / *_ _ 10^
I 7 *\ 9 "\7*7
r
J (e-i
The integral is solved by breaking up the fraction
into three fractions
C
6 -
~ h i i T ~h
The constants can be readily found and the integration carried
out without the slightest difficulty.
A becomes -=
a 2
B becomes r
a 2
and C becomes
a
The result is:
Nk rr (1
rr l + /dz
loe; e : h
La e ir
10 8 a La " e - ir 1 + M J
where a = r -\- eki.
In this case a simple solution which is quite accurate is obtained
if the last term in (73) be omitted since an inspection will show
it to be small as compared with the second term.
We have then
di _ . JVMQ-8 di
Separating the variables we get:
di W*t
? (74)
(e - ir) (1 + fet) Nk
again 1 A B
A +Be + i(Aki - Br)
(e - ir) (I +
Since the left-hand member does not contain the unknown i
and since the constant term is 1, we get
A + Be = 1
Aki = Br.
CIRCUITS OF RESISTANCE AND INDUCTANCE 59
Thus A = _ _J1_ and B = fei
r + efci r + eki
The intergral is thus broken up into two simple integrals
r_ <** r ^ _ , C-
J (e- ir) (1 + kj) J (r + efci) (e - ir) ^ J (
r + *i)(H-
^
i fci
fci
r + eki * e ir
If it is desired to find the value of i at any time after the circuit
is closed then i = for t =
(75)
e ^r e
The curve connecting z and t can conveniently be obtained by
assuming different values of t and solving for the left-hand mem-
ber of the equation. The value of i can then, of course, be easily
determined.
Curve a in Fig. 25 shows the relation between the exciting cur-
rent and the time for the field current of a direct-current generator
having the following constants:
e = 100 volts = voltage impressed on the field.
r = 100 ohms = field resistance.
TV = 4000 = total number of field-turns in series.
01 = 1 megaline with 1 amp. excitation.
02 = 0.6 megaline with 0.5 amp. excitation.
From FROLICH'S equation follows then:
1 = and 0.6 = ' k = L5
i .i
ifci = 0.5.
It is instructive to verify this curve.
Curve b gives the corresponding values if the saturation curve
had been a straight line, i.e., if the flux were 1 megaline for 1 amp.
excitation, and 0.5 megaline for 0.5 amp. excitation.
60 ELECTRICAL ENGINEERING
In that case the inductance L would be constant and would be
4000 X 1,000,000
L= 10' X 1 "
and the equation e = ir + L -r would be 100 = lOOf + 40 -^
in which case
i=l- e- 2 - 5 ' (76)
^
f *
7
s
/
^
^ ^
/
.0
/
' /
/a
7
/
/
/
g
.
/
2J -6
7
/
3
7
g
/
/
,
//
3 .4
X
;
/
W
A
//
K
.2
j
1
/
/
n
f
1.2
Time in Seconds
FIG. 25.
2.0
It is interesting to see that the field flux builds up considerably
slower than would have been the case if L had been constant.
The reason for this is that, while at the final value i = 1, the
inductance is the same in both cases, for all smaller values of
current the inductance is greater because the flux is greater
for the same current. When the saturation can not be expressed
by a simple equation, there is no better method than to calculate
step by step.
Let Fig. 26 represent such a saturation curve. Determine
the rise of current when a constant impressed e.m.f. of 100 volts
is impressed on a coil 4000 turns having a resistance of 100 ohms.
Thus
N
CIRCUITS OF RESISTANCE AND INDUCTANCE 61
Using differences instead of differentials:
N A0
6:=ir + WAi
or 10 8 (e - ir) M
= 0.25 X 10 7 (1 - i
(77)
(78)
1.4
1.2
| 1-0
I * 8
c
5 .4
.2
.4 .8 1.2 1.6
Current in Amperes
FIG. 26.
1.U
.9
.8
.7
a
Q>
3.6
___
^~~ f
-^
z~-~
"
^
^
"'
'
^
-<^
'"*
^.
/
/
'/
/,
'
/>
'
/,
/+
/ f
/
5 .
p .4
/
/
/
/
w
.3
.2
/,/
//
/}
/
-
/
/
}
'/
"0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 1.1 1.2 1.
Time in Seconds
FIG. 27.
If the values of current are determined every one tenth of a
second A = 0.10.
.'. A<j> = 0.25 X 10 6 (1 - i).
The actual flux at any time is of course SA< the corresponding
relation as obtained by the use of the differential equation.
62
ELECTRICAL ENGINEERING
The method of calculating is illustrated in the table given below
and the results plotted in full-drawn lines in Fig. 27.
First approximation
Second approximation
t
A*>
SAy>
i>
i - 1
A?
SA<p
t
0.10
0.25
0.25
0.182
0.818
0.205
0.205
0.146
0.20| 0.205
0.41
0.316
0.684
0.171
0.376
0.286
0.30; 0.171
0.547
0.447
0.553
0.138
0.514
0.413
0.40
0.138
0.652
0.555
0.445
0.111
0.625
0.525
0.50
The starting of an alternating current in an inductive circuit
containing iron is of special interest since almost all electrical
devices used with alternating current have iron. Unfortunately
the equations are very complex and are not subject to solution,
even with long and elaborate treatment by series. Even in the
simplest case, when the saturation curve can be represented by
FROLICH'S equation, an accurate solution is not possible, although
to be sure it is not difficult to bring the relation into the form of a
linear differential equation. The problem in that case can be
solved as far as a mathematician is interested; but the engineer,
and indeed the mathematician, can not use the solution for any
practical purpose.
To illustrate this assume that an alternating-current e.m.f. is
impressed on a magnetic circuit having N effective turns per
ki
phase, and a saturation curve represented by <j> = .. , , .
Assume that the resistance of the winding is r ohms, and that the
impressed e.m.f. is a sine wave. At any instant the following
relation exists:
E sin ut = ir + L -r. -\- i -r:'
dt at
But L = j-fTjp where N = number of turns
1 i is read off the saturation curve or since in this particular case a satura-
tion curve which can be expressed by FROLICH'S equation has been assumed
for the relation i =
1.5
1.5
1 + 0.5*
CIRCUITS OF RESISTANCE AND INDUCTANCE 63
again neglecting the last term.
, di
. . E sin ut = ir + - : : . -7-
1 + kii dt
where a = JVfc 10~ 8
substituting for 1 -f- k\i = ;
y
y = e~ ~T d'
r r + ki E sin o>< dt
J 1 1 i
(8o)
dto* 1 d
2/i i?/
r r a dy
k\y ki kiy dt
dy
kiy E sin ut = r ry a ~~
dy (r + fci E sin cup _ r_
dt+ y ~ a 'a
r + ki E sin ut
--J- .
i E sin w< d< "]
(82)
the solution for i is found by a simple substitution.
Unfortunately, however, the solution is not in a simple form
and can not be simplified; and thus, while mathematically the
problem is solved, practically it is unsolved. In cases like this it is
necessary to proceed by a step-by-step method.
Consider then the case of an alternating-current impressed
upon a magnetic structure having a saturation curve of any
shape. Let it, for instance, be expressed by
ki
= r+i^i
The following relation exists at any instant :
E sin orf = ir + g (83)
64 ELECTRICAL ENGINEERING
where r is the resistance
N
/. E sin cot eft = ir dt + j^ d< (84)
PI V 10 8 7> ID 8
*' d0 = sin ^ d * ~ "" d * (85)
If, with full-load effective current I e the resistance drop is p per
cent, of the rated voltage, then
7 e r = j^Q r=, and for any other value of / as i, ir = ^ ^- (86)
' 2
or since , cos coi ,
d = sin cot dt
E X 10 8 r , cos co pidt
d * = ~N L~ ~u TooT T ~
It is usually more convenient in alternating-current problems to
introduce 6, the phase angle, instead of co.
In that case = ut and dt =
Referring to, (127)
_ E X 10 8 pin 6 dB _ pidB 1
~ ~~N~ co " I e 100 X/2COJ
OT
In most problems E, N, $ max and the frequency are known,
so that numerical values can directly be substituted in the above
equation. Since, however, there is a relation between them, one
or more of the quantities may be unknown.
The most general aspect of the problem is given by eliminating
the numerical value of the impressed voltage, turns and fre-
quency, and specifying the maximum value of the flux : <j> maximum
= $.
We have the following well-known relation between $, N, E
and co.
2irf 3>N
10 8 10 8
.'. - (88)
CIRCUITS OF RESISTANCE AND INDUCTANCE 65
When in an inductive circuit the resistance is very small compared
with the reactance so that the impressed and counter e.m.f. are
equal numerically.
Substituting this value in (87)
1.0
/
NS
/
\
/
\
/
\
\
'
/
\
/
f
^
/
v
\
1
l\
\
/
/
k
^
1
I
\
\
\
1
/
\
7
/
i
\
\
\
1
/
/
/
/
\
\
/
/
\
\
/
/
\
\
7
/
i
\
\
\
\
j
c
)
$
3
1(
<?
]
50
ll
2
2
270
I
'i
irr
e
\
\_
t *
1
\
\
1
V
[
I
\
;
L
\
1
V
\
/
,-
/
\ L
\
i
\
I
\
/
/
\
/
/
s
-1.0
/
\
^. ^
FIG. 28.
Substituting differences
A0, A cos and A0 instead of differentials, the equation becomes:
66
ELECTRICAL ENGINEERING
If the ratio between flux, current and phase angle is deter-
mined every 10 then A0 = 10 = 0.175 radians.
cos 6 + 0.00124
pj
\
Numerical Example. Determine by "step-by-step" method
the current in an iron-clad inductive circuit when an alternating-
current e.m.f. is impressed thereon. Assume that the saturation
\
\
ssaodtu y ut ^uooanQ
FIG. 29
curve of the magnetic circuit is represented by Fig. 26 and
equation :
<t> = i ,'HKV megalines.
Assume that before the switch is closed the remanent magnetism
is zero as is practically the case when the magnetic circuit con-
tains an air gap. Assume further that under normal conditions
CIRCUITS OF RESISTANCE AND INDUCTANCE 67
of operation the maximum flux is 1.4 megalines, that normal
effective current is 1.7 amp., and that the resistance drop is
3.91 per cent. Then
A</> = -- 1.4 [A cos + 0.00286i]
- 1.4 A cos - 0.004?;
The total flux is obviously ZA<. If the switch is closed when
the e.m.f. passes through zero and is rising, the normal flux at
that instant would be a maximum in the negative direction as
shown in Fig. 29. As it has been assumed that the flux really is
zero it is evident that there is a transient stage in the mag-
netization before permanent condition is reached. It is evident
also that if the switch were closed when the e.m.f. was a maxi-
mum no transient condition would result, because the condition
then demands zero flux, and the flux is assumed to be zero.
In the numerical example it is assumed that the switch is closed
when the e.m.f. wave passes through zero.
The method of using the above equation is best shown by the
table given below.
No. 1
No. 2
No. 3
No. 4
No. 5
No. 6
No. 7
No. 8
No. 9
9
Cos 6
A cos
- 1.4 A cos
SA<
i
0.004i
SA<
i
1.0
10
0.98
-0.02
0.028
0.028
0.01884
0.000075
0.027925
0.01875
20
0.94
-0.04
0.056
0.0839
0.0576
0.00023
0.08367
0.0573
30
0.87
-0.07
0.098
0.18167
0.1289
0.000516
0.1811
0.1288
40
0.77
-0.10
0.14
0.32115
0.2398
0.000959
0.3202
0.2395
50
0.64
-0.13
0.182
0.5020
0.402
0.001608
. 5004
0.4010
60
0.50
-0.14
0.196
0.6960
0.604
0.002416
. 6935
. 6020
70
0.34
-0.16
0.224
0.9175
0.881
0.00352
0.9139
0.8740
80
0.17
-0.17
0.238
1.1519
1.247
0.00497
1.1469
1.238
90
0.00
-0.17
0.238
1.385
1.712
0.00684
1.3781
1.699
Column No. 1, phase angle; No. 2, the cosine of the phase angle; No. 3, difference in the
value of the cosine between two successive steps, for instance cos 20 cos 10; No. 4 is
self-explanatory; No. 5, first approximation of the flux (sum of No. 8 of the preceding line
and No. 4 on the line under consideration") ; No. 6, current as obtained from the saturation
curve or the equation if such is given; No. 7, ohmic drop; No. 8, second approximation to
the flux which takes into consideration the ohmic drop (the algebraic sum of No. 5 and
No. 7) ; No. 9, current corresponding to the last approximation of the flux column, No. 8.
CHAPTER IV
CHARACTERISTICS OF CONDENSERS
The charge q of a condenser is proportional to the voltage;
or q = Ce, where C is the capacity the value of which depends
upon the mechanical construction, dimensions, etc., of the con-
denser, and 6 is the voltage.
The charge q is expressed in coulombs or ampere-seconds.
Thus the charge dq given in a time dt when the current is i amp.
is:
dq = idt.
The capacity is expressed in farads, a very large unit; so large
indeed that in actual practice it is never used. The capacities
of condensers are almost always given in microfarads, that is,
in a unit which is one-millionth of a farad. Nevertheless,
in all formulae involving capacity, C stands for farads, not
microfarads (m-f.) unless stated to the contrary.
To give an idea of the capacity of condensers used in engi-
neering, it may be of interest to know that the ordinary paraffine
paper and tinfoil 500-volt blocks of the size of the average
text-book have a capacity from 1 to 2 m-f. In a high-potential
transmission line the capacity of one wire against neutral is
about 0.016 m-f. per mile. The capacity of underground
cables is relatively high. Depending upon the voltage and
type of cable, etc., it must obviously vary much. It is usually
less than 2 m-f. per mile and more than }/{ Q m-f. The capacity
of an ordinary Ley den jar is extremely small a very small
fraction of a microfarad.
The fundamental equations for the condenser are as stated
above
q = Ce (1)
and dq = idt (2)
From these follow: q / Q \
= C
dq = Cde (4)
and . = | (5)
68
CHARACTERISTICS OF CONDENSERS 69
Substituting (4) in (2)
dp
Cde = idl ori = C ~ (6)
(it
or e, the voltage across the condenser = -~J*idt (7)
The rate of energy supply or power is ei
or from (6),
Cde de
or from (3) and (9), . _ ? . _ q_ dq
~ C ^ " C dt
The energy stored in a condenser, which is the same as that
required to charge a condenser to a voltage E or to a final
charge Q, is therefore the rate of energy multiplied by the time.
It is:
f
Jo
Jo
e
o2
/~n , /
or
cedt = C - C
2C
Equations (10) and (11) are obviously identical, since at any
instant
q = Ce thus for e = E when q = Q
Q = CE, which, substituted in (11), gives
CE 2
2C 2
As in the case of inductance, the calculation of the capacity
of any but the simplest circuits is difficult. It will be dealt
with in later chapters.
Of particular interest to engineers,
however, are a few simple forms of con-
densers, the approximate capacity of
which are given by equations which are ^ IG 39.
well known.
Thus the capacity between parallel plates, Fig. 30 is:
c = ' in microfarads
70
ELECTRICAL ENGINEERING
where K, the specific inductive capacity is approximately 1 for
air, 2 for paraffin paper, 3 for rubber, 5 for mica and 6 for glass.
A, the effective area is given in sq. cm. and d, the thickness
of the dielectric, in centimeters.
The capacity between concentric conductors (Fig. 31) is:
0.0386LK
C
log
in microfarads
io
where the length I is given in miles of cable, K
is the specific inductive capacity, D the inside
diameter of the outside conductor, and d the
diameter of the inside conductor. This is the
capacity between the conductors, not the
capacity to neutral or ground. The capacity of
one conductor 1 mile long to neutral is twice as great.
The capacity between transmission lines is:
0.0386?
FIG. 31.
C =
in microfarads
where I is expressed in miles and the capacity is that of one line
against neutral. D is the distance between wires, center to center,
and r the radius of wire. The charging current is thus
. 2irfCe
10 6
where e is one-half of the line voltage in the single-phase system
and 58 per cent, thereof in the three-phase system.
Circuits Containing Concentrated Capacity and Resistance
Consider at first the case of a constant e.m.f. E impressed
upon a circuit of resistance r and
capacity C, Fig. 32. After the cir-
cuit is established a current flows
and energy is delivered to the re-
sistance and the condenser. In the
resistance heat is developed and
in the condenser an electrostatic
field is produced. The energy given by the source of supply of
power is fEidt. The energy supplied to the resistance is
fftrdt
FIG. 32.
CHARACTERISTICS OF CONDENSERS 71
and the energy supplied to the condenser
Thus
fEidt = fi*rdt + Cq ~? . (12)
Eidt = i*rdt + q ~l
Ei = i z r + - Jj which is the power equation (13)
and ,-, . q dq
E = * r + Ci dt
or substituting for dq = idt
E = ir + -~ which is the voltage equation (14)
C
Obviously the voltage equation could have been derived directly,
since ir is the e.m.f. consumed by the resistance and ~ the voltage
across the condenser.
The condenser voltage is thus ei = E ir; but
or de, I = E
dt ' Cr ? " * r
Referring to equation^! = Ae c r -\- E (15)
where A is the integration constant: The current is readily
CA _JL< A _!,
c cv' -e c, (16)
r . ei
found, since % = C ~r
The charge g is = Cei - CAe~cr' + ^C. (17)
Special cases:
(a) Condenser charge.
At time t = 0, e\ = 0.
.'. referring to equation (15), A + E .'. A = E
(18)
Referring to equation (16)
t---&' (19)
72
ELECTRICAL ENGINEERING
Referring to equation (17)
EC\l -
(6) Condenser discharge.
In this case the impressed voltage E
Referring to equation (15)
i
and e = A
i .
(20)
for t = 0, ei = e .
= At cr l
(21)
1 that is in opposite direction to charging current (22)
2000 Ohms.
""* -WV\A-
FIG. 33.
q = Ce e~cr (23)
2oom.f. Referring to the e.m.f. of the con-
denser rather than to the impressed*
e.m.f., the current becomes positive
since the discharge current
,-.-- -C^
dt
dt
_
C Cr
(24)
Cr * " ' r - (25)
In order fully to understand the action of condensers it is not
sufficient to follow the equations given above, but it is essential
and indeed necessary to figure a number of numerical examples.
h
2|
||.05
S.2.04
JU
If 02
jfiOl
3
oo
.1 .2 3 .4 .5 .6 .7 .8 .9 1.0 1.1
Time in Seconds
FIG. 34.
90 1.8
80 1.6
6o|l.2|
40^ .8
300 .6
20 .4
10 .2
For this reason Figs. 33 and 34 are given. The curves shown
there should be checked numerically by every student. They
CHARACTERISTICS OF CONDENSERS 73
are calculated under the assumption that a constant impressed
e.m.f. of 100 volts is impressed on a circuit of 2000 ohms resist-
ance and 200 m-f. capacity, as shown in Fig. 33.
An interesting problem in connection with the charging and
discharging of condensers, is to consider the flow of current be-
tween two Ley den jars of different
capacity and voltage (Fig. 35) . The
energy stored in condenser A at volt-
age E is J^CE 2 . The energy stored
in condenser A at voltage e is
The energy stored in condenser B at FIG. 35.
voltage Ei is %CiEJ. The energy
stored in condenser B at voltage e\ is J^Ciei 2 . While current
flows between the two condensers, a readjustment of energy takes
place.
The energy equation is obviously:
0.5CE 2 + 0.5Ci#i 2 - 0.5Ce 2 - O.SCi^ 2 = fi z rdt.
By differentiating this equation, the following results:
- Cede - Cieidei = i*rdt (26)
As it is assumed that the voltage of A is higher than that of B,
the latter being charged; thus
where e\ is the voltage of B at any time. Equation (26) contains
three variables, e, e 1} and i, which, however, are dependent upon
each other.
At any instant the following relation exists between the e.m.fs.
e = ir -f- ei
Thus de_ di^ dei _ d 2 ^ dei
dt ~~ T dt "" dt dt 2 ~ ~dt
Substituting in (26)
ClCir-rr 4-ei) (Ctf-tt + -37) dei-j^ = Ci 2 r (-57)
\ eft / V di 2 d^ / rfi \ dt I
or
6 , dei
ei \ n ei lr r ei j
757 ' Cl ~di( Cir ~dt + ei
or dei \ / dei rr d 2 ei deA
) l Cl ~di + CCir W + c ~dt) =
74 ELECTRICAL ENGINEERING
Since Cir -57 + 61 can not be zero
at
+ C-0 (29)
Integrating (157)
or dei C + Ci K
Kl
ei = K 1 + K 2 e- (3D
where CCi
/ /-? i /^
C + Ci
The integration constants KI and X 2 are determined from the
initial condition that for t = o, ei = EI and e = E
.'. Ei = KI + K 2 or K 2 = Ei - K l
i+ (E, -Kfc c
(33)
l dt~ C r"
but e = ir + e\
/.-- ^(^i - Xi)e " ^ + K!
Co
for t = o e = E
:. E =- ^(E l - K,) + K
Co
.'. K, = E, + ^
and Co
The problem can be solved in a simpler way if it is realized
that the total charge in the system, is not changed after the
switch is closed. Thus
Qo = EC + E, d = q + qi (36)
Where q and qi are the charges at any time in jars A and B
respectively.
CHARACTERISTICS OF CONDENSERS 75
In that case e = ir + e\\ or since q = eC and qi = e\C\,
Assuming E>Ei, then jar A is being discharged thus
dq
or g i
dt ' Cdr q ~
where ^ CC\
Co ~ C + C
for J
Since condenser B is being charged
K EC si , si Co(E
Since condenser A is being discharged
,__* + z*j.-,-.
at r
The voltage across condenser A, which is being discharged is
for t = o e = E
n
, , (
76 ELECTRICAL ENGINEERING
= O = O C (E E } ~ ~Cor
1
TJ C (E EI
1 E - E^
/"* {s QT<
Ci r
i
for t = o,
and C , _. ,, x C
With a slight modification of this equation it is seen that for t
= oo the final voltage between the coatings of the Ley den jars is
E = Q
Numerical example: condenser A has a capacity of 1 m-f. and
is charged to 1000 volts; condenser B has a capacity of 2 m-f.
and is charged to 500 volts; the resistance is 10,000 ohms. Find
the current after the switch is closed.
The original charge in A is then 1000 X T = 0.001 cou-
500 X 2
lomb; the charge in B is I-^G = 0.001 coulomb also.
E - EI = 500
2 X 10 6 2
3 X 10 12 3 X 10 6
and 1
C + Ci = 3 X 10- 6 c V c = 0.667 c _^ Ci = 0.333
*' = i<w-- 150 ' = - 05 - 150 '
CHARACTERISTICS OF CONDENSERS
77
e = 1000 - 0.667 X 500(1 - e- 150 <)
= 500[2 - 0.667(1 - e- 150 0]
i = 500 + 0.333 X 500 (1 - - 150
= 500[1 + 0.333(1 - e- 160 ')].
14
12
,10 1000
i
I
I 8 80 S
i6 > 600 1.060
S <j
| 4^1400-5.040
| 2 W *200 g.020
>
000
.140
.120
10 1000 .100
I
! 8 5 800 | .08
> S
i 6.SCOO-J .06
ir* -^
^ 4-
. 4 3 400 g .04
? 2 200 O .02
>
-2 -200 -.02
-4 -400
\
Tim
Case No.2 Voltages H
-500
)00
.005
.010 .015
Time in Seconds
FlG. 36.
025
.025
For t = oo , e Q = e i 0.667 volts which is the final voltage of
the two jars.
Fig. 36 gives the result of these calculations.
Harmonic E.m.f. Impressed upon a Circuit of Resistance and
Capacity in Series. Let e = E sin cot be the impressed e.m.f., r
the resistance, C the capacity and q the charge at any instant.
78
ELECTRICAL ENGINEERING
Then, referring to Fig. 37,
E sin wt = ir -{-
Differentiating
= ir -\- -^ \ i
- \ idt
di
u cos u>t = r-r + -^
di i Eos cos cot
(1)
E Sin
FIG. 37.
Thus,
_ i t [ C +- 1 -tEu J
i = e Cr \ I Cr cos cotdt + K
The integration is readily made and the result is:
E
i = -y sin (cot + j8) + -KV
where
and
Z = r'
a^ r
tan /8 =
r
The voltage across the condenser is:
= ~ sn
fldt +
^ fe c
At the moment of closing the switch e c = 0.
Thus for t = t 1} e c = 0.
(2)
(3)
thus,
or
t = | [sin (8
cos
cos
(4)
(5)
CHARACTERISTICS OF CONDENSERS 79
As an interesting application of these equations and the corre-
sponding equation for inductive circuits, consider the nature of
the current supplied to a tuned circuit, Fig. 37, when the resist-
ance is small.
x = - x C jZ = \/r 2 -}- x' 2 = -\/r 2 -\-x c 2
The line current at any time is the sum of the currents in the
two circuits.
Q (0 - 8) + sin (0 + 8} - e ~x (e ~ ei}
sn 1 - -
= ^| 2 sin cos j3 e r ' ' cos (0i /?)
XC. ""I
r e ' cos (0i -f )8) (7)
The line current is a combination of a sine wave of form - ~
sin and two exponential or logarithmic curves. Since r is
^_ fa a \
small compared with x or x ej one of the logarithmic curves e *
M *^ c fa a
sin (0i ft) dies down at a slow rate, whereas the other -V 7 l
cos (0i -f- (3) dies down with extreme rapidity.
x
In the limiting case when - is large and ft therefore approaches
90 the permanent term disappears since cos = 0,
and Er -J-(g- gi ) x c _ x f(0_0,) . ~|
= '7? I * COS 0i H r sin 0i
ZL r J
for #
r = OLto = ~~ cos QI
x
Thus the interesting situation occurs that if an alternating
voltage is impressed on a tuned circuit as shown in Fig. 37 and if
the resistance is zero the line current is a steady unidirectional
current having the value:
. _ E cosd i
x
If the switch is closed when 0i = 0, then the direct current is a
maximum and is . At any other time it has a smaller value.
x
80
ELECTRICAL ENGINEERING
In Fig. 38 is shown a series of curves which illustrate this in
the case where the resistance is considerable and the circuit is
closed when 0i = 0. The constants for the circuit are:
E = 1, r = 0.05, x = x c = 1
z'o is the total line current, the dotted sine wave is the impressed
e.m.f., IA the current in the condenser circuit, and IR the current
in the inductive circuit. As is seen, i Q is a unidirectional current
l.U
(
^
Q
^
x
^^,
__,,
---'
~~
^
'o
5
K.
M.
i 1 .
*
^^
^'
f
\
/
\
/
^
/
(
1
1
\
i
*
11 a
1 i
\
-*
-~-
A
i^
-^
<"4
)
Q
1
)
11;
^
?,
()
^-
^
i()
4
g
(1-
- ,
ft
)
_^-
1
-.1
\
/
\
.2
(
.3
\
/
N
J
1
N
,
/
^
/
\ '
t- A
g
\
/
\
1
7
\
/
\
1
g
1.9
E
M
F,
5
1 8
/"
>
\
4
1 7
\
2
V
l r
/
/
\
\
/
^
1 fi
t
\
c
/
\
/'
2
/
\
/
\
\
/
/
\
\
/
1 4
\
\
/
/
V
,
72
1 3
91
;/
li
9
2
70
p
30
4
/
b
ip
\
Gl
/
1 9
V
\
1
\
N
1
/
o
1 1
A
\
\
/
q
1
q
N
I
7
A
/
4
9
/
A
i
^
-5
/
\
1
\
\
6
7
/
\
/
\
\
7
g
/
\
/
\
1
\
5
J
^i
K
\
I
._
/
9
4
N
\
/
y
/
1
3 .
\
\
;' r
/
/
\
\
/
\
I
\
/
N
\
/y
\
i
/
12
)
9
1
^p
2
70
\
3
30
/
4
oO
5
IP
6
\
I
2
V
f
\
\
i
Q
x^
^}
\
4
\
/
\,
/
\
,
^^
e
-.6
FIG. 38.
of slight pulsation, slowly decreasing in magnitude. After a
number of cycles it would become a small alternating current, as
shown in the curve marked Final i .
This feature of a tuned circuit might be of practical importance
in connection with problems of rectification charging of storage
batteries from an alternator by occasional interruption of the
CHARACTERISTICS OF CONDENSERS 81
current and starting it at the time when normal current in either
branch would be a maximum.
Circuits of Inductance and Capacity. While practically such
circuits can never exist they offer much interest from a theoretical
point of view since their study
represents an introduction to oscil-
lating circuits, which are of much
importance in electrical engineering.
In Fig. 39 is shown such a cir-
cuit. In practice the condition
there indicated is approached FIQ
when a very low resistance over-
head transmission line supplies power to a cable net work, which
case, however, is fully treated in a subsequent chapter.
The following relation exists at any time.
where E is the instantaneous value of the impressed voltage and
e\ the voltage across the condenser.
But . _ n dei
- dt
thus di _ d*ei
=
dl*
thus LC^ + ei = E
or d*ei , _fi_ J^
dt 2 ~^ CL~ CL
The solution of this equation has been given, it is
ei = E + Aie mi '
mi and ra 2 are the roots of equation;
m. - f i . ,
<CL
m
82 ELECTRICAL ENGINEERING
But e } ' at = cos at + j sin at
-jat = cos a t j s [ u a t
.'.A** ' *
+ j sin at(Ai A 2 )
= Ao cos aZ -f- Bo sin at = A sin (a +
where A and 5 are integration constants.
The current
The integration constants are determined from the knowledge
of the initial conditions.
Assume for instance that it is desired to know the current and
the voltage across the condenser at any instant after the switch
is closed:
At the time t = 0, i = and e\ = 0;
.'. = E + A smB
and 1C D
= A -J cos B.
Thus cos B must be:
.'. B = thus sin B = 1 and A = - E
and
The voltage across the reactance is L -j- it is
= cos
CHARACTERISTICS OF CONDENSERS
83
In Fig. 40 are shown the voltages across the condenser and
inductance as a function of time. It is seen that there is no time
at which the voltage across the inductance is greater than the
impressed voltage and it is also seen
that at the instant of the first half
period the voltage across the con-
denser is twice as great as the im-
pressed. Thus the condenser will be
subjected to double voltage during
each cycle.
The maximum value of the current
is, as seen;
FIG. 40.
> f>
/
-'"
'*
V 1
1.2
.8
.4
/
7-)
\
SI
7
v
/
s
\
/
\
.
-.4
\
/
s
i
-
-1.2
^
^**s
< ^T=2if\TcI. Seconds >
The frequency of the alternating current / = - ~/TT * s c
the natural frequency. 1 It is almost always much higher than
the frequencies used in commercial alternating-current circuits.
The Discharge of a Condenser through an Inductance. Re-
ferring to Fig. 41. Let E be the voltage of the condenser before
the circuit is closed and i the current at any instant after the
switch is closed.
Then
But the discharge current
e = L
thus
and
or
i = - C
di
di
dt
de (
dt
FIG. 41.
L dt CL dt 2
eo = -CL' 1
di 2
CL
the auxiliary equation becomes :
m 2 + -== =0 or m =
1 In case of a transmission line the natural frequency will be shown to
1
84
thus
and
ELECTRICAL ENGINEERING
cos
cos
The integration constants A and B are determined from the fact
that at t = 0, i = 0, and e = E . Thus:
EQ = A sin B
= -
C D
cos B
'. cos B = and B ^
and
.'. sin B 1 and A
t
= En sm
cos
2
E
I = En COS
sn
s
It is seen that the discharge frequency is the same as the
frequency at charge, and that the maximum value of the current
P ft
E \L-
As another application of this will be considered the condition
when a short-circuit is suddenly
opened and the large current in-
stantly interrupted.
This condition is diagrammat-
ically illustrated in Fig. 42. S
represents a switch which short-
circuits the condenser and is
opened at the instant under consideration. (In practice this
switch may represent a short-circuit across the cables opened by
the magnetic effects of the current.) The current I in the
short-circuit is evidently the same as the current in the inductance ;
therefore the energy stored in the magnetic field is 0.5LJ 2 .
At any time after the switch is opened the current i flowing
through the condenser, inductance and generator (all assumed as
FIG. 42.
CHARACTERISTICS OF CONDENSERS 85
having zero resistance) is governed by the condition that the
energy stored in the condenser and inductance is the same as
the original energy.
.'. 0.5Li 2 + 0.5Ce 2 = 0.5L/ 2
LI' 2 - Li* = Ce*
but . _ ~de
1 ^7.
dt
thus /de
,- 2 = C2 /^V
\dt)
:.LP - LC*(~} 2 = Ce\
Differentiating or/^2 ^ 2g
~ 2LL dt 2 dt =
. d*e I
-+ 6 '-
and . / t %
e '-'- A sin + B )
for t = Q,i = I,e =
Thus = A sin 5; and sin 5 = 0, and B =
1C
=
i = I cos
It is interesting to note that while at the instant of opening the
switch, or the short-circuit, the voltage e across the condenser is
zero, one-quarter of a period later (period being here the natural
period which is extremely short) the voltage is a maximum and
is
6max =
These equations are instructive in that they show that the
maximum voltage obtained in opening a short-circuit in a cable
86
ELECTRICAL ENGINEERING
or transmission line is independent of the length of the line and
depends only upon the constants of the circuit per unit length and
the current at the time the circuit is interrupted.
They also show that when the circuit is closed on a transmission
line of considerable inductance and capacity, the maximum rush
of current is also independent of the length of the line and depends
only upon the value of the e.m.f. and the circuit constants.
Harmonic E.m.f. Impressed upon a Circuit of Inductance and
Capacity but Negligible Resistance. This strictly theoretical
condition is chosen for two reasons. The solution of the equa-
tions introduces some mathematical operations which have
hitherto not been considered and the problem from the electrical
point of view illuminates in a relatively simple way what happens
in the extreme case in switching high-potential circuits.
The general equation obviously becomes:
E sin
ei
(1)
where e\ is the voltage across the condenser, Fig. 43.
But
^ dt
.'. E sin
thus L -7
at
CL
or
FIG. 43.
d 2 6i 61 E
W + CL = CL
sln
(2)
It is seen that the right-hand member of the equation is a
function of L To solve such equation the solution of the comple-
mentary function is first found; that is, zero is substituted for
the right-hand member:
V ' r CL ~
i 2 +-== = or m = j .J
CL/
(3)
CHARACTERISTICS OF CONDENSERS 87
Ai +A/ cE' + AvT^m,' = A^ at + A#- iat (4)
where
The equation, as has been shown previously, can be written
61 = A sin (at + #)
Thus the complementary function is
e Q = A sin (at + B) (5)
The next step is to eliminate the sine function from the general
equation (2) by two successive differentiations:
Substituting the value of Ea 2 sin coi from equation (2) and
arranging the equation in the order of the derivatives:
(7)
The complete solution of (7) is obtained in the usual way:
m 4 + (a 2 + a> 2 )ra 2 + coV =
W 2 (m 2 + a 2 ) + a> 2 (ra 2 + 2 ) =
or (m 2 + w 2 ) (m 2 + a 2 ) =
.*. m = + jb)
m = + jo.
Thus d = Ai sin (at + #0 + A 2 sin (a + 5 2 ) (8)
By referring to (5) it is evident that A 2 and 5 2 must be the same
as A and B.
Thus i = e Q + AI sin (o> + BI)
The integration constants AI and BI are determined from the
fact that the expression A\ sin (co + BI) must be a particular
solution of (2).
Thus i = AI sin (o> + 1)
-7- = AICO cos (&t -}- BI)
d z d
- = -- Aiw 2 sin (cot + BI)
88 ELECTRICAL ENGINEERING
Substituting these values in (2).
Thus
- Aico 2 sin (ut + Bi) + Ai 2 sin (coZ + BJ = Ea* sin ut (9)
or A^a 2 - co 2 ) sin (at + BI) = Ea 2 sin ut.
Thus equating the coefficients of similar terms:
Ea 2 = Ai(a 2 - co 2 )
* --^.-5%; (10)
or or c a:
and 1 = /. from (8) (11)
ei = EXc sin ut + A sin (erf + 5) (12)
3?c 32
The second term in this equation may in this case be more
advantageously written :
AQ sin at + Bo cos a
/. ei = - - sin coi + A sin a^ + Bo cos at (13)
3?c E
= 7a: c sin coi + A sin a^ + Bo cos at (14)
Where / is the maximum value of the permanent current,
that is,
/ - ~^ (15)
X c - X
i = C 3 - = Ix c u cos ut + A Q a cos at B a sin CK (16)
at
Considering the problem of starting a current in such a circuit :
when t = ti, i = and e\ = 0. If these values are substituted
in equations (13) and (16) it is readily found that
BQ = Ix c sin ati cos coi sin at\ cos cd\ (17)
AQ = Ix c sin ati sin wt\ -\ cos at\ cos co^i \ (18)
Substituting these values in the equations for the current and
e.m.f. across the condenser we find:
ei = - sin at cos coZi sin a(t ti)
Xc X L a
- sin coii cos a(t - fi)l (19)
CHARACTERISTICS OF CONDENSERS 89
and
i = C -TT = ~ cos at cos coti cos a(t ti)
at Xc x L
1C 1
+ x c ^Ij- sin coti sin a(t - fc) I (20)
While as a rule equations of the form given in (2) having a
sine function or exponential function on the right-hand member,
can be solved by the method given above, it may be opportune
here to call attention to another well-known method of more
general application.
The differential equation is:
fJZf,
<L + eia 2 = Ea 2 sin cot
This expression is given in symbolic factors as follows:
(D - mi) (D - m 2 )ei = Ea 2 sin cot (21)
where D is an abbreviation of -r and mi and mz are the roots of
the complementary function
d 2 ei
777- -f- eia 2 or m 2 + a 2 =
at
.'. m = ja or mi = + ja
and mz = ja
Equation 21 can thus be written
(D + ja) (D - ja)ei = Ea 2 sin cot
Let u = ei(D ja)
.'. (D -f ja)u = Ea 2 sin cot
^ + jau = Ea 2 sin coi (22)
This is a linear differential equation of the first order and its
solution is.
u = e -i*t | 6 +/ j^o:2 s in w^ ^ + Ci~^ (23)
/. . . . ,. . , / ja sin cot co cos coA
e iat sm cot dt = e ]at [ J
\ co 2 - a 2 /
cot-^co^cot)
90 ELECTRICAL ENGINEERING
Since u = e^(D ja)
dt
i = e +3 ' at e-* at udt
I e-
e ] ' at I ~ } ' at - - (ja sin cot co cos
J to 2 a 2
t | e -
+ iat e -jat C lC -J dt + C #>*' (25)
These four integrals are solved independently below :
First,
s 3 .*f.
- 2 ' J f
sn
( la sm co* co cos co *\ EC
, jat J I A
co + a 2 / J co 4 +a 2
(ja sin co* + co cos co*) (26)
Second,
r
~ iat cos co^ dt =
a* I co" a*
2 -
I -jat u s ^ n ^ ~ J a cos ^ \ ^ a?co
\ C " co 2 + a 2 / : ~ J co 4 -a 4
(a cos co* jco sin co*) (27)
Third,
r _. at _ _. at r _ 2 . a<
O 1C I C (Zt' vy l6 I C ttt'
J J
" 2ja e ' a< ~ 23at = 2ja e ~ } ' at = C ^ at (28)
Fourth, C 2 e>' at (29)
The last two terms can be written
C 3 e-' a < + C 2 e>< = C 4 sin (a* + C 6 ) (30)
In the general equation it is seen that the second integral is
negative thus:
co 2 sin co + a 2 sin co^ jaco cos co^ (31)
= C 4 sin (at + C 6 ) H -- ri jaw cos
+ co 2 si
n - / *
= C 4 sin (at
sin (
This equation is identical with (12).
= C 4 sin ( + C B ) + -- sin co* (32)
X c X
CHARACTERISTICS OF CONDENSERS 91
As an application of these formulae will be considered a 100-
mile 60-cycle transmission line supplying power to a cable net-
work of 50 miles. For the sake of simplicity and for the sake of
later instructive comparison the resistance of the cable and of
the overhead line will be neglected in this particular investigation
and it will also be assumed that the inductance of the cables and
the capacity of the overhead lines are so small as to be negligible
when compared with the inductance of the overhead line and the
capacity of the cables.
While the inductance of a line, of course, depends upon the
size of the conductors and the distance between them, in reality
it is not subject to a great deal of variation in ordinary lines.
It is about 0.002 henrys per mile of single conductor.
The capacity of a cable system is, however, subject to great
variation, depending upon the nature of the cables. Assume
that in this case it is 2 m-f. per mile of single conductor, when
referring to neutral voltage:
C = 4 farads and L = 0.2 henrys
= 26.4 ohms
x = 2irfL = 75 .4 ohms
= 27T/ = 377.
If the circuit is closed when the impressed e.m.f. is zero, that
is, when ti = 0, then equations 209 and 210 become:
ei = - 0.54#[sin 377* - 1.69 sin 223*]
and i = - 0.0204#[cos 377Z - cos 223*]
The time for one complete cycle of the fundamental wave is
gg = 0.0166 sec.
If, therefore, the circuit is closed when the impressed e.m.f.
is a maximum, that is, when
h = t = 0.00416
then the equations become:
d = - 0.54#[sin 377* - cos 223(* - 0.00416)]
i = - 0.0204#[cos 377* + 0.59 sin 223( - 0.00416)]
92
ELECTRICAL ENGINEERING
These curves are shown in Figs. 44 and 45 when the impressed
e.m.f. is 100 volts.
The curve e\ in Fig. 44 shows the e.m.f. across the condenser,
the curve i the current when the switch is closed at zero value of
4
^-~,
/
\,
/
\
2100
^_
/
\
/
N.
^
>
/
^^
s c i
Y
"X
'
V
,
m
/
/
/
V
?
V
\
/
/
\
V
/
\
s
N
\
55 1 50
/
/
/
\
\
\
v/
cY
/
\
/
/
\
\
^0 *0
/
.X
\
\
\
/
\
/
\
/
"- .
\
o >
i
^0
\
2
70
\f
30
^
oO
b
ty
6
JO
\
I
8
LO
/
yu
a 1 50
\
\
i
A
/
a.
t
A
A
/
\
\
$/
\
/
/
\
/
\
x
i
>
v \^
^
\
/
\
/
/
U 2 100
\
V
/
\
/
V.
^x
\
/
4
s^
FIG. 44.
2 200
150
.S oO
50
1 100
150
2 200
90
isp
270
\
/450
720
9(0
FIG. 45.
the impressed e.m.f. The corresponding lines in Fig. 45 show
the same quantities when the switch is closed when the impressed
e.m.f. is a maximum. In both figures the dotted sine wave is
the impressed e.m.f.
CHARACTERISTICS OF CONDENSERS
93
Circuits Having Resistance, Inductance and Capacity. Con-
stant Impressed E.m.f. Let E, Fig. 46, be the constant impressed
e.m.f.
r, the resistance
L, the inductance
and
C, the capacity in farads.
L
c:
Then
FIG. 46.
T di ,
Differentiating di d*i i
= T + L +
or
dH > r dii _
dt* + L dt + CL "
The solution of this equation has been shown previously to be :
Where mi and ra2 are the roots of the auxiliary equation.
where
and
_
2L
4L 2 CL
CL
2L ~ =
= -
-
4L
94 ELECTRICAL ENGINEERING
Three conditions are possible:
(a) r 2 -T^ is positive.
(b) r* - -~r is negative.
4L.
(c) r 2 -fT is zero.
Considering first the Case (a),
^ ~^r (positive).
Then i = ^[A*? + A*-'* 1 }.
The e.m.f. across the condenser is:
ei = E -ir -L d ~ = E - re- at [A^ 1 + A*'* 1 ] - L[((3 - a
The integration constants for starting the current in this circuit
are determined from the fact that when t = 0, i = and e\ = 0.
Thus E
Ai = Ai and AI = ^j
ZLp
2L/3
2 4L
\ r '^ :=>S >
If
then . El -l'-=^)i -f'J^\ t \
*-( - )
By differentiating i and substituting its value and that of i in
the equation of the voltage of the condenser:
e\ = E ir Jl -j-.j
dt
we get
[1 / r ~ s t r+s t\~]
i if._i_ cr\ o7~ i (^ Q\ f 9r~ t \
1 ^.^ I (r +- o) 2L IT ^Jc *^
^o \ / J
The equation for the discharge current can readily be proven
to be exactly the same as that of the initial current except for
reversed sign.
CHARACTERISTICS OF CONDENSERS 95
During the discharge the voltage is
r-f S
where E is the voltage of the condenser before the discharge.
Case (6).- r - 4 negative.
In this case + r 2 ~- can be written j -\\-~ r 2
\ C \ C
Si where S x == - r 2
It has been shown previously that this equation can be written :
(1)
where A and 5 are integration constants. The e.m.f. across the
condenser is
Substituting the values of i and - as obtained from equation (1)
Counting time from the instant of closing the circuit, then for
t = 0, i = 0, and e\ = 0.
From equation (1) it is found that A sin B = 0; .'. B = 0,
2#
and from equation (2) A = -~-
01
. 2E * t . Sit
t = ST 6 2L Sm 2L
where tan 7 =
In a similar way is found the equation for the discharge
current which will be identical with the charging current, and
the voltage across the condenser, e a = E Q e ~2L l sin ( ^F + T ) >
where EQ is the original voltage across the condenser.
96 ELECTRICAL ENGINEERING
Case (c).
4L
r 2
In this case, as has been shown previously,
i = e m i* (A + Bx)
r
i = e 2L (A + Bt)
or
= E
re ~2L l (A + Bt) - LBe 2L l + ^~-
If the time is counted from the instant of closing the switch,
then for t = 0, i = 0, and ei =
A =
and - " - ^ hi
E - LB, or B
and
The equation for the discharge current is the same as that of
the initial charging current, the voltage across the condenser
during discharge being:
As an application of these equations
will be considered the case of starting
a direct current at 500 volts in a 20-mile
concentric cable having the dimensions
given in Fig. 47.
In this instance it will be assumed
that the capacity of the line can be
represented as that of a condenser at
the end of the line taking one-half of
the charging current.
It will be assumed that the specific
inductive capacity is 3, the diameter
of the inside conductor 0.5 in., the
inside diameter of the outside conductor 0.7 in., and the' out-
side diameter 0.86 in. The resistance of the 20 miles of cable is
8.8 ohms.
FIG. 47.
/
CHARACTERISTICS OF CONDENSERS 97
The capacity per mile of concentric cable previously given is
n o^sft Jf
Cmf = - ~ = 0.795 m-f. per mile.
^og-
Thus the capacity of the 20 miles of cable is 15.9 m-f. and the
equivalent capacity at the end of the line is 7.95 m-f., or 7.95
by 10~ 6 farads.
Since the determination of the inductance of a concentric
cable involves the general method applied to other systems, it
will be given below, although such determinations do not come
within the scope of this treatise.
The inductance is recollected to be numerically equal to the
interlinkages of the turns and flux per unit current.
In general if the m.m.f. acting in a circuit is M then the flux
4irM X area of magnetic circuit
produced is , -r * ~r- = ^ The interlinkage
length of magnetic circuit
factor is that fraction of the total current which is enclosed by
the flux and
L = y-2J flux X turns X interlinkage factor.
Consider first the flux in the inside conductor due to the assumed
uniform distribution of the current.
At a distance x from the center see Fig. 47, the m.m.f. is ^ /
where / is the total current. The area of the flux per centimeter
of length of conductor is dx and the length of the magnetic circuit
is 2irx.
x 2 dx x
... dv , 1 = 4 ,_ / _ ==2J _ & (
TTX 2
This flux interlinks with ^ of the total current, and hence the
x 2
interlinkage factor is ^.
1 C r x 3
.'. LI = I 27 dx = % (assuming /x = 1)
Between the conductors the flux interlinks with the whole
current, and hence by a similar reasoning we get
98 ELECTRICAL ENGINEERING
The current in the inner conductor interlinks with the entire
flux which is inside of the outer conductor but which is caused by
the difference in m.m.f . in the inner and outer conductor.
At a distance XQ the m.m.f. is thus
_
R 2 -R 2 * *R Q 2 -R 2
The interlinkage of this flux with the current in the inner con-
ductor is of course unity; therefore
L * = I J ~*o flo 2 ^!^ dx = R 2 - R 2 log ^ " " l
The inductance of the outer conductor should be added to give
the total inductance of the cable.
The m.m.f. is _ r RQ ~ X Q
Ro 2 - 1
The interlinkage factor is
xp 2 - R 2
-R 2
,2 _
' L * - - 7/
z (Ro 2 -R 2 )
^ R Q 2 - R 2 + (R 2 -R 2 ) 2 log ~R
The total inductance L = LI -j- L% + L s + LI which is readily
proven to be:
This inductance is expressed in the absolute system of units.
By dividing by 10 9 the inductance is expressed in henrys.
The combined inductance L = 0.0039 henrys; thus r = 8.8,
L = 0.0039, and C = 7.95 m.f.
r 2 = 77.5 1960 is thus negative.
c
Therefore this problem comes under the second case and
t = ^/%- - r 2 = 43.4.
^ = 5550 ~ = 1130 tan 7 = - = 4.93r = 78.5
LU ZLi T
= 1.37 radians
/. i = 23e- 1130 ' sin 5550*
and e c = 500[1 - 1.02 e - 1130( sin (5550* + 1.37)].
CHARACTERISTICS OF CONDENSERS
99
The frequency of the oscillation is
5550
27T
= 885 cycles, and the
time for one oscillation 0.00113 sec.
The maximum value of the current is determined by differen-
tiation. It occurs when t = 0.000246 sec. When 5550^ =
78.5, the current is 17.1 amp. The next maximum value occurs
when t = 0.000246 + 0.00113 = 0.001376 sec.
The maximum value of the voltage across the condenser is also
determined by differentiation. It occurs when t = 0.000565
sec., when e c = 763 volts. The next high value occurs obviously
at t = 0.001695 sec.
These curves are shown in Fig. 48.
0004
Starting
V; Condenser Volt; K
urrent
m-
.001
odd!
0012
0014
0016
.0018
400
200
,002
FIG. 48.
It is of interest to note that for a given distance of transmission
the capacity, and therefore the charging current, is several times
as great in the case of the concentric cable as in the case of the
cable with parallel wires.
Similarly the inductance is several times as great in the case of
an overhead line as in the case of the cable. As a second numer-
ical application of these equations will be considered: 100 miles
of overhead transmission line supplying energy to a cable network
50 miles in length.
It will be assumed that the cable system consists of a large
number of short cables projecting in different directions from the
terminal substation, as would be the case when a high-tension
7
100 ELECTRICAL ENGINEERING
line supplies energy to a city lighting network. The resistance
of the cable system can therefore be neglected. It will be
assumed that the high-potential line is three-phase and consists
of No. 00 B. & S. wire, having a resistance per 100 miles of 40
ohms and an inductance of approximately 0.2 henry. Hence the
capacity of the overhead line is very small compared with that of
the cable and it will be neglected.
The problem is to determine the values of the current and vol-
tage across the condenser when a steady e.m.f. of E volts is
applied at the generating station.
E = 100
r = 40
L = 0.2
C = 0.0001 farad.
/. r 2 = 1600
-Q = 0.8 X 10 4 = 8000.
4L.
.'. r 2 = -^ is negative.
Therefore there is an oscillation when the switch is closed, and
the constants are to be obtained from case (6).
= 2= 80 ' 7 = !o = 2 " tan ^ = 2
and r 40 Si 80
V - 6 3.5 2L = ^4 =: 10 > 2L = 04 =
:.i = 0.025#e- 10 * sin 160*
and a = E[l - e" 100 ' 1.12 sin (160* + 63.5)].
The time for a complete cycle is T^T = 0.0392 sec., corresponding
to a natural frequency of ^ = 25.5 cycles per sec. It is inter-
esting to see that the effect of the resistance is to lower the
natural frequency, since if the resistance is neglected it would be
7= = 35.5 cycles.
CHARACTERISTICS OF CONDENSERS 101
Circuit Containing Resistance, Inductance and Capacity in
Series. Harmonic E.m.f. Impressed. Fig. 49. From previous
discussions it is evident that the general equation is
di
dt
E sin cot = ir -\- L ^- -f- ~- I idt (1)
r E sin = ir + x~ + x c (idd (2)
FIG. 49.
The latter form is preferable when dealing with alternating-
current phenomena, but of course it must be remembered that x
and x c refer to the impressed frequency and not to the natural
frequency of the system; that is,
x = 2-nfL and x c =
The solution of (2) can best be obtained by differentiating
twice,
di d 2 i
E cos 6 = r -TT + x -7T- 2 + x c i E sin 6
cttf
d z i d*i di
dl ' idB (3)
Differentiating (3) and rearranging the equation:
d*i dH d*i di]
Xcl =
The auxiliary equation is:
xm* + rra 3 + (x c + x)m 2 + rm + x c =
ra 2 (zra 2 + rm x c ) + xm 2 + rm + x c =
/. (m 2 + 1) (xm 2 + rm + x c ) =
Vr 2 -
Let r
2x
102 ELECTRICAL ENGINEERING
then mi = + j
W2 = j
m 3 = a + ft
1714 = -- a ft
and i = A, sin (0 + A 2 ) + ^'(A*-* 8 + A,^ e ) (6)
The integration constants A \ and A 2 could be found by methods
outlined in the chapter on circuits of inductance and capacity.
It is possible, however, for students familiar with elementary
electrical engineering to determine them at once.
Apparently the first term represents the permanent and the
second the transient condition. In permanent operation the
current leads or lags behind the impressed e.m.f. by an angle </>
which depends upon the numerical values of the two reactances.
The final value of the current becomes, then,
-T7
i = ^ sin (0 + 0)
^o
if tan 4 = ^ and Z == vV + (x c - x) 2 (7)
.'. i = ~ sin (6 + </>)+ <r a V 3 /' + A#- ft ) (8)
&Q
The integration constants A 3 and A 4 depend upon the terminal
conditions.
Before proceeding farther it is well to discuss the possible con-
ditions, namely:
(a) r 2 - 4xx c is positive.
(6) r - 4xx c is negative.
(c) r - 4xx c is zero.
Case (a).
r 2 4xx c is positive.
Here ft is real and the solution of i is given in equation (8).
Case (&).
r 2 4zz c is negative.
In this case \/V 2 xx c = j V4xx c r 2 (9)
Let
then i = lj- sin (6 + <l>) + AiT'to sin (0J + j) (11)
^0
where
CHARACTERISTICS OF CONDENSERS 103
and , A B As + A 4
7 = tan" 1 -r- = tan" 1 7-1
A 6 (A 3 - A 4 )
a S t = ~- sin (0 + </>) + ~"*[A 6 cos 0i0 + A 6 sin faO] (12)
(c). r* -
If (12) is true, then m 3 = w 4 , and we do not have a general
solution; that is:
- a + = - - since = (13)
A general solution is obtained by letting
w 4 = ra 3 + h (14)
where h is very small.
Here w 3 = a and ra 4 = a + h
then (see also equation 42, Chap. II),
E
i = -yr sin (0 + 0) + A 3 e~' ' -f A 4 e ( ~ (15)
which may be written:
i = sin (0 + 0) + e~ ae [A 8 + A 9 0] (16)
where A 8 = A 3 + A 4 and A 9 = A 4 /i.
Each case will be considered independently.
Case (a). r 2 4xx c (positive).
Since r 2 4xx c is positive, is a real number and the solution
of the differential equation (4) is :
i = ^- sin (0 + 0) + A 3 e~ ( * 6 -\~ A 4 e~ ( (17)
Let ( - 0) = K and (a + 0) = K (18)
By differentiating equation (17) and substituting in the
equation
d = E sin ir X-J-Q, (19)
the voltage across the condenser is:
1?7
ei = E sin - -^ sin (0 + + ) + A 3 e~ J
^ lfl (^ia: - r) (20)
where .,. x (21)
(22)
104 ELECTRICAL ENGINEERING
If the problem is to find the current and the condenser voltage
at any instant after the circuit is closed, and if the circuit is
closed when = 0i, then i = and ei = 0.
Substituting these conditions in equation (23), it may be
written :
A s e~ Kei + A,*-* 1 ' 1 = -- |^sin (0! + 0) (24)
Also equation (20) can be written:
EZ
Ax' 1 (Kx - r) + A 4 e~ J 9l (Kix - r) = -^- sin
(0i + + ^) - E sin 0i (25)
Solving equations (24) and (25) for A 3 and A 4 we have:
E
A 3 = K01 [Z sin (0i 4- <f> + ^) +
(Kix r) sin (0i -f 0) Z sin X ] (26)
and,
E
^r e Kl01 \Z sin (0i + d> + ^) +
x - r) sin (0i + 0) - Z sin Oi] (27)
Case (6). As stated before the expression \A* 2 xx c is
imaginary and from equation
(28)
Equation (11) may be written:
E7
z = ^- sin (0 + 0) + e~ a0 [A 5 cos |8i0 + A 6 sin j3i0] (29)
-^0
where A 6 = A 3 + A 4 and AG = j(A 3 A 4 ).
If the switch is closed at = 0i, e\ = and i = 0. From
these conditions
A 5 = [- f- sin (0! + </>) - A 6 e- afll sin /3 1 1 ]-^- (30)
L Z/o J COS Piwi
From these relations and the equation (19)
can be written
-
- r) cos 0i0] + A 6 [- $ix cos 0i0 + (ax - r) sin
di
e\ = E sm zr x ->
EZ
= E sin - -=r- sin (0 + + ^) + e~ a0 [A 5 [/3iZ sin
CHARACTERISTICS OF CONDENSERS 105
where
Ecosp,d } ae [Z .
A 6 = - TToT- e r sm 0i - cos (01 + 0) - 01
and [# sin (^i + 0)e a<?1 + A 6 ^o sin ffifln
Z cos /3i0
Case (c).
(r 2 = 4a:x c ), or the "critical case."
Equation (16) may be written:
E
i = -y- sin (0 -f- <p) -{- A%e ' -f- ^4.9^e ' (31)
If as before th'e switch is closed when = 6 1 then i = and
e\ 0. From these relations and the equation (19)
di
ei = E sin 6 ^r x -^
uu
the condenser voltage is found to be:
777
e, = E sin 6 - -^- sin (0 + + ) + e~ a5 [^ 8 (o!X - r) +
^ 9 (ax - a: - r)] (32)
where,
77 r^ n
A 9 = -=r e" 01 " sin (9 T - cos (B l + <j>) - a sin (0i + </>) (33)
Z/o L ^
A 8 = - ~ e^ 1 sin (19! + 0) - A 9 (34)
CHAPTER V
A CIRCUIT CONTAINING DISTRIBUTED RESISTANCE
AND INDUCTANCE
An aerial transmission line with negligible capacity and leakage
conductance is an example of such circuit.
Fig. 50 represents an aerial transmission line with negligible
capacity and leakage conductance and with a load having an
impedance of \/Ri 2 + Li 2 co 2 .
Measuring x from the receiving end, consider an element of
the line dx.
Let the resistance of the line be R ohms per unit length of the
conductor and the inductance L henrys per unit length.
Then the resistance of the element is Rdx and its inductance
Ldx, and the voltage across the element is de. Therefore,
or
de = Rdxi
de
TJ di
MX df
di
As the same current flows in all parts of the line i is not a
function of x, thus equation (1) is readily integrated, it is:
106
RESISTANCE AND INDUCTANCE 107
where T di
x = 0, e = R]i + Li -T ,
. = di
f n . . T di\ r di
. . e = (/fa + L--i- iz -f 72 1* + Lrr (2)
\ d< / d/
where x = I, e = E or E sin o>Z, depending upon whether the
generator voltage is constant (a) or alternating (6), therefore,
(a) E = Ri + LJJ I + R,i + Lr = (fl/ + R,)i +
(JH+Zu)^ (3)
(6) sin co* = (Rl + 720* + (^ + LI)^ (4)
/?? + Ri is the total resistance and LI -f ^i the total inductance
of the circuit. Hence, neglecting capacity and leakage conduct-
ance, a circuit of distributed resistance and inductance may be
considered as if the resistance and inductance were concentrated
as far as the determination of the current is concerned.
Case (a). Unidirectional voltage impressed on the circuit.
For the current, solution of (3) gives
E f Rl + Ri\
(1 LI+LI* (see equation (17), Chap. I) (5)
i \ /
P 7 , P
til -f rt
Substituting (5) in (2)
n
\
From (5) and (6) it is seen that at the moment of closing the
circuit, that is, when t = 0, i = and e = TJIT 1 E. In the
case of non-inductive load (when LI = 0), e = jE and e = for
x = 0.
It is interesting to note, that, while resistances consume no
voltages when i = 0, inductances do consume voltages when
i = 0, provided -^ = 0. When t = , that is, when the current
and the voltage reach their permanent conditions,
77? J?T \ 7?
*' = ~ e = " E ' the expected results -
108 ELECTRICAL ENGINEERING
Problem. Assume reasonable values for the constants, and
plot a series of curves for the voltages at various values of t and x.
Case (b). Alternating voltage impressed on the circuit.
From equation (39) in Chap. I, the solution of (4) is found to
be:
E r Rl + Rl < t , n
i = | [sin (co* - 13) - sin ( w *i - /3)e " T+T7 (t (7)
where ti is the time at which the circuit is closed.
Z = V(M + Ri) 2 + (LI + LJW
and _ , (LI + Li) co
(Rl + Ri)
Differentiating (7),
cos (/ - 0) + fj^-f^ sin co/! - |9)e " ^TiT w ' J (8)
Substituting (7) and (8) in (2), or,
sin (ft' 0) sin (wti 0)
(9)
where
-Li) 2 co 2 , and j8 r = tan -1
Referring to (7) and (9) it is seen that the current is the same
as if the resistance and inductance were concentrated, but the
voltage is different at different points, being modified in magni-
tude, and displaced in time phase.
It is noticed that no transient component in the voltage or
current exists at any point of the line, if the circuit is closed at
R
ti = -, or in other words when sin (wti (3) = 0.
CO
If sin (co/i |8) is not zero, the transient voltage appears at all
values of x except x = I, for ft' can not equal (3, or ^ ZTT?
(LI + I/O co
can not equal to P7 , 5 unless x = I.
11 + K\
When t becomes large, that is, many cycles after the circuit is
closed, the exponential term approaches zero and the whole cir-
cuit becomes free of the transient, and (7) and (9) become :
i = | sin (ut - |8) (10)
EZ'
e = --sm (co* + j8' - 0) (11)
RESISTANCE AND INDUCTANCE 109
Z f
where x = I, that is, at the generating end. -^ = I
L
&' = ]8, therefore,
e = E sin ut, as assumed.
The voltages at other values of x can readily be computed
from (11) and it is seen at once that the amplitude is proportional
7'
to -;= and the phase is ($' /3) radians leading the impressed e.m.f.
The effective value of the voltage at any value of # is:
(12)
where e e /f = the effective value of the voltage at x, and E e f/ that
at generator, that is, at x = I.
When the line is open, that is, when Ri = oo , then (7) and (9)
become : i = 0, and e = E sin co for all values of x. No current
flows, which is to be expected.
Grounding the receiving end of the line, Ri = and LI = 0.
.'. -jj = -y and &' = j8, hence (7) and (9) become:
/ i
i = 277[sin (ut - 0") - sin (wti - /3")e v "'J (13)
and Ex .
e = - sin at (14)
where ., / , ,.,, _, Lo>
A = \/ R 2 -f- L 2 co , and p = tan "5"
It is interesting to note that in this case the voltage has no
transient component and is in time-phase throughout the line.
PROBLEMS
1. Assume reasonable constants of the circuit for equation (9) and plot e
against t for (a) x = 0, (6) x = ^'
2. When an accidental ground occurs on an aerial transmission line the
voltage 10 miles away from the generator station is found to be 60 per
cent, of the generator voltage. Determine the point of grounding.
CHAPTER VI
CIRCUIT CONTAINING DISTRIBUTED LEAKAGE CON-
DUCTANCE AND CAPACITY
A low-voltage cable may be considered as an approximate
representation of such a circuit, since it contains distributed
leakage conductance and capacity but usually low resistance and
inductance. Since the resistance and inductance are considered
negligible as a limiting case, it remains to consider a system of
parallel conductance and capacity. The voltage may be con-
sidered the same at all points of the circuit, that is, independent
of x.
Let i in Fig. 51 be the current at x, then i + ---dx is the current
at x + dx. Let C be the capacity in farads per unit length of the
conductor against the ground or neutral, and G the conductance
'i
-hdz
->
-
-,*
1
1
>
1
J
1
r\
= :
i -
-- :
_ <
> ~
<
<
*
> -
>
1-
>
<
! i
> ^
:=
> -
rr
\
'
1
'
FIG. 51.
in ohms per unit length of the conductor to the neutral. Hence
de
the current in the path of the capacity is Cdx-?-, and the current
in the path of the conductance is Gdxe. The difference in cur-
rent between the two sides of the element dx is dx. Therefore
dx
T- dx = Cdx -JT + Gdxe f
or
(1)
110
DISTRIBUTED LEAKAGE 111
This equation is similar to (1) in Chap. V with i for e, e for i,
C for L, and G for R.
As e is independent of x, equation (1) integrated gives:
K (2)
It is of no interest to consider short circuit of the cable, since
the resistance and inductance are neglected, for it would mean
a dead short-circuit on the generator. Therefore consider the
case of switching the generator on the open cable. Thus, where
x = 0, i = 0. .'. K = 0, and (2) becomes:
(3)
Case (a). Unidirectional voltage impressed on the cable.
In this case, it is assumed that e = E from i = to t = ,
but just before t = 0, e = 0. Therefore it is assumed that
= co just before t = 0, and -77 = 0, just after t = 0; that is,
equivalent to assuming that the fictitious condensers were charged
with an infinitely large current during an infinitely small period.
In reality, the rise of the impressed e.m.f. takes time, though
extremely short, and the resistance and inductance of the circuit
limit the initial value of the current and lengthen the period of
charging.
These assumptions thus do not allow a study of the transient
condition. The equation indicates simply that i = <*> , for t = 0.
de
For the permanent condition we have -rr = 0, and (3) becomes:
i = GEx (4)
which is expected.
Case (b). Alternating voltage impressed on the cable.
Let the impressed e.m.f. be e = E sin wt, and the time of apply-
ing to the cable be fa, thus e = E sin u>(t ti). Hence,
de
- -J7 = Eu COS 0>(2 ti).
Substituting these in (3)
i = [Cw cos u(t - ti) + G sin w(t - ti)]Ex,
112 ELECTRICAL ENGINEERING
or i = #zVC 2 u> 2 + G 2 sin (ut - wi + 0) (5)
where ft = tan- g^
This equation represents the permanent values of the current,
and shows that the current is proportional to x at all values of t,
and leads the impressed e.m.f. by /3 radians at all values of x.
The transient component does not appear in this equation, as
explained in case (a).
The transients will be studied in the following chapters, where
the capacity and inductance are considered.
CHAPTER VII
CIRCUIT CONTAINING DISTRIBUTED RESISTANCE AND
CAPACITY
In the study of the problems involving distributed inductance
and capacity, and the simpler problems involving the penetra-
tion of current or flux in conductors, etc., where alternating cur-
rent of sine shape is assumed, a certain differential equation,
given below is met.
Its solution is of importance to the engineer and deserves
consideration.
The equation is
^ - * ^ (i)
dx*~ ' at
A general solution which can readily be verified differentia-
tion is:
y = A Q + 2Ae ax+bt sin (ax + # + 7) (2)
or
y = A + Aie 0lX+6 " sin ( ai x + ft it + 71)
+ A 2 e a * x+M sin (a 2 x + fat + 72) + . . .
The evaluation of the different constants is accomplished
partly from the known conditions at some points of the system,
and partly by solving for the constants by differentiation and
substitution.
In most problems, y or its derivative is known at some point,
where for instance after permanent condition has been reached
y = Y sin ut.
If the point happens to be where x = x\, then equation (2)
becomes
Y sin ut = A + SAe ox i +w sin (aXi + fit +. T)
= Ao + SA'c M sin (#+/)
= Ao + A\e blt sin fat + T 'i) + A'*** sin fat + y'*) +. . .
113
114 ELECTRICAL ENGINEERING
Writing the equivalent of sines in terms of e
Jut _ f -jwt j(0it + y'i) _
'
Y-
J(P*t + y',) _ -j(ft*t + y'd
-- -+. .
it - j(0,t + y't)
+ . . .
Thus, since the left-hand member contains the imaginary only,
and the right-hand member a constant and the complex imagin-
ary and the two sides must be equal for all values of t it is evident
that AQ and the b's are separately equal to zero.
.'. y= 2Ae ax sin (ax + ft + 7) (3)
^ = 2Ae ax [a sin (ax + ft + 7) + a cos (ax + ft + 7)]
oX
*~ = 2Ae ax (a 2 - a 2 ) sin (ax + ft = 7) + 2aa cos (ax + ft + 7)
oX
^f = 2Ae ax p COS (ax + # + 7)
at
Substituting these values in (1) and equating the coefficients
for similar trigonometric terms,
a* - a 2 = 0, and 2aa = k z fi (4)
.*. (a -\- a)(a a) = 0, thus a + a = 0, or a a = 0, or both.
For a + a = 0, or a = a, the second equation of (4) gives,
2 2 = fc 2 /3, which is evidently impossible. Thus there remains
only a a = 0ora = a.
Then 2a 2 = k 2 P, or, a = a = fc -v/^ where a and a must
have like signs.
The general solution then becomes:
y = A,e ax sin (ax + ft + 71) + A 2 e~ ax sin ( - ax + ft + 72) (5)
and,
~ = A ie ax a[ sin (ax + # + 71) + cos (ax + ft + 71)]
- A z e~ ax a[ sin ( - ax + ft + 72) + cos ( ax + ft + 72)]
= V2 a [A ic * sin (ax + ft + 71 + | )
- Aae'^sin (- ax + # + 72 + ) (6)
RESISTANCE AND CAPACITY
115
Application of these equations will be found in the case of a
circuit of distributed resistance and capacity but negligible
inductance and leakage conductance, such circuit being approxi-
mately represented by the cable in Fig. 52.
Let R and C be the resistance and capacity respectively per
unit length of the cable. Let the distance be counted from the
receiving end of the line.
Let the voltage at B be e, and the voltage at A e -f- dx.
ox
.'. the voltage consumed in the line element is:
. de j de j
e + dx e = dx.
dx dx
di
Let the current at B be i and the current at A be i + dx.
ox
Receiving End
of Line
A
FIG. 52.
Thus the difference in current on each side of the line element is
.
i -f dx i = dx.
ox ox
de
or,
.'. dx = iRdx,
dx
de
= ^R
dx
(7)
The difference in current on each side of the element is the
charging current of the element.
or
. di , nj de
. . dx = Cdx
dx dt
di = de
dx dt
(8)
116 ELECTRICAL ENGINEERING
di 1 d 2 e
From equation (7) we get: - = >, ~~ 2
oX it oX
l_ d 2 e _ d^
' ' R dx 2 ~ di'
or,
d 2 e de
Referring now to the general equation, it is seen that k 2 = CR,
[RC0
and y = e, and a = a = \l~cT-
.'.e = A^s'm&x + ft + 71) + A 2 e~ ax sin(- ax + ft + 72) (10)
and,
1 de A/2a r . / TT\
* = Rdx= -ir[^ a *^(+ fl + Ti + y)-
A 2 e-*sin(- ax + ft + 72 + ^)] (11)
Case (a). Alternating current supplied to a circuit of distrib-
uted resistance and capacity.
Example No. 1. If the voltage at the generator end of the
line (x = e) is e = E sin cot, and if the cable is open at the
receiving end, then i = for x = and all values of t, and
e = E sin ut for x = e. From (10)
E sin ut = Aie al sin (al + ft + 7i) + A 2 ~ al sin (- al + at +72)
= [Aie al cos (al + 71) + A<*r al cos (- al + 72)] sin ft +
[Aie al sin (aZ + 71) + A 2 ~ al sin (- al + 72)] cos #
.'. Aie al cos (aZ + 71) + A 2 e~ al cos ( al + 72) = 1?
and |8 = w,
and Aie a/ sin (aZ + 7i) + ^2- z sin (- al + 72) = (12)
For x = 0, i for all values of t from (11),
= Ai sin (# + 71 + j) - A 2 sin (# + 72 + |
Since this must hold for all values of i,
Ai = Az = A and 71 = 72 = 7
Then from (12) we have:
E
t al cos (al + 7) + e~ al cos ( - al + 7)
and, e al sin (al + 7) -f e~' sin ( al + 7) =0
or,
RESISTANCE AND CAPACITY
f = tan" 1 I al _ al tan al , and
117
A =
e = A [e ax sin (ax
and.
= ^-~ A [e a * sin (
E
(13)
2al + -2aZ + 2 (COS 2 ttZ - SHI 2 (rf)
-f- 7) + e-*sm (- a* + a>* + 7)! (14)
(15)
I ax + a>c + 7 +
e~ ax sin f ax + a>2 + 7 +
Where A and 7 are given in (13) and
o- +^/^
If the voltage at the receiving end of the line were known rather
than the voltage at the generator then :
e = EQ sin ut for x = 0,
EQ being the maximum value of the voltage at the receiving end
of the line.
Thus,
EQ sin ut = AI sin (/ft + 71) + A 2 sin (/ft + 72)
= sin /ft(Ai cos 71 + A 2 cos 72) -(- cos /ft(A x sin 71 -f A 2 sin 72)
.'. AI cos 71 + A 2 cos 72 = EQ, fi = u and, AI sin 71
+ A 2 sin 72 = (16)
For x = 0, i = for all values of t (assuming again an open line).
.'. AI sin \8t -\- 7i + T) = A 2 sin (/ft + 72 + j)
AI sin /ft cos (71 + ^rj -f AI cos fit sin (71 + T) =
A 2 sin fit cos (72 H- jj + A 2 cos fit sin (72 + jj
In order that this shall hold for all values of t, the coefficients
of the similar trigonometric terms of t must be the same.
.*. AX cos (71 + T) = A 2 cos (72 + T-J and,
or.
AI sin (71 + jj = A 2 sin (72 + -
.'. tan (7! + j) = tan (72 + | j ,
7i = 7? = 7-
(17)
118 ELECTRICAL ENGINEERING
Then from (16)
(Ai + A 2 ) cos 7 = EQ and (Ai -f A 2 ) sin 7 =
.'.7 = and AI + A 2 = #o.
From (17) A , . A - --
^-1 /12 A . . A - ~
Therefore
771
e = __[ a* gm ( aa . _f_ ut ) + -ax gm (_ a:C
and,
where a = ~\~ \l~~n & n d .E'o is the maximum value of the e.m.f.
at the receiving end.
In the examples, both 1 and 2, the current leads the voltage
by 45 at all points of the line.
Let CQ be the voltage at the receiving end and e\ that at the
generating end.
From (14)
2E sin (co + 7)
e Q = 2A sm (wt + 7) = ,
V>' + *-** + 2 (cos 2 al - sin 2 al)
and
ei = E sin
From (18)
e Q = EQ sin
and
TJ
ci = ^[c ai sin (al + + e~ a ' sin (- aZ + )]
TTf
= -n[(e al + ~0 cos aZ sin co^ + (c ai e~ al ) sin ai cos
= 2^o e 2 ^ + ~ 2 ^ + 2 (cos 2 aZ - sin 2 al) sin (co* - 7).
Hence both equations show that, (a) the voltage at the re-
ceiving end leads the generator voltage by an angle 7, and (6)
the maximum voltage at the receiving end is:
_ 2 _
V* 201 + e- 20 ' + 2 (cos 2 aZ - sin 2 aO
times the maximum generator voltage. In fact examples (1)
and (2) refer to one phenomenon, but one terminal condition
already known and one terminal condition to be determined are
interchanged in the statements of the examples.
RESISTANCE AND CAPACITY 119
Example No. 3. The same phenomenon may be studied in
still a different way, namely, measuring x from the generator
end, that is, x = I refers to the receiving end of the line.
When the generator is taken as the point from which the dis-
tance is measured, then, as the voltage and current decrease as
x increases, we have:
dx = iRdx,
C/Jv
and di , de
- dx = Cdx
dx dx
which by a similar transformation, also resolves in the differ-
ential equation :
*_!?. _ rp de -
dx* ~ K dt
.*. e = A^ ax sin (ax + &t + 71) + Ax~ ax sin (- ax + pt + 72)
and,
For x I. i = for all values of t.
.'. A** 1 sin (al + fit + 71 + j) = Ax~ al sin (- al + pt +. 72 + |
or,
= A! ai [sin pt cos (al + 71 + ^) + cos # sin (+ al +
= A 2 e- ai | sin ^ cos ^ a^ + 72 + ^j + cos pt sin f a/ + 72 + 4) I
As this must hold for all values of t,
.'. Aie al cos (al + 71 + ^) = A*-" 1 cos (- al + 72
and,
Ai6 fll sin (a/ + Ti + |) = A 2 c- ai sin (- al + 72 +
.'. tan (al + 71 + ^j = tan ( al -f- 72 -f- |j
/. 72 = 7i + 2aZ,
and it follows that,
At = Aie (20)
120 ELECTRICAL ENGINEERING
For x = 0, e = E sin ut ;
.*. E sin cot = Ai sin (fit + 71) + A 2 sin (j3t + 72)
= (A i cos 71 + A 2 cos 72) sin pt + (Ai sin 71 + A 2 sin 72) cos fit.
In order to make this hold for all values of t,
|8 = CO
AI cos 71 -f- A 2 cos 72 = E,
and
Ai sin 71 + A2 sin 72 = 0.
From (20)
Ai sin 7! + Ai 2oZ sin (71 + 2al) =
.*. sin 71 = 2aZ (sin 71 cos 2al + cos 71 sin 2aZ),
- e 20 * sin 2aZ
' ^ = tan 1 + e-cos^
Let 7 = 71 + al, then 71 = 7 al, and 72 = 7 + aZ. And
let A = Aie z = A 2 e- z . Then,
pi
e~ al cos (7 - al) + c aZ cos (7 -f aZ) = -T (22)
From (21) and 71 = 7 a, we have:
sin (7 al) e 2aZ sin 2al
tan 71 = tan ( T - = COS(Y _ oi) - j + cos 2al >
.'. 7 = tan
i
r-?l- ( -l
_ e0( + e - ol
which is the same as that in example (1)
Substituting the value of 7 in (22)
V ^ al + e~ 2ai + 2(cos 2 al sin 2 aZ)
also the same as that in example (1).
Hence,
e = A[e~ a(l ~ x} sm( al x + wt + y) + c^^^sin (aZ x + ut -f 7)]
sn a
which are identical with the equations obtained in example (1),
only with (Z x) in the place of x.
RESISTANCE AND CAPACITY 121
It is noticed that at any particular point of the line the current
and e.m.f. waves are sine waves.
The wave length X is found when, x \^~ =
.*. X = X i = = 2ir\hnr
CRa) \CRu
mr/~ \fCR'
The time required for the wave to go one complete wave length
is H-
Thus the velocity of propagation is:
distance / w I Ifw /2o>
Thus the velocity of propagation is proportional to the square
root of the frequency.
Higher harmonics travel faster than the fundamental. The
third harmonic travels 73 per cent, faster, etc.
But while the higher harmonics travel faster than the funda-
mental their attentuation is greater as will be seen.
When the wave has traveled one complete wave length, that
is, when x = X = ^\J~ (
The exponential term becomes:
2
CRu
= e~ 2v = 4- = 0.0019.
That is, the wave is only 0.2 per cent, of its original value. It
has reached = = 0.368 of its original value when
CRu
~~ = l r x ~- =
2 / 2 /I
cT^ = VCRW = \Cte
Thus the third harmonic has decreased to 37 per cent, of its
original value in a distance which is only 58 per cent, of that
required by the fundamental to be reduced to 37 per cent, of its
original value.
122 ELECTRICAL ENGINEERING
To find the time for the wave to decay to - of its original
value, we have:
distance / 1 //TT
time = ; r? = -\/ ~^ -j- 2 A/T^: =
velocity \CRirf \CR
Thus the time required for a given decay varies inversely as
the frequency. The third harmonic requires only one-third of
the time of that of the fundamental.
Instance. A concentric cable 100 miles long. Assume a
capacity of 1 m-f. per mile to the neutral.
Using the mile as the unit of distance,
C = 10 6 .
Assume the cable to have a resistance (of one conductor) of 1
ohm per mile.
Then R = I
At 60 cycles, / = 60 and co = 377.
/2 X 377
.'. Velocity of propagation = A/ 6 = 27.500 miles per
sec. The velocity of the triple frequency wave would be,
\/3 X 27,500 = 47.500 miles per sec.
The main wave is reduced to 37 per cent, of its original value
after = = 0.00265 sec.; and the triple frequency wave is
reduced to the same fraction in one-third of the time or 0.0009
sec. In the first case the wave has traveled 73 miles; in the case
of the triple harmonic 42 miles.
Problem. Develop the equation of the voltage and the cur-
rent in a closed cable under alternating impressed e.m.f.
Case (6). Direct current supplied to the cable.
Example No. 1. Consider the line open at the receiving
end (x = 0).
Assume,
e = K + 2A ax+bt sin (ax + ft + y)
where x = 1, e = E for all values of t,
This is evidently only possible if
2Ae al+bt sin (al + pt + 7) = and K = E.
RESISTANCE AND CAPACITY 123
As this must hold for all values of t, p = 0, and al + 7 = nir. 1
7T
It is found convenient to let 7 = ~, that is, to make
z
cos al = 0, and al = mr + = i^L+ii*:
where 7 TT STT STT 7?r
ai== 2 ' "2" T T' etc "
Thus, e = E + ~SAt ax + l " cos
f) f*
. 2 = a ae ax cos ^ _ aae ax sn
ot
Substituting these values in the general equation
d*e de
.W Tt
and equating the coefficient of similar trigonometric terms we
get:
a 2 -a 2
~CR~
and ax = or a = 0, since a. can not equal zero.
'eosia+M? (23 )
1 sin v ' ' (24)
K dX K n = o 2i
When * = 0, x < 1, e =
n =
1 In this equation appear several constants, some of which are determined
by the terminal conditions, others by mathematical transformations. It is,
of course, possible to do a certain amount of choosing as long as the choice
satisfies the differential equation as well as the known conditions which exist
in the problem. So, for instance, we may assign an arbitrary value of
7 and carry the calculations through when we may find that the final expres-
sion is simple or too complicated to be of practical value.
It is reasonable that in the first trial 7 may be assumed as zero. When
the problem is worked out on this basis it is seen that the answer is not sus-
ceptible to a simple equation. The trial will suggest another value, most
likely 7 = - |. This is therefore used.
124 ELECTRICAL ENGINEERING
In order to determine the values of A n , multiply both sides
of (24) by cos - dx and integrate between and e, thus
C
Jo
e 7r(l + 2k}x n =~ . 7r(l + 2n)z,
cos ^ J - 2, A n cos- -7^7 - dx
21 n =o 21
Each term on the left-hand side equals zero except that one
which has n = k, and hence this particular value is used, and we
have
A, cos * * <fc = - E
Integrated,
coB (25)
TT n = i n 2/
sin - -^-- (26)
Lli n = i ^t
The voltage at the receiving end is:
For i = 0, < = _ 4""" (-
, TT2?
which is zero, in accordance with the assumption made in develop-
ing the equation.
Therefore, incidentally, we get
which is a known interesting series from which the value of TT
can be computed.
The current at the generating end is
(27)
when t = and x = I,
RESISTANCE AND CAPACITY 125
which is a limiting value never reached, since with the slightest
increase in t the series converges very rapidly.
7T 2
For the sake of briefness, write ra for ~p7^r 2 then, for x I,
2E
i = Jft(e~ mt + e~ 9w ' + *~ 25m ' + e~ 49w< + . . .) (28)
From (28) it is readily seen, that, when t has any appreciable
value, the current dies out approximately according to the
exponential, "*'. When the line is very long, the initial large
current will remain during a considerable length of time. When
I is very small, the limiting case is that of concentrated capacity.
As I = (28) approaches:
-*-*
HI is the resistance and Cl the capacity of the entire line.
In the case of concentrated resistance and capacity it has been
shown that
E - l t
Comparing the equations it is seen that the transient current
can be fairly well approximated by assuming that the line capac-
ity is concentrated in the middle of the line.
Example No. 2. In case the line is grounded at the receiving
T^
end, the permanent voltage is -T--
Thus, _ Ex
I
7 may in this case be conveniently taken as zero, thus,
Ex -^ t
- 2Ae CR sin ax
for x I, e = E for all values of t.
Thus sin al = 0, and al = rnr
/. e =^ + T^-"^' S in^
1 n =
For t = 0, e = for all values of x < L
Ex
sm T T
126 ELECTRICAL ENGINEERING
A n is determined as before by multiplication and integration
and we get finally:
. sin (29)
(30 )
J-l/lr il/l/ n _, J
The current at the receiving end is:
1 =s CD
* -,
2 (31)
CHAPTER VIII
DISTRIBUTED INDUCTANCE AND CAPACITY
Permanent Condition. Let L and C (Fig. 53) be respectively
the inductance and capacity per unit length of the circuit.
The voltage consumed in the line element dx is:
de , , di ,
- dx = L dx (1)
ox dt
_J Receiving End
of Line
The difference in current between two sides of the element is :
(2)
Differentiating (1) with respect to x
d 2 e di
_ _, j ^
dx* ~ dtdx
Differentiating (2) with respect to t,
~dxdi == C ~di*
d%_ _ M
' dx* di
dx~* = LC W
127
Similarly,
(3)
(4)
128
ELECTRICAL ENGINEERING
In this problem there is, therefore, encountered an equation of
the following type:
=
dt 2 dx 2
(5)
An often successful procedure for finding particular solutions
of simple partial equations of this or similar simple types, is to
assume the solution to be:
y=UV (6)
Where U is a function of t only and V is a function of x only.
Differentiating, we get:
" = U
dx 2 dx 2
d 2 y _ d 2 U
dt 2 ~ dt 2
Substituting (7) in (5)
^_^^Z or J_^
dt 2 " U dx 2 Uk 2 dt 2
1 d 2 V
V dx 2
(7)
(8)
Since the left-hand member is a function of t only and the
right-hand member a function of x only, it follows that each side
of the equation must be equal to the same constant. Let that
constant be a 2 .
and
d ' V 2T7
- = a 2 V
dx 2
(9)
The following trigonometric terms evidently satisfy (9) .
U = sin akt or U = cos akt
V = sin at or V = cos at
Thus the solution is:
y = K -j- S[Ai sin ax sin afcZ -f- A 2 sin ax cos fccrf +
As cos ax sin aAtf + A 4 cos ax cos
Where AI, A 2, A 3 , A 4 and K are to be determined, and k =
and the S sign refers to summation with all possible values
(10)
(U)
1
of
a.
INDUCTANCE AND CAPACITY 129
Consider now the specific case of an open alternating-current
line of negligible resistance and leakage. Determine the values
of the current and e.m.f. at any time at any point of the line after
the permanent condition has been reached.
If the distance is counted from the generator end the generator
voltage is e E sin ut
then di de
-dx dx = +C di dx
and, de di
The final differential equation becomes, the same as equation
(3).
The conditions for open line are:
for x = 0, e = E sin ut
for x = I, i = for all values of t.
Since we are dealing with permanent condition the current and
e.m.f. vary with fundamental frequency and the solution is
therefore :
e = A i sin ax sin kat + A 2 sin ax cos kat +
A s cos ax sin kat + A 4 cos ax cos kat
i = A 5 sin ax sin kat + A 6 sin ax cos kat +
A-j cos ax sin kat + A 8 cos ax cos kat (15)
These are related by the equation:
de _ di
dx~ ~ d~t
- = a[A i cos ax sin kat + A 2 cos ax cos kat
(j 30
AS sin ax sin kat A 4 sin ax cos kat].
L = LkalAz sin ax cos kat A 6 sin ax sin kat +
of
A? cos ax cos fca A 8 cos a sin fcorf].
Equating the coefficients for similar trigonometric terms, of t,
AI cos ax As sin a = Lfc[ A 6 sin ax A 8 cos ax]'
and,
cos ax A.4 sin ax = Lk[A b sin ax + A 7 cos ax]
(A)
130 ELECTRICAL ENGINEERING
Since these must hold for all values of x, we can substitute
ax = and ax = ~>
:. A l = LkA B (16)
A 2 = - LkA 7 (17)
and, -A 9 = LAA 6 (18)
-A 4 = - LkA b (19)
For x = I, i = for all values of t,
.' . = A 5 sin al sin kat + A 6 sin al cos fcotf +
A 7 cos sin fccx -f A. 8 cos cos fc (B)
.'.A 5 sin a? + AT cos a/ =
.'.A e sin al + A 8 cos al =
.'.A 7 = - A 5 tan Z (20)
/.A 8 = - A 6 tan al (21)
.'. e = LkAs sin ax sin fcotf LfcAr sin ax cos fcZ -
LkA 6 cos ax sin kat + LfcA 5 cos ax cos fcaL
For a; = 0, e = E sin ut
.'. # sin ut = LkA & sin kat LkA$ cos /ca^,
W
.' . E = LkA & , or A 6 = yr-' A 5 = 0, and co = ka.
From (19) and (19) A 3 = E
and A 4 =
From (20) and (21) A 7 =
W
and, A 8 = yy tan al
From (16) and (17) A l = 7 tan aZ
and A 2 =
Therefore, A^ = E tan al
A 2 =
A 3 = #
A 4 =
A 6 =
A
A 7 =
A 8 = TT ^ an a ^
i>/C
e = E tan a? sin ax sin fcatf + E cos ax sin
INDUCTANCE AND CAPACITY 131
' i = -- v-r sin ax cos kat + -f-p tan at cos axkat.
LK LK
Substituting, , _ 1 _ _
" VLC *
e = #[tan ul-\/LC sin co\/LC sin co + cos u\/LC x sin coj],
rr
v-~
or,
e = E sin w<[tan ul^/LC sin u\/LC x + cos u\/LC x],
/ _ _
i = E-l~ cos co^ftan coZ\/LC cos u\/LC x sin w\/EC a;
rr
t = v-~[tan wl-\/LC cos w\/LC x cos co sin u^/LC a; cos coZJ,
or, cos co \-LC (Z a;)
e = # sin coi - , (22)
cos co v LC t
C ^sin
cos coi - (23)
The voltage at the end of the line is :
E sin co^
Example. If the receiver voltage, instead of the generator
e.m.f. is known and if the distance had been counted from the
receiving end of the line, then
i = 0, for x =
and, de _ di
d~x~ ' L dt
Thus the signs for A 5, A 6 , A^ and A*, in equation (A) would
have been reversed.
:.Ai = - AsLk.
A 2 = + A 7 Lk.
A 4 =
Equation (B) would have been:
= A 7 sin kat + A 8 cos kat, .'. A 7 and A 8 =
.'. e = LkAs cos ax sin kat LkA & cos ax cos fc
For c = 0, e E Q sin ut.
.'. EQ = sin co = LkA 6 sin kat
.'. EQ = LkA 6) and A 5 = and o
.". e = EQ cos ax sin fca
Z = s * u ax cos ^ a ^
132 ELECTRICAL ENGINEERING
or, e = EQ cos to\/LC X sin coi (25)
i = Eo+Lr sin u\/LC X cos WT (26)
\L
Therefore the generator voltage is:
e = cos u-\/LC IE Q sin cot = E sin co,
cos co \/LC I
and (25) becomes:
cos co \/LC x
e = E sin coi - , ?
cos co \/LCl
which is obviously identical with (23) as obtained before.
It is seen at once that the receiver voltage is
1
cos co \/LC I
times that of the generator e.m.f. As the cosine is always less
than unity except I = 0, the receiver voltage is always greater
than the generator e.m.f.
Therefore the receiver voltage would approach infinity, when
2wfVLCl = I
, J_ _1 JL
_
4(Ll)(Cl) 4L Co
that is, when the natural frequency of the line and the frequency
of the impressed e.m.f. coincide.
The wave length is X = /
WV.L/O Co
X 2 ^/ 1
Thus the velocity of propagation = = /, - == /, ^
1 coV^o^o V-^o GO
If the inductance inside of the conductor is negligible, then the
velocity becomes that of light = 188,000 miles per second. In
reality it is somewhat less.
So for instance in a transmission line consisting of No.
B. & S. wires, 18 in. apart,
L = 1.6 X 10~ 3 henrys per mile.
C = 0.019 X 10~ 6 farads per mile.
-4= = 182,000 miles per sec.
INDUCTANCE AND CAPACITY 133
For short distances,
sin co -\/LC x = u-\/LC x
cos co \/LC x = I
.*. e = Eo sin ut
Ic
i = Eo -^1 co -\/LCx cos
jC/o
cos co = cos co
X c
where x c is the capacity reactance of length a of the cable. It is
seen that the current in time phase leads the voltage by 90.
Transient Condition. When a steady voltage is impressed
upon the circuit.
DR. FRANKLIN in his book on waves and his paper before the
A. I. E. E. of April, 1914, has approached the subject from a most
simple and instructive point of view and has been able to make
some generalizations which are of great value.
He shows that whatever the distribution of the current or
e.m.f. in a travelling wave along a transmission line there must
be a fixed ratio between the instantaneous values, which ratio is
C
Y when the line resistance and leakage reactance are negligible,
and it can be represented by a somewhat more complicated ex-
pression when they are taken into consideration.
His reasoning is briefly as follows:
If the current in an element of the line is i the magnetic flux
in the area a, b, c, d, Fig. 53, is Lidx.
If the current wave progresses toward the right with a velocity
V the time required for the flux to sweep past be is -y] thus the
e.m.f. induced along be is ^ x = LiV.
y
Similarly if e' is the voltage in the line element then the charge
on ab is e'Cdx.
This charge flows past the point in time y, where V is the
velocity of propagation of the e.m.f. distribution; thus
., _ Q _ Cdx
1 t = '' dx :=CeV '
V
134 ELECTRICAL ENGINEERING
In order then that these distributions shall sustain each other,
i = {', e = e r and V = V.
*
.'. e = LiV and i = CeV
or
and
Ce' - Li- or,^ = + -
e ' \L
The -f sign belonging to outgoing waves, arid, the sign to
the reflected waves.
. i 1C , i' 1C i i'
. . - = + A / T and -SB - /-- or-= -
e \L e \L e e
where index ' refers to the reflected waves.
When the line is open at the receiving end the sum of the in-
coming and reflected current waves must be zero, thus i -\- i' =
, ., e i . c' = + e
. . e' = i' -. = -f- -e = + e .
i i
When the line is short-circuited at the receiving end, e + e' =
.'. e f = e. Thus i' = i.
When the receiving circuit is noil -inductive and of resistance/^,
e + e' = R(i + i'}
but e 1L e
substituting these values above, then,
(R - a) . a - R
e = e ~^r- t and i = i ^*
R + a a + R
It is seen that the reflected current and e.m.f. waves may be
positive, zero, or negative, depending upon the relative values
of R and .*/-
In DR. FRANKLIN'S American Institute Paper (April, 1914) is
given a very full discussion of the nature of these reflected waves
and some highly instructive diagrams are shown.
For example, when the receiving circuit is inductive the line
acts at the instant of reflection as if it were open circuited, since
the current can not rise instantaneously in an inductive circuit.
After some time the condition becomes that of a non-inductive
INDUCTANCE AND CAPACITY 135
receiving circuit, discussed above (since we are dealing with
direct-current voltage). Between the two periods of time the
current and e.m.f. change according to a simple exponential
law.
Of special interest is the condition of the waves when the line
constants change. DR. FRANKLIN illustrates this condition in
the case of an overhead line connected to a cable system.
Let i y i r and i t be the instantaneous values of the outgoing
current, the reflected current and the transmitted current, and
let e, e r and e t be the corresponding values of the e.m.f.
Then e + e r = e t
i + i r = it
e e r e t
* - = - = a ~ = b
+ I Ir It
From these equations are found
b -a
It is of interest to apply these simple relations numerically.
Assume that the inductance and the capacity of a cable sup-
plying power to an overhead line are: L = 0.0002 henrys and
8
C = ~Q farads per mile, and that the corresponding constants
for the overhead line are LI = 0.0015 henrys and C\ = ~T^
farads per mile.
dllCl -, ; \J*\J\J1.U /, ~ v_ i
If therefore such cable-overhead line combination is connected
to a source of steady e.m.f., e, the voltage at the junction as the
548
wave reaches it will be e t = oo? o = 1-88 times that at the
.6OC7.O
136 ELECTRICAL ENGINEERING
generator. Should the overhead line be open at the receiving
end the voltage will be doubled as the reflected wave starts on
its journey back. Thus as a maximum at the junction the volt-
age would equal 3.76 times the impressed value.
The mathematical solution of the problem is given in equation
(11) which can be written in the following way:
e = K + ZA sin ( ax + kat + 7) (27)
where
=
+ a applies to the waves issuing from the generator and a
to those going toward it. From the expression ax + kat, it
is seen that the waves of all frequencies travel with the same
velocity, +k or k where the signs indicate the direction of
motion.
It will be shown that in the case of an open line connected to a
source (of negligible resistance) of undirectional voltage, four
waves have to be considered before the cycle repeats itself.
First the outgoing rectangular wave of value E which begin-
ning at the generator progresses toward the open end of the line.
Second the reflected wave also of strength E which returns from
the open end toward the generator which with the initial wave
gives a wave of double voltage. Third a negative wave of
strength E which progresses from the generator toward the
open end of the line, which wave is necessary in order to maintain
the generator voltage E. Fourth the reflected wave of the
negative wave which is of strength E and which progresses
toward the generator.
Consider now what happens at a point located say at one-fourth
of the length of the line from the generator.
If the time required for the wave to reach the end of the line
is T, it is evident that during Y T there is no voltage at the point.
After that time the voltage remains constant at a value E until
the first reflected wave arrives. This occurs evidently when t =
1%T. Thus between t = ^ and t = 1.757 7 the voltage at the
point is E.
From that on it has a value of 2E until the negative gener-
ator wave reaches the point which occurs when t = 2T -f-
T
- = 2.25 2 T . After that time the voltage has a value of 2E -
INDUCTANCE AND CAPACITY
137
E = E until the reflected wave of the negative wave arrives
T
which is when t = 4T - -j = 3.75 T 7 . Then the voltage = 2E -
2E = 0, and it remains zero until a time t = 4.25T 7 when the
voltage again equals E and the cycle is repeated.
The result is the wave shown in Fig. 54.
E.M.F, Wave
E
',.,
5 1
2
.0 1
5
2
2(1
-X)
2
5 3
2
3
X
.5
J.O
2(1 -x)
4
5
k
A
V
fc
Q.
r,i
A
k
A train of waves would pass the point indefinitely since we
have neglected the energy loss in resistance. The wave length
is evidently four times that of the open line.
Consider now the current wave of Fig. 55.
As successive equal elements of the line are being charged to
voltage E a constant current has been shown to flow from the
generator while the voltage wave progresses toward the end of
Current Wave
l-x
Jc
1+
X
k
3i-
X
**
t
Ti
31 4
X
k
-J
FIG. 55.
the line. At the end the current must be zero, therefore the
reflected current wave must be equal but opposite to the incoming
wave. The reversed current reaches the generator after a time
2T, when the current becomes zero. After that time the genera-
tor supplies E voltage and a negative wave of current flows
until it also is neutralized by the reflected current which occurs
when t = 4T.
Consider the current at the particular point mentioned above.
138 ELECTRICAL ENGINEERING
T
From t = to t = -j no current flows. After that the current
is constant until the reflected current reaches the point (at t =
1.75T) when it drops to zero. It remains zero until the negative
current issuing from the generator reaches the point fat t = 2T
T\
-f -T) . Then it becomes negative and remains negative until the
negative reflected current now positive reaches the point It =
T\
4T -j] t when it again is zero and so forth.
In general centering our mind on a particular point x, from the
receiving end of the line there is no e.m.f. or current at that point
until t = JT . After this the voltage is E, the generator voltage
for some time. At t = T, the waves reach the end of the line and
I + x
reflect, therefore after t = j- , e = 2E for a period of time. At
21
t = -j-, the waves return to the generator end. In order to keep
the voltage at the generator end constant at E the generator must
now begin to supply E. Therefore after time t = ]T~~> e at x
becomes 2E E = E] and after t = -W , e at x becomes 2E
41
2E = 0. At t = -r the generator reverses its voltage from E
to +E again, and the voltage at x repeats its cycle again and
again.
Referring now to equation (27)
e = K + SA sin (ax + kat + 7) + SA' sin (- ax + kat + 7') (28)
when x = I, e = E for all values of t.
:. 2A = SA' and sin (al + 7) = - sin (al + 7')
*v' 'Y ~4~ TlTT
= sin (mr al + 7') thus a = - ^
where n = is an odd number (29)
and K = E
:. e = E + SA[sin(aa? + A;a/ + 7)+sin(- ax + kat + y')] (30)
INDUCTANCE AND CAPACITY 139
At the receiving end of the line (x = 0).
e = for all values of t which are less than T.
But when t = T the voltage is 2E.
/c
Thus t = is a transition point, a point of discontinuity, e is
either or 2E. Substituting the two values in equation (30)
we get:
+ E = 2,4 [sin (al + 7) + sin (al + 7')] respectively (31)
At the open end of the line where there is complete reflection
the incoming and outgoing waves are identical
.'. 7 = 7'.
Thus from (29) a = -~j where n is an odd number.
.'. (31) becomes + E = 22A sin ( + 7) (32)
In the development of the trigonometric series it is found that :
4 n=c sin nO
+ 1 = - - S - where n is an odd number (33)
where the negative sign refers to values of 6 between ?r and 2w and
the positive sign to values between and TT or TT < 6 < 2ir for
negative sign. < 8 < TT for positive sign.
See "BYERLY'S FOURIER'S Series and Spherical Harmonics"
(page 51). Comparing equations (32) and (33) it is evident that:
_ E 4 1
An ~ 2 IT n
and nw
y + T = nfl.
It remains to determine the value of 6. The two series have
TT and 2-K or in common. It remains to choose the proper value
of these.
If TT were chosen the signs in (32) and (33) would be reversed,
if however, or 2ir is chosen, the signs are satisfied. Thus
mr nir
+ 7 = 2n?r or . . 7 == -
140 ELECTRICAL ENGINEERING
Thus equation (30) becomes:
e = E + j- -S^Fsin ^ (x + kt - 1) + sin ~ (- x + kt - I)] (34)
a Tt H L 1 &l J
and since de di^
dx ~ dt
2 i[sin ^ (a + fc - - sin ^ (- s + & - Z)] (35)
The curves drawn in Figs. 54 and 55 may be verified as follows :
=
for t <
x = T
Z -x
k
x -\- kt I < x -\- I x I or smaller than
.'. 6 in the trigonometric series lies between TT and 2ir.
Therefore S - sin (a; + kt - Z) = - .
Tl 4
Consider with the second term in (34), x + kt I < x+l
31
x I, thus smaller than 2x or smaller than 2 + -r or 1.51
thus 6 is again negative and the series of the second term in (35)
-
-7
kt l = X-\-l
adds up to -
.'. e = E -\- ^ - ( -7 jj =0 which agrees with the curve.
When k
= -- = =
we are in the first quadrant
and - x + kt - I = - x + I - I = - x = -
6 lies in the fourth quadrant
.*. 2J- sin ( x -\- kt 1) = 2"
.'. e =1 which agrees with the curve.
07
F or j = _ k + kt-l=*x + 2l l = x + l= 1.751.
6 lies in the second quadrant- thus:
INDUCTANCE AND CAPACITY 141
and - x + kt - I = - x + I = 0.25Z.
6 lies in the first quadrant, thus :
Z^sin (-x + kt - Z) - +
TV 71
.'. e = 2# which agrees with the curve.
The current wave may similarly be checked. When for
instance t = T, it is readily seen that the algebraic sum of the
trigonometric terms become j
E 4 1C TT 1C
i = ?r ~\ IT o = E \ T which agrees with the curve.
A 7T \-L/ A \ Li
It is thus seen that when considering the outgoing waves only the
i JC
relation between the current and e.m.f. waves must be - = \ T>
e \L
the equation also shows that when considering the reflected waves
The effect of the line resistance is to taper the waves so that
instead of their being represented as a ribbon of parallel sides
the sides slant toward each other; thus the reflected e.m.f. wave
is not as great as the original wave, and the line soon reaches a
state of permanent condition.
In reality the wave front is not vertical but slants and the
corner is rounded off, due to the skin effect of the conductors.
The higher harmonics of the current meet a much higher resist-
ance than do the lower, and hence the resistance is not a constant
quantity but different resistances should be assumed in connec-
tion with the different harmonics.
The mathematics involved becomes, however, altogether too
complicated for any practical application. The important
point is that if the values of the waves are determined in a cir-
cuit having no resistance, the most pronounced variations in
current and e.m.f. are discovered.
A circuit having no resistance and no leakage is said to produce
pure waves the characteristics of which are, as has been shown,
such that
142 ELECTRICAL ENGINEERING
That is the electric energy is always equal to the magnetic
energy.
The wave may, however, be pure even if there is resistance and
leakage but in that case the energy dissipated in heat per unit
length of line must be equal to the energy dissipated by leakage
in the electric field.
e 2
.'. i 2 R = 77- where R i is the leakage resistance per unit
length.
/?/? e2 L
' ' RHl == i* C
A line in which this relation exists is called a distortionless
line.
For a full discussion of such circuit the reader is again referred
to DR. FRANKLIN'S book on waves.
CHAPTER IX
DISTRIBUTED RESISTANCE, INDUCTANCE, LEAKAGE,
CONDUCTANCE AND CAPACITY
Let R, L, G and C, Fig. 56, be the line constants per unit
length of the line, K being expressed in ohms, L in henrys, G in
ohms and C in farads.
The voltage equation is evidently
or,
or,
de . di
dx = Rdx i -\- Ldx --
ox ot
de . di
dx dt
dx = Gdxe 4- Cdx TT
ox dt
(1)
(2)
R L
FIG. 56.
Differentiating (1) partially with respect to x and (2) with
respect to t and combining the results with (1) and (2) we get:
^ = LC d ~ + (RC + GL) ~ + RGe
OX" Ol" Ol
and,
(3)
(4)
where a' and & may be positive or negative, real or imaginary,
simple or complex.
143
The general solution of these equations is
144 ELECTRICAL ENGINEERING
Substituting the general solution in (3) or (4) and equating
the coefficient, we get
a' 2 = LCp'*(RC + GL)ff + RG (5)
Substituting a + ja for of and b + JP for ff in (5) and separat-
ing the real and imaginary terms, we have
a 2 - a 2 = LC (6 2 - 2 ) + (RC + GL) b + RG (6)
and 2aa = 2LC6/5 + (RC + GL)/? (7)
A slight consideration shows that the exponential solution
given above can be written
e = k + ZAt* ax bt sin (pt ax + 7) (8)
If now for the sake of simplicity only the permanent condition
is considered we get
e = k + SAc ax sin (pt a* + 7) (9)
If as a further limitation the current and e.m.f. are assumed
to be simple sine functions, depending in time upon the impressed
frequency, then p has only one value <o. From (6) and (7) follows
then that only two values of a. and a exist, one being positive the
other negative
.'. e = Ait ax sin (pt + ax -f 71) + A 2 e~ ax sin (pt - ax + 73) (10)
In this equation, one term represents the sum of the outgoing,
the other the sum of the incoming waves.
If the line is open at the receiving end then the beginning value
of the reflected waves must be identical with the final value of
the incoming waves when x = 0.
Thus under this condition for x =
A l sin (pt + 71) = At sin (pt + 72)
Since this must hold for all values of t
' 7i = 72 and A\ = A%
If the voltage at the generator end is E sin o>/, then
E = sin ut = A 1 [e al sin (pt + al + 71) + t~ al sin (pt - al + 7^]
which by simple transformation becomes
E sin at = A i (sin pt [e al cos (al + 7i) + e~ al cos (- al - 71)]
+ cos pt[e al sin (al + 71) + e~ al sin (- al + 71)]}
+ c- 20 ' + 2(cos 2 Z - sin'aZ) sin (pt + B)
DISTRIBUTED RESISTANCE 145
where , _ _ e al sin (al + 71) + e~ al sin ( al + 71)
al cos (al + 71) + e~ al cos ( al + y^
Thus 13 = co and =
#
= V^M- <r 2 <" + 27cos 2 ~al - sin* of)
Since =
e al sin (al + 71) + <r al sin (- aZ + Tl ) =
which gives e al e~ al
71 = tan" 1 aT^~^. ^ an ^
Equation (10) is now completely determined.
-Q- + Gc = Ai{Cw[e oa! cos (w + 0-0: + 71) + e~ ax cos (co/ ax -f-
71) + (7 [e a * sin (co^ + ax + Tl ) + ~ oa; sin (co - ax + 71] }
= AI \/C 2 oo 2 ' -|- G 2 [e ax sin (co^ -f- ax -j- 71 ~h ^) ~h
e -ax g j n ^^ _ ax _|_ ^ i _|_ ^J Qjj
where . Cw
<p = tan" 1 ^
Let i now be the permanent component of the current, and
assume :
di>
= Biae ax sin (cot + ax 4- ^>i) Bia*~ ax sin (co a
ox
+ Biae ax cos (co^ + 0:0; + <^?i) Biat~ ax cos (coi ca
z sin (co ax + <^ 2 + <r) (13)
,
" 1 -
a
According to (2), (11) and (13) must be identical, and hence,
where ,
<r = tan" 1 -
a
Pi -f <r == 71
To sum up,
E' [ e ai sin (co^ + ax + <p t ) + ~ oa; sin (ut -{- ax + 71)]
A/c 2ai + ~ 2al + 2 (cos 2 al - sin 2 aZ)
(15)
'-*>pyf?
[ e az sin (w^ + aa; + 71) e~ oa; sin (coi ao: +
V^ + e~ 2al + 2 (cos 2 al - sin 2 ^)
146 ELECTRICAL ENGINEERING
where
[ e al f al
^r+7^ tan a
1
a = tan" 1
a
. Ceo
<p = tan- 1
(Pi = <P + 7i
and,
a =
+
the latter two values being determined from (6) and (7) by let-
ting 6 = and (3 = w,
These solutions apply when the transient terms become negli-
gible, i.e., when t is large enough to make e~ bt comparatively
small.
Case (6). Direct-current distribution in an open line. Con-
sider the line open as before. For the permanent component of
the solution, i.e., a solution which applies after the line has been
switched to the generator for a sufficient length of time, the equa-
tion can be derived as follows:
Referring to equations (6) and (7), b is zero, when only the
permanent component is considered, and ft is also zero, as there
exists no periodic phenomenon, when the impressed voltage is
constant and when the starting phenomenon is reduced to negli-
gible magnitude.
Substituting 6 = and ft = in (6) and (7) we get:
a 2 - a 2 = RG and 2acx = 0,
from the latter, either a or a must be zero, while from the former
a can not be zero, since a itself must not be imaginary.
/. a = and a = \/RG.
Let e f be the permanent component of the voltage and assume:
e r = A^ x + A 2 e~ ax (18)
where x = I, e' = E
l (19)
DISTRIBUTED RESISTANCE 147
Let i be the permanent component of the current. According
to (2),
Substituting (18) and (20)
-
Integrating, ^ = G ^ ox _ A ^_ ax ^ + R (21)
According to (1) de' . di'
Tx~ Rl+L ~dt
Substituting (18) and (21) in (2),
(A * - A e )
a
Since a 2 = RG, K = 0, and (21) becomes:
I---WI--A* )
where x = 0, i r = 0, .'. AI = A 2 .
From (19) . . E
1 2 ~ ai I e ~ al
Therefore, ^ __ ^ e ' + ~ aj;
^ j-^ (24)
where a = + *\/RG, these equations apply when the transient
terms become negligible.
10
CHAPTER X
PERMANENT CONDITIONS WHEN ONE OF THE FOUR
CONSTANTS, R, L, G, AND C IS NEGLIGIBLE
I. R = 0.
Case (a). Alternating current: The solutions are given by (14)
and (15) in the previous chapter, but in this case,
a=
andj
Case (6). Direct current: Referring to (23) and (24) in the
previous chapter,
a =-- ^/RG =
/. e' = E.
Under this condition the equations deduced give $ in the case of
the permanent current. Thus they do not lend themselves to the
determination of the current.
II. G =
Case (a). Alternating current: With
a = + \/-7r 1+ \/L 2 6o 2 + R z Leo],
and
equations (14) and (15) in the previous chapter give the solution.
Case (6). Direct current: From (23) and (24) in the previous
chapter,
e' = E and i = 0.
III. L = 0.
148
PERMANENT CONDITIONS 149
Case (a). Alternating current: In this case
and
1+ VG 2 + C 2 co 2 - <?]
Case (6). Direct current:
f ax I -02;
e' - E--
* -
and,
al ~ 6~
IV. C =
Case (a). Alternating current: In this case,
and
Case (6). Direct current: Same as III.
CHAPTER XI
THE DISTRIBUTION OF FLUX OR CURRENT IN A CYLIN-
DRICAL OR FLAT CONDUCTOR
The general reasoning and the mathematics involved in the
study of flux or current distribution in conductors is very simi-
lar to that involved in the study of propagation phenomena in
transmission lines. It is therefore included in this part of the
book even though it is again and more fully considered in a later
chapter, where the subject is approached from a different point
of view.
Distribution of Flux in Cylindrical and Flat Bars. When a
cylindrical bar is magnetized by a winding surrounding it, the
etizing Winding 1
FIG. 57.
flux of final flux density corresponding to the external m.m.f.
appears at the surface nearest to the magnetizing winding.
At a distance from the surface of the bar, the flux density is
less than that at the surface, because as the flux penetrates the
inner layers of the bar, it induces a voltage in the outer layers,
which causes a flow of current that produces m.m.f. of a direction
more or less in opposition to the external impressed m.m.f.
Referring to Fig. 57, consider a concentric tubular element of
150
DISTRIBUTION OF FLUX 151
thickness dx and mean radius x, then another of thickness dx but
mean radius x + dx.
Let <{> be the flux in the tubular element of radius x, and <f> + d<fr
that of radius x + dx. Thus d</> is the increment of flux in the
tubular element, as x increases from x to x + dx, but the total
flux in the tubular element is 0.
</> is the result of the external m.m.f . and the m.m.f. (demagnet-
izing) due to the current between x and XQ', </> + dcf> is the result
of the external m.m.f. and the m.m.f. due to the current between
x + dx and XQ. Therefore d<j> is caused by the decrement of
demagnetizing m.m.f. due to the current between x and x + dx,
i.e., within dx.
Let i be the current density at x, then the current within dx is
ildx, and the m.m.f. due to it is also ildx, as the number of turns
is unity (7 being the length of the cylinder).
Let B be the flux density at x then dB the increment of flux
density as x increases from x to x -\- dx. Thus d(j> = 2irxdxdB.
m.m.f.
Since flux = 0.4 TT
and the reluctance in this case is
-=
reluctance
I
2ir X dxu, *
OAirildx
We get d<p = 2>jr X-d X dB = - ,
thus dll = . ,^
If p is the spec, resistance
then the resistance that the current within dx meets is I 7.7
and the e.m.f. consumed by the resistance = ildx I yr~ = 2-jrpxi.
Let e be the e.m.f. induced in the circle of radius x, and e + de
that in the circle of radius x + dx.
As no external e.m.f. is applied around the circle of radius x
the sum of the consumed and the induced e.m.fs. is zero, thus:
e + Zirpxi = (2)
Substituting (2) in (1)
UlJ v/.i/iyuc/ /Q\
" - ^r W
152
ELECTRICAL ENGINEERING
e is induced by all the flux within the circle of radius x, and e -f-
de by all the flux within the circle of radius x -f- dx, thus de is
induced by all flux in the tubular element 2irxdx, which is
1 d<t>
according to our notation. Hence de = ^c s -77 or using partial
differentials,
de -
10 8 dt
or
or,
0.2-n-xdxdB
de
10 8 dt
2irx dB
io 8 l>i
(4)
Equation (3) may be written:
=
Differentiating with respect to x,
d 2 B dB
X r -f - =
dx 2 dx
Combining (4) and (5), dividing by x
d 2 B I dB 0.47rju
dx 2 " + x dx = : T0 8 P
Q.47TM de
+x>
dx
dt
(5)
(6)
A long thin flat bar may be considered as a
cylindrical bar of infinitely small curvature or
infinitely large radius, thus x considered as the
radius becomes infinity andeauation (6) becomes:
d*B
dx 2
0.47T/Z dB
10 8 P dt
Wmdmg while dx and x take the meanings as shown in
FIG - 58 ' the Fig. 58.
Equation (7) may be directly derived from consideration of a
flat bar in place of a cylindrical bar.
Distribution of Current in Cylindrical and Flat Bars. Reason-
ing as in the previous paragraph, but considering the current and
flux interchanged in their places, not only similar but also iden-
tical equations will be derived for the distribution of current.
DISTRIBUTION OF FLUX
153
Let B in Fig. 59 be the flux density at x, i be the current density
at x, and i + di the current density at x + dx.
The flux in the tubular element dx is Bldx.
The reluctance of the flux path is r-r-
0.47T m.m.f .
Ihus since rlux =
reluctance
2irBx
m.m.f. = m = j^-r
2 ATT p.
(8)
(Current
I
1
Impressed
i 1
E.M.F.
1
FIG. 59.
As x increases from x to x + dx, the m.m.f. increases from what
is within the circle of radius to that of radius x + dx
dm = 2irxdxi, or -T- = 2irxi where i is the current density at
distance x
Differentiating (8)
dm
dx
Equating (2) and (3)
27T
0.47TM
(+')
f.
dx
, o. 4 ,,,-
x
(9)
(10)
(11)
As x increases from x to x + dx, the increment of current
density is di, and the increment of current in the tubular element
is 2Trxdxdi. , The resistance of the material that this increment
/
Therefore the increment of the
of current traverses is
consumed e.m.f. is:
2irxdx
2irxdxdi
2-irxdx
P Ui.
154 ELECTRICAL ENGINEERING
Hence the decrement of the induced e.m.f., de is pldi. This
-de is caused by the flux in the tubular element, viz., Bldx.
Therefore
de= pldi = -- ^ -r.Bldx,
J. \J Ct'L
using signs of partial differentials, and re-arranging, we have :
dx = W~p dT
Differentiating (11) with respect to t and (5) to x,
d 2 B I dB di
+ - = 0.47T/*- (13)
(14)
a* ~ lov at
The solution of equation (15) is somewhat difficult and is
therefore delayed until later (see Skin Effect in Cylindrical
Conductors).
Equation (16), however, which shows the flux distribution in a
lamination is readily solved when the impressed m.m.f. and there-
fore, at least in non-magnetic materials the flux density is a sine
function of time.
Let B = B m sin cot
-
The effective value of the first expression may
be represented by vector OA = B, and that of
B A the second expression in the derivative by OM =
FIG. 60.
dxdt ' x dt " dt
and d 2 i 1 d 2 B
dx 2 10 8 p dxdt
Substituting (-12) and (14) in (13)
d 2 i \ di 0.4717* di
dx~ 2 ~^~ x dx = 10 8 p ~di
For long or thin flat bars,
di
Thus dealing now with effective values we can write:
DISTRIBUTION OF FLUX 155
where
P 10 8
To solve this equation we write,
d 2 B
- v 2 B = /. m 2 - v 2 =
.'. m = v
and B = A^ vx + A^e~ vx .
Since the density must be the same at equal distances from the
center line
A^ vx -f A 2 e~ vx = Ai~ vx + Aze Lvx
which requires that A i = A 2 = A
2 _ 04^
plO 8 '
It is readily seen that if v = (1 + j)a
V 2 = 2ja 2
2j P 10 8
- oa: ~ Jaz
or B = A[e ax e jax + e- oa: c
Substituting trigonometric expressions and combining the
real and imaginary terms we get:
Since e :iax = cos ax j sin ax
B = A[(e ax + e~ ax ) COS ax + j(e ax e ax ) sin ax].
If B i is the effective value of the surface density then B = BI
for x = d.
If, furthermore, it is assumed as an approximation that e~ a&
is small compared with e aS then,
B l = Ae a5 (cos ad + j sin a 8)
and flic"*'
cos a8 + j sin ad
t * + e " X) COS + " - ") n ]
156 ELECTRICAL ENGINEERING
and B Bie~ a V e 2ax + e~ 2ax + 2 cos 2 ax
where as given above
<0.27r2/^7
: \ p io 8
For iron p is approximately IO 5 and /* may be anything up to
18,000.
1
For copper p is approximately
and i = 1.
PART II. PROBLEMS IN ELECTROSTATICS
CHAPTER XII
FUNDAMENTAL LAWS
Coulomb's Law. The fundamental law upon which our know-
ledge of electrostatic or electromagnetic phenomena rests was
found experimentally by COULOMB. It is similar to NEWTON'S
law of gravitation and is:
F = c^orF = c^ (1)
iv>i ntZ
where F is the force acting upon the point charges Qi and Q 2 , or
the magnet poles of strength mi and w 2 , c is a constant depending
upon the system of units employed, and r is the distance between
the charges or magnet-poles, respectively.
In the electrostatic system of units, as well as in the electro-
magnetic system of units, c is taken as unity when the medium
is air, or rather vacuum,
.:r-3&.,**r-*F- . (2)
where F is expressed in dynes.
Thus two unit charges or two unit magnet-poles repel each
other with a force of 1 dyne when separated 1 cm.
If the medium has a specific inductive capacity K, then
1
=
K r*
If the magnetic medium has a permeability /z, then
The strength of unit poles is then measured assuming that
it were possible by the repulsion between two similar poles.
When the force is 1 dyne and the distance is 1 cm., the poles have
unit strength.
157
158 ELECTRICAL ENGINEERING
Field Intensity. Surrounding electric charges or magnet-poles
is a field, and the intensity of the field at a point is defined as the
property of the space, which is measured by the force exerted by
the field on unit charge or unit pole located in that point, when
electric and magnetic fields, respectively, are considered.
Because of this definition, it must not be inferred that the
intensity of the field is a force; it is not a force, but merely a
space function just as the gravitational field intensity is a space
function. The force acting on a certain mass at a certain point
may have any value, depending upon the particular mass used
in the experiment.
Important Theorems. While COULOMB'S law forms the basis
on which the theories rest, the progress in the art would probably
have been slow were it not that a number of theorems have been
worked out more or less directly from that law. These theorems
are:
GAUSS'S theorem, the divergence theorem, GREEN'S and
STORE'S theorems, etc., all having important bearing on prac-
tical problems.
Surface Integral of a Distributed Vector. As a preliminary
to these theorems the surface integral of a distributed vector
will be defined.
It will be assumed that an electric field exists due to some
charge and that lines of force or tubes of force radiate from the
charge in all directions. It is desired to find the number of lines
that go through a surface, say a cap that is placed in the field.
In Fig. 61, AB may be assumed to be, for instance, the inter-
section of the plane of a loop of wire, over which the cap is made,
with the plane of the paper.
If the surface of the cap were divided up into a number of
elements and the direction and the intensity of the field at every
point were known, then it obviously would be possible to calculate
the total number of lines (the flux) that crosses the cap or the
surface.
The sum of the fluxes normal to each element of the surface
is called the surface integral of the normal field intensity over the
cap, or the total outward flux through the cap. (The normal to
the elementary surface is always understood to be drawn out-
ward from the surface. On account of the sign of trigonometric
function a normal drawn inward will lead to a negative surface
integral.)
FUNDAMENTAL LAWS
159
If another diaphragm or cap (Fig. 62) were stretched across the
wire loop AB, it is evident that a certain amount of flux would
enter the space between the diaphragms and a certain amount
leave it.
It will be shown in this case that the total normal outward flux
from the space enclosed by the two caps will be zero, as long as
no charged particles are enclosed by the diaphragms.
Consider then a distributed vector field (Fig. 63), and let R
be the value of the vector at the small surface element dS,
making an angle 6 with the normal to the surface element. R
represents the electrostatic or electromagnetic field intensity.
The field intensity along the normal is then R cos 6, and the
flux going through dS is: dif/ = R cos 6dS, i.e.,
df = R cos (N t R)dS.
.'. \l/ = total outward flux through the surface,
RcosddS,
where N is the component of the field intensity normal to dS, i.e.,
N = R cos 0.
This can also be expressed in rectangular coordinates by
vector analysis, provided that dS represents not a surface dS,
but a vector at right angles to dS of size dS. (See appendix for
dot product.)
Let
and
R = X(x,y,z)i + Y(x,y,z)j + Z(x,y,z)k
dS = dS x i + dS y j + dS z k.
160 ELECTRICAL ENGINEERING
Then,
ffR-dS = ff(XdS x +
~ ,~ ' ^,N / //T?J j where obviously dS z = dydz.
YdS y + ZdS.) =ff(Xdy-dz _ y
+ Zdxdy),
Another way of expressing the surface integral of a distributed
vector field is:
ff(Xl + Ym
In these equations X, Y andZ are the components of the vector
along the three axes and I, m, and n, the direction cosines of the
normal to the surface.
Thus: I = cos (N,x),
and, m = cos (N,y),
n = cos (N,z).
.'. IdS = dydz,
and, mdS = dzdx,
ndS = dxdy.
.'. ff(XdS x + YdS y + ZdS.) = ff(Xl + Fw + Zn)dS.
Gauss's Theorem. According to GAUSS'S theorem the total
normal outward flux from a closed surface containing a charge
Q is = 47rQ.
Let N be the component of the field intensity R normal to an
elemental surface of the bag dS.
The theorem can be expressed mathematically by:
ffNdS =- 47rQ.
Let dw, Fig. 64, be the solid angle at A corresponding to dS or dSi,
which is perpendicular to R
or, dSi = r 2 w.
ffNdS = SfRcosedS = ffRdSi = f fRr*dvu.
But COULOMB'S law gives:
_QQ
or since by definition
F _i
r 2
F = R,
when
FUNDAMENTAL LAWS
Qi = unity.
161
(1)
FIG. 64.
the integral to be taken around the entire surface, that is over the
complete solid angle, which is 47r,
Thus,
= 47rQ.
ffNdS = 47TQ,
or if there are many charges in the envelope,
ffNdS = 4irSQ.
It is seen that the total flux radiating from a point charge Q or
a magnet pole m is <p = 4ir Q and (p = 4irm respectively.
It will be shown that while the conception of lines or tubes of
force is very much the same, both serve to map out a field, by
convention a tube includes 4?r lines.
From (1) it is seen, that the intensity at a point distant r from
a point charge Q is 2 -
By a similar reasoning, it is readily seen that in magnetic
problems,
tp = magnetic flux = 4?rm;
and, m
H - r ,,
where m is the strength of the pole causing the field, and H is the
intensity of the magnetic field.
162
ELECTRICAL ENGINEERING
But, to return to GAUSS'S theorem, it is readily seen that the
shape of the bag is immaterial. Assume so, for instance, that
the shape is that shown in Fig. 65.
FIG. 65.
FIG. 66.
The vector R cuts the surface three times. The outward
normal flux is positive at A, negative at B, and again positive
at C. Thus the net result is one positive outward flux (Fig. 65).
Were the charge outside of the envelope, then the flux cuts the
bag two, four, six or an even number of times, so that the total
outward flux is cancelled by an equal total inward flux (Fig. 66).
The net result then is, that
ffNdS = 0,
when the bag does not contain a charge.
Potential. The electric potential is similar to the potential
energy of matter; it is a space function.
The electric potential at a point is defined as the work done
in bringing a unit positive charge from a place of zero field to the
point under consideration.
The magnetic potential is defined in a similar way, substituting
unit pole for unit charge.
Path of Unit
Charge
FIG. *67.
Referring to Fig. 67 R is the intensity at a point of the path of
the unit charge in its journey from infinity, where the field is
zero, to the point P, where the potential is to be determined, then
V = - f R cos 6ds,
FUNDAMENTAL LAWS 163
the minus sign being adopted by convention; but
dr = ds cos 6, .' .ds = - ,
cosd
and,
r r = rp C rp o ii rp o
V - Mr- -I -jUr = Q-M ..
Jr = oo J oo I I \ oo ' p
In general, the potential at a point due to several point charges
It is interesting to note that the potential is not dependent
upon the path chosen in the journey; 'it depends only upon the
point charge at A and the distance between P and A. It is
strictly a space function.
The potential is the same on any surface the elements of which
have the same distance from the point charge.
Thus the potential of the surface of a sphere having a point
charge in its center, and influenced by no other charge, is:
where r is the radius.
Since by definition the capacity is C = y, we note that the
capacity of an isolated sphere is C = r.
The capacity in the electrostatic system of units is in centi-
meters. A sphere of 10 cm. radius is said to have a capacity
of 10 cm.
It will be shown later that to convert the capacity to farads
V 2 (3 X 10 10 ) 2
involves a division by ^ = ~ ~Tnir~ = ^ X 10 n . Thus in
this case the capacity of the particular sphere would be C =
10 t
~ farads -
Line Integral. The intensity R of the electric field has not
only a definite numerical value, but also a definite direction.
Let the components of R along any three rectangular coordi-
nates be, X, Y and Z, and let the components of the distance ds
on the respective axes be dx } dy and dz. Then, since the poten-
11
164
ELECTRICAL ENGINEERING
tial is the same, no matter what path we may take, by travel-
ling along the axes, we get:
fx,y,z
V J- '
Xdx + Ydy + Zdz.
This integral is called the line integral of the distributed vector
along the path.
Using vector analysis (see appendix) we get:
V = J*R-ds, the integral of the dot product,
where R = iX + JY + kZ,
and ds = idx + jdy + kdz,
.'. R-ds = Xdx + Ydy + Zdz.
Differentiating, we get:
dV = - (Xdx + Ydy + Zdz) (1)
Recollecting that if V is a function of the space coordinates x, y
and z only,
d
dV
It is evident, since it has been shown that the potential is a
function depending only upon the space coordinates, that
~
do;
and
and
ay
dn
= - z
= - R
(2)
where means differentiation along the lines of force, T is
dn dn
usually denoted by G, the potential gradient.
Equation (1) must be a complete differential, the criterion
of which is that,
ar _ dx
~dx ~ dy = '
_
dy dz
FUNDAMENTAL LAWS 165
and aX ^ dZ^
dz ~~ dx ~~
dY a 2 7
~dx ~ ~ dydx
and dX d z V
dy ~ ~dxdy
Thus, aF dX
. = 0, which was to be proven
Gauss's Theorem in Term of Potential Gradient. From equa-
tion (2) it is evident that GAUSS'S theorem can be expressed in
yet another form.
Since R = -_ . it is evident that
dn }
dS is the total outward flux.
^
Thus,
if the envelope contains a charge; and
- dS = 0, if the envelope contains no charge.
dn
In both cases, T- means differentiating along the normal to
on
surface, dS.
On account of the similarity between the electric and magnetic
definitions, we obtain, by reasoning identically with that given
above, for the magnetic potential,
fx,y,z fx,y,z
V = - I H cos dds = - I (Ldx + Mdy + Ndz),
where
dV *, dV Ar dV i 17 dV '
L = > M = N = --> and H =
a^ a^/ a^ an
and,
7=2; I J dS = - 47rZra, if the envelope contains
magnetic particles; and I I - dS = 0; if the envelope does
J J on
not contain magnetic particles.
166
ELECTRICAL ENGINEERING
Equipotential Surfaces shown in Dotted Lines Around Two Point Charges
Separated 5 cm.; Q l = +1Q.; Q 2 = - 5.
FlG. 68.
Lines of Force shown in Fine Lines and Equipotential Surfaces shown in Heavy Lines
Around Two Point Charges Separated 5 cm. ; Q t = + 10 , Q 2 = - 5
FIG. 69.
FUNDAMENTAL LAWS 167
In these equations, H is the intensity of the magnetic field,
L, M and N are the components of the intensity, H, along the
three axes.
V
Application of the Formula, V = 2 - In Fig. 68 is shown the
equipotential surfaces between two point charges, Qi = +10 and
Qz = 5, separated 5 cm.
The potential at any point, P, is obviously :
V P - ' + $.
ri r 2
The lines of force can not well be shown in a plane, but a fair
idea of their shape can be gained from Fig. 69.
The direction of each line of force is obtained by combining
72 1, the intensity at a point due to Qi, and R 2 , the intensity due
to Q 2 .
CHAPTER XIII
METHOD OF IMAGES, APPLIED TO THE PROBLEM OF
POINT CHARGES +10 AND -5, SEPARATED 5 CM
In plotting the equipotential surfaces of the problem given
above, it is readily seen and proven that the surface of zero
potential, Fig. 70, is a sphere, and that the following relations
obtain:
(1) a =
FIG. 70.
25
_ 50
cm. and p = -
Substituting, we get a = =~
X 5 = 3.33 cm.
It is evident that the field distribution will not be affected if
a grounded metallic sphere of radius, p, at a distance (D + a)
from the positive charge is surrounding the negative point charge
at B.
And it is also evident that the potential will be the same ( = 0),
if the charge at B is removed altogether.
168
METHOD OF IMAGES
169
It is thus evident, that, reversing the line of argument, the
potential distribution in a system involving a point charge at A
and a grounded sphere of radius p with center at a distance L from
the point charge, can be determined without the laborious deter-
mination of the distribution of the induced charge on the sphere,
simply by using two point charges at A and B.
The location of B and the charge which must be assumed at the
non-existing point B can be determined from the following rela-
tions which are easily proven:
L P (L - D) = P 2 , or D = L - p *>
L
and ~ p
Potential Distribution between a Point Charge and a Metallic
Sphere. While it is evident then that the field can be determined
without much labor in the case of a grounded sphere, the problem
becomes quite involved if the sphere is insulated and kept at a
certain potential, V, which is not zero.
To calculate the potential distribution in that case, it is neces-
sary to study the distribution of the surface charge.
FIG. 71.
Consider first the case of the grounded sphere. The intensity
of the field at a point is the resultant of the intensities due to the
charges at A and 5, the so-called inverse points.
It must be expected that the direction of the resultant field
is perpendicular to the surface at all points, thus we can draw the
170 ELECTRICAL ENGINEERING
diagram in Fig. 71. Remembering that R } the intensity is
the vectorial sum of
-4 and v or of Ri and
ri 2 r 2
The intensities R\ and R 2 are resolved along the radius CP ana
along a line parallel to AB. It is seen that PE and PG are equal,
and cancel each other, so that the resultant intensity is in line of
CP and is PF + PH, algebraically.
Let the radius of the sphere in Fig. 71 be p. By similar tri-
angles,
PF PC PC p.
~R~ = PA ' =/*/! p^ : = *ti ,
and PH _ PC _ T PC
Po " P#' "
But
and
And it has been shown that Q 2 = Qi j-'
Li
Since La = p 2 , from the figure, it is seen that,
n L p
' C 2 72\ C1\
p7? (p - L)
By Coulomb's theorem, 47r(T = .R, where o- is the surface density
of charge, or charge per square centimeter,
- 2 ) (2)
METHOD. OF IMAGES
171
When the radius is very large, the surface of the sphere ap-
proaches a plane, Fig. 72, and a approaches p. Thus, if d, in
Fig. 72, is the distance of the point charge from the plane of
zero potential, we have:
L = P + d,
which, substituted in (1), gives:
R = 3 (p2 ~ (p
=: ( ~ 2pd ~
or, since d- is small compared with 2pd,
and the surface density of charge is:
Qid
27rri 3
FIG. 72.
The surface density of charge decreases inversely as the cube
of the distance from the point.
Assume now that the sphere is insulated and without charge,
it will then have some potential not zero.
It was shown, that, when the sphere is at zero potential, it
acts as if it had a charge Qz = Q\Y ^ ^ ne inverse point B
of point A. In order that its charge shall be zero, we have to
apply mathematically, somewhere in the sphere, a charge =
Q 2 = _|_Qiy. Then the total charge obviously is zero.
Li
Since the resultant potential of the external charge Qi and the
internal charge Q\j gives zero potential of the spherical
surface, in order to maintain a uniform potential V all over the
sphere, the assumed charge must be applied in the center of the
sphere.
Thus we deal with three charges, which combined cause the
external field.
First. The field due to the external point charge Q\.
Second. The field due to the charge Q 2 at tne inverse point.
Third. The field due to the charge Q 2 in the center of the
sphere.
172 ELECTRICAL ENGINEERING
The charge Q 2 gives a uniform surface density of
Q 2 (?i
47T P 2 *~ 47T P L
The combined effect of Qi at A and Q 2 at 5 has been shown to
give a surface density of
=
Thus the actual surface density is:
Equation (3) then gives the distribution of the surface charge
on an insulated sphere without any independent charge. The
equation must, and does show, that a is positive on one side and
negative on the other side, in order that the total charge be zero.
The potential of the sphere is obviously,
charge _ Qip _ Qi
radius pL L
This is of interest, in that it shows that the potential of a sphere
due to a point charge Qi situated L cm. from the center is -j^-
This can be proven in a more general way as follows :
Assume that a non-conducting sphere be placed in an electric
field caused by a number of point charges, a, 6, c, etc. Let the
potential of a small element of the sphere be V. The value of
V changes from point to point of the surface of the sphere.
The average value of the potential V is:
VdS
where dS is an element of the surface.
Referring to Fig. 73 :
dS = r sin 6d4>rdd
.'. V m ^^ffVs
and the average potential gradient along the radius is:
* Cf
irr 2 J J
dS.
4irr 2 J J dr
Since is the intensity as well as the gradient it follows that
METHOD OF IMAGES
173
dS is the flux diverging from the sphere. This is zero as we
have assumed that no charge exists in the sphere.
Thus -~^ and V m is a constant for all values of r.
We conclude then that the average potential of a sphere is
the same as the potential at the center.
rdO
FIG. 73.
Suppose now that the insulated sphere had a charge Q Q .
In order that the surface of the sphere shall be an equipotential
surface, this charge also should be considered as placed in the
center, and its surface density should be added to those given
above.
L n 3
The potential of the sphere will obviously be the sum of the
potentials due to its own charge Q and due to the point charge
Qo , Qi
or,
Usually V is known rather than
It is of interest to find the attractive or repulsive force between
the point charge at A and the sphere.
174 ELECTRICAL ENGINEERING
The force is, by COULOMB'S law, proportional to the product
of charges and inversely proportional to the square of the
distance.
The following conditions therefore exist:
First. A charge + Qi at A. Fig. 74.
Second. A charge Q\ y- at B.
Li
Thisd.A charge + #1 ~ + Q at C.
Li
-< D
^ r
V - - :
FIG. 74.
Thus the force between the sphere and the point charge is:
F =
L 2 p 2
But it has been shown that D = j >
Lt
p = ~ Qi 2 L 2 jo Qi 2 p QoQi
" (L 2 -p 2 ) 2 " L~* L 3 L 2 '
which, by transformations, becomes :
QoQi ,p 3 (2L 2 -p 2 )
L 2 " Vl L 3 (L 2 - p 2 ) 2
Example Qi = 1, Q Q = 10.
L = variable,
p = 10.
For L = 100,
10 1000 20,000 - 100 1
10 4 ' 10 6 (10,000 - 100) 2 " * 1000 yn
For L = 11,
10 1000 (242 - 100)
F - 121- 1330 (121 -10Q) 2= - 0.158 dyne attraction.
It is thus seen that a lightly charged particle may be repelled, if
far away from a charge of the same sign, and may be attracted
when near. If, however, the charges are of opposite signs, the
charges attract always.
METHOD OF IMAGES
175
Problem. Construct the equipotential surfaces between an
insulated charged sphere and a point charge, when
p = 10 cm.,
L = 20 cm.,
Qi -- i,
and _ 21
~ 20'
Potential Distribution between Two Spheres. Let sphere A
in Fig. 75 have a pot. V and a radius R] and sphere 5 have a
pot. FI and a radius
FIG. 75.
Calculate first the charges at A and B and the location of
these charges, when A is at potential V and B is at zero potential.
Then reverse the operation, and calculate the charges at A and B
and the location of these charges, when B is at a potential Vi
and A is at zero potential. Then add the charges and potentials
respectively, and the desired solution is obviously obtained.
(1) Calculation of the charges on A and B when the potential
of A is V and that of B is zero :
The first approximation is obtained when the potential of A
alone is considered. We have then, since in general Q = VR, a
charge in the center of A of value Q = VR, and we may, for
completeness, say that its distance a from the center is zero.
VR
This charge affects B by giving B a potential, which is -y
Since, however, the potential of B must be zero, it is necessary
VR
to supply B with a charge which gives a potential - -j-. This
charge, which may be called Q'i> has previously been shown
/7? \
to be Q'i = VR(-J-}, shall not be placed in the center of
176 ELECTRICAL ENGINEERING
the sphere, but at a distance 61, which is obtained by the relation
previously proven:
(radius) 2
~ distance from charge to center of sphere
\T~D~D
But the charge Q'i or -- j at 61 affects the potential of A,
so that its potential is no longer V, but
V + f -- r -j -s- (distance from charge Q'i to center of A)
V- VRRl
L(L - &0*
To bring the potential of A back to V, A must be supplied with a
charge, which is:
As far as the external action of the charge is concerned, it is
located at a i} where as before
This charge at a\ affects sphere B and induces a potential which is
VR 2 Ri
L(L-bi) (L-ai)'
In order to bring the potential back to zero, a charge Q'% has to
_ T/P2P
be added to B, which gives a potential of r /r _ ^ \ (T~I )'
and this charge, as far as external influences are concerned, is
located at a point 62, where
, RS
62 =
Continuing the process, the necessary additional charge on A to
balance the effect of Q'% at 62 is found to be :
Qi= _^_^_ =
and
Again, n ,
v 3 L(L - 61) (L - ai) (L - 6 2 ) (L - at)'
and , R\ 2
3 " (L-a 2 )'
METHOD OF IMAGES
177
The total charge on A is Q A = Q + Q l + Q 2 + . . . ;
The total charge on B is Q B = Q'i Q'z + Q'z + . . . .
But, it must be remembered that in order to find the intensity
of the field at any point, the position of the charges has to be
considered.
, (2) By an identical method, a new set of charges are obtained,
when A is kept at zero potential and B at its potential V.
The total charges on A and B are the sum of all the charges so
calculated.
Assuming, for instance, that the potentials of A and B are
both positive.
The first set of calculations will then give a number of positive
charges in A, all of which, except the first, located at points, riot
its center, the charges in B will all be negative, and all be located
at points not its center.
The second set of calculations (not shown above) will result in
a series of negative charges in A, all of which are located at points
not its center, and a set of positive charges in the sphere B } the
first of which is at its center. Thus the total charge in either A
or B is a sum of a series of positive and negative charges.
Simple Case. For two similar spheres, one at zero potential
and the other at a potential, V, we have:
On the sphere of pot. V
On the sphere of pot. zero
Q = VR
a =
o' VR2
Ql = T
h R2
bl== ~L
o VR3
o' VRi
L(L - 6x)
R 2
L(L - bJ(L - a x )
t ^
ai ~ (L - 60
(L - at)
Q VR *
Q's -
VR 6
L(L - 6 X )(L - ai )(L - 6 2 )
R 2
L(L-6 1 )(L-a 1 )(L-6 2 )(L-a 2 )
b> - RZ
* (L - 6 2 )
(L - a,)
178 ELECTRICAL ENGINEERING
The total charge on the sphere of potential V is :
QA = Qo + Qi + Q 2 + . . . ;
and that of the sphere of zero potential is :
QB - Q'l + Q' 2 + Q'z + . .
To study the sphere gap, the following problem has been
solved to show more particularly, that, while the difference in
potential between two gaps may be the same, one gap may break
down with considerably lower potential difference than the
other.
Air at atmospheric pressure appears to sustain, as a maxi-
mum, a density of about 100 lines per sq. cm., or a potential
gradient of 100, electrostatic units or in practical units 30,000
volts per cm. If, therefore, the potential to ground is high, the
air may well break down around the spheres, even though the
potential difference between the spheres may be comparatively
low.
When the air breaks down, corona appears. Then the effective
dimensions of the spheres are increased and the gap length
correspondingly lowered.
The following three cases are calculated, and the results are
tabulated below.
Diameter of the spheres, 25 cm.
Distance between surfaces, 14 cm.
Potential difference 1000 electro static units or 300,000 volts.
In the first case, sphere A has a potential of 1000 and B is
at zero potential, in the second case the spheres are at potentials
+500 and 500 respectively, and in the third case they are
at potentials +1500 and +500 respectively. In the example
the potential gradient G is calculated at the surface of the sphere
of highest potential on the center line between the spheres al-
though it may, of course, be greater at some other points. In
general G = S .,. . 2 - The gradients due to the two spheres
should obviously be added if the charges are of opposite potential.
Since the intensity of the field is in the same direction at the point
considered.
METHOD OF IMAGES 179
Summary of the first case:
For sphere A,
a = Co = 12,500
ttl = 4.46 Ci = 1,430
a 2 =4.53 Q 2 = 186
a 3 =4.53 Q 3 = 24.4
For sphere B,
60 = Q'o =
b l = 4 Q'i = - 4,000
62 = 4.5 Q' 2 = - 516
6 3 = 4.53 Q' 3 = -- 67.3
6 4 = 4.53 Q\ = - 9
In general G = ^
.'.G = - S ~ = - 114.6, or, - 34,500 volts per cm.
Thus the sphere probably begins to glow.
Summary of the second case :
ao = 6 = o Qo = Q'o = 6,250
Ol = bi = 4.01 Qi = Q'i = 2,000
a 2 = 6 2 = 4.45 62 = Q' 2 = 714
a 3 = 6 3 = 4.51 Q 3 = Q's = 258
a 4 = & 4 = 4.53 Q 4 = Q\ = 93.5
a 5 = 6 6 = 4.54 Q 6 = Q' 6 = 34.2
(^ = -100.2 or about -30,000 volts per cm. The spheres
ought to be just about on the point of glowing.
Summary of third case :
Qo = 18,750 Q'o = 6,250
Q l = - 2,000 Q'I = - 6,000
Q 2 = + 2,140 Q't = + 714
Q s = - 258 Q ; 3 = - 775
#4 = + 280 e ; 4 = + 93.5
Q b = - 34.2 Q' B = - 102.6
The a's and 6's are the same as above.
G = -128 or -38,400 volts per cm.
Thus the spheres glow undoubtedly, and if "ground" is under
the spheres the potential gradient may be slightly higher below
the line connecting the centers of the spheres.
12
CHAPTER XIV
APPLICATION OF THE POTENTIAL FORMULA V = 2 -
TO SOME MAGNETIC PROBLEMS
The magnetic potential at a point in a magnetic field is, as
has already been stated, the work done in ergs in bringing a unit
pole from infinity, or a point of no magnetic field, to the point
under consideration.
By GAUSS'S theorem the outward normal flux from a pole of
strength m is 4irm. Thus the intensity of the magnetic field, H,
at a distance, r. from the pole is | ^ '>
or, H = m- f
r 2
tlll(l) . -|-T flv 7 "V.
or in general, V = 2
Obviously, a magnetic pole can not exist alone; there is always
a north pole and a south pole in every magnet. Thus to get
the potential at a point, at least two poles of opposite signs must
be considered.
The potential of a small magnet at distance large compared
with its dimension is:
V = - ' S > where 6 is the angle the axis of the magnet makes
with the radius vector to the point.
This is readily seen, if the magnetism be assumed as con-
centrated at the poles of the magnet.
Referring to Fig. 76, the potential at P is:
_ m m m
V = AP + ~BP~
m , (1)
180
THE POTENTIAL FORMULA 181
If r is large, compared with I then V =
m m
A/r 2 + lr cos \/r 2 Ir cos
The square root can be expanded by the binomial theorem.
We have,
= |l -- J^ - cos -h . . . .' 1 J - cos . . .1
= % cos 6 (approximately) (2)
Aba.
FIG. 76.
It is seen from (1) that the magnetic potential at P is in times
the difference in -, as we go from one pole to another, where r is
the distance from a pole to the point P. Let I = ds, then the rate
of change of - along ds, is:
- (-) . thus the total difference is (-)ds,
ds W ds W
.
dr
If I', m', and n f are the direction cosines of the magnetic particle
at (x, y, z), we can then also write,
>""[> I +'sC') + i OK
Magnetic Shell. A thin piece magnetized at right angles to its
surface is called a magnetic shell. It can thus be assumed as
182
ELECTRICAL ENGINEERING
made up of a large number of small magnets as shown in Fig. 77.
Let the total pole strength in Fig. 78 be m and the area S, then the
pole strength per unit area is -. Let the thickness of the shell be
I, then the potential at P due to the shell is from equation (2).
v
S
where is the solid angle at P subtended by the surface of the
shell.
(Recollect that the solid angle, doi = -y cos 0.)
FIG. 77.
FIG. 78.
ml is called the magnetic moment, the strength of a mag-
netic shell or the moment per unit area is usually denoted by 0,
ml
and, V = grco.
WEBER proved experimentally that a small circuit in a plane
carrying current produces the same kind of a field as a magnet,
and that the potential at a point depends upon the area A of the
coil, the current 7, and the distance to the point, by a relation :
_. KAI cos
V = y >
r 2
from which the electromagnetic unit of current can be deter-
mined by making k unity.
AI cos ml cos
thus, and 7 = r- = g, the strength of the magnetic shell sur-
rounded by the circuit or coil.
THE POTENTIAL FORMULA
183
Since we have proven that V = #co, we get the following simple
relation between magnetic potential and current:
V = 7co, where co is the solid angle subtended at the point by
the surface of the coil. It is evident then, that, as long as we do
not tread the circuit, and as long as we return to the starting
point, the work done in moving a pole in the field is zero.
To illustrate this, the potential at a point on the axis of a cir-
cular wire carrying I abs. amp. will be determined.
First let the point be at the center of the coil, Fig. 79, then co =
27r, and, V = 2?r/.
FIG. 79.
FIG. 80.
If the point is on the axis, but a distance x from the face of the
coil as shown in Fig. 80, then the solid angle is:
co = 2ir(l - COS a) = 27r( 1
and, - / x
' The magnetic field intensity along the z-axis, which is the
direction of the magnetic field, is:
dx ~ (R 2 + x*)* '
and the force in dynes on a pole of strength m is :
2TrR 2 m
= (R 2 + X*)* '
for x = 0, that is if the point is in the plane of the coil and in its
center,
H =
27T/
R '
The work done in bringing unit pole once or several times
through a loop carrying a current / will now be investigated.
184 ELECTRICAL ENGINEERING
Referring to Fig. 81, before the journey starts, the potential at
P has been shown to be 7<o.
When the journey has covered 1 revolution, the solid angle
has changed from to 4?r. Thus, after n revolutions of the unit
pole the potential of it is :
Iw + 4irln = I (co + 47rn).
It is evident then, that, when a magnetic-pole of strength m is
moved around in a field, and returned to the starting point, work
will be done every time the circuit is treaded. If it is treaded n
times, the work is:
The magnetic potential is thus a multi-valued function of the
space coordinates.
p
Path of Unit Pole
FIG. 81. FIG. 82.
dV
The intensity of the magnetic field at the point, H = -^
depends, however, only upon the term involving the solid angle
co, not upon the term involving 4?m.
Consider now a straight infinitely long wire carrying current I.
Let the wire form the y-axis and let the point be in the x-z
plane (Fig. 82). The cone subtended by the plane of the current
(x-y plane with ?/-axis as one edge) which goes out to infinity and
the point P has a solid angle, 2(ir 0).
NOTE. If the angle in the x-z plane had been TT, the solid
angle would have been 27r; in this.case the former is (TT 0), the
latter is 2(ir - 0).
.*. V = /( + 4arri) = (27r - 20 + 47m).
The direction of the lines of force which are circles around the
y axis are along the arc, rdB, then
ff = ~S = T'
an equation very often used in electrical engineering.
CHAPTER XV
R (read del dot R},
V is sometimes called LAME'S
DIVERGENCE OF A VECTOR, POISSONS AND LAPLACE
EQUATION
It has been shown by GAUSS'S theorem that the total flux
entering and leaving a closed surface in a vector field is zero,
unless the (closed) surf ace, contains some charge Q, in which case
the outward flux equals 4?rQ.
This charge may be a single charge, or it may consist of a large
number of small charges throughout the interior of the surface.
The divergence of a vector is the excess of outgoing flux over
the incoming flux per unit volume of the space enclosed by the
surface; it is the number of lines which diverge per unit volume.
If the excess of flux in a small volume dv is d\f/, then the diver-
d\L>
gence of the vector is -T-
It is written div. R, div. (X, Y, Z) or V
where V stands f or + ^r + ^
dx dy dz
differential parameter.
It is evident, from what has
been said above, that unless
some charges are enclosed in
the small volume, there can be
no divergence. If there are as
many units of positive charge
as of negative charge in each
small volume, there can also be
no divergence, i.e., div. R = 0.
The divergence is positive, if
there is an excess of positive
charge; it is negative (sometimes called convergence), if there is
an excess of negative charge. The presence of divergences
involves the presence of charges. In hydraulics the presence of
divergence means either the presence of some source of fluid in
the element or some change in density.
Consider a small volume represented by a cube, in Fig. 83 for
the sake of simplicity. This cube is assumed to be a small part
185
T Axis
FIG. 83.
186 ELECTRICAL ENGINEERING
of the total volume enclosed by the envelope that contains the
charges.
Let X, Y and Z be the components of the field intensity R
parallel to the coordinate axes and at the center of the surface
a, b and c.
If R is a continuous function, which depends upon the space
coordinates only, and if the edges of the cube are dx, dy, dz then
the value of the ^-component of the field intensity at Ci = Zi =
Thus, the incoming flux at c is: Zdxdy, the outgoing flux at
is [Z + -T- dz\ dxdy.
Consequently, the difference is
'dZ
(dz}dxdy;
Similarly, for the other sides,
and /
- dy } dzdx.
dy
The total diverging flux is thus:
Hence by definition
div. R = V - R~ = V-R.
dv
If p is the charge per unit volume or the volume density, then
the outward normal flux is 4?rp.
ax ar az
dx dy dz'
A vector field is said to be solenoidal, if there is no divergence.
Such a field is, for instance, the electric field in free space or the
field of force of gravitation in free space.
The divergence theorem connects the surface and volume inte-
grals and states that the surface integral of the normal outward
flux of a distributed vector is equal to the volume integral of the
divergence taken throughout the volume. It is one of the forms
of Green's theorem.
ELECTRICAL ENGINEERING 187
It is
ffR cos SdS =
Using the notation of vector analysis, we get:
ffn-RdS = fffRdv,
where n is the unit normal vector.
This theorem is subject to rigid mathematical proof, but can be
understood without advanced mathematics, if the volume
enclosed by the surface is assumed to be divided up into a large
number of small volumes, each fitting tightly against the others.
As we add the normal outward fluxes of the different elemental
volumes, all will cancel, except those on the very outside surface,
since every wall separating two elements is integrated over twice
with normals in opposite direction.
The outward normal flux is J* J* R cos Ods. Since the excess
of outgoing flux over the incoming flux in the element of volume,
dxdydz, is:
-- h ~^~ + -Q-J dxdydz, it follows that the total outgoing
flux is:
III f-r + -.J h -Q-} dxdydz, which is equal to I I R
cos 6dS.
Poisson's equation is:
d*V\
" = " p -
This becomes: dX dY dZ
If X } Y and Z are gradients or intensities of a scalar point
function V, so that
X = - - V - Y = - d ~- Z = -
d# ' dw ' 62
ax
ay
-ay =
az
62 = dz 2 '
188 ELECTRICAL ENGINEERING
and
dx* " dy 2 " dz* = 47rp >
where p is the density of electrification or charge per unit volume.
This equation then applies, when the region of the electrostatic
field under consideration contains positive or negative charges, or
sources and sinks as some writers call them.
Laplace's equation is:
VV VV dF _
dx* ~ dy 2 " dz 2 '
or, as it is often written,
V 2 F = 0,
(Read del square V = zero) and refers to a region in which there
are no charges, or to a solenoidal field.
By means of LAPLACE'S equation it is possible to determine the
potential at any point in the dielectric surrounding a charged
body. If the body is unsymmetrical in every way the equation
becomes very involved, but if, as is almost always the case in
practice, there is some axis of symmetry and particularly if the
body has circular symmetry then the potential distribution can
usually be calculated fairly , easily, especially if a table of
LEGENDRE'S coefficients is available.
CHAPTER XVI
LEGENDRE'S FUNCTION
The potential at points outside of the bodies having circular
symmetry, such as circular discs, circular rings, etc., can be
determined very readily by means of a certain function,
viz., LEGENDRE'S function, which has been worked out and is
tabulated much in the same way as trigonometric functions.
LAPLACE'S equation
g-O (1)
dx 2
can be used as has been shown in exploring the space surrounding
charged body.
With circular symmetry of the charged body it is obviously
advantageous to express the equation in spherical coordinates
(see Appendix heading Partial Differentiation). Thus,
rd 2 (rV) _J_J)/. dF\ 1 d 2 V
dr 2 sin 6 dd \ Sm BO/ sin 2 6 d<? 2
With z-axis as the axis of circular sym-
metry, the potential will be the same for ,--
all values of <, as long as r and 6 are con-
stant, as is readily seen in Fig. 84.
Equation (2) becomes:
rd 2 (rV) , 1 <
(2)
dr 2
(3)
This is then an equation of two inde-
pendent variables, r and 0. The general
method of solving such equation is to FIG. 84.
assume the solution to be:
V = R'6', 1 where R r is a function of r only, and 0' is a function
of 6 only.
Substituting in (3),
l NoTE. See Byerly's "Fourier's Series and Spherical Harmonics."
189
190 ELECTRICAL ENGINEERING
or, * ., d 2 . R' d 30'
re'^(rR'}= -^^sin*-,
or, r a 2 (rR'} I d I . dO'\
W ~^~ ~ W^e de ( sm e W < 5 >
The left-hand term is a function of r only, the right-hand term
of 6 only.
In order then that this shall hold for all values of r and 0, each
term must not only be a constant, but must be the same
constant.
Let this constant which is entirely arbitrary, be a 2 ,
-'-0 (6)
and 1 d /sin 6
Equation (6) becomes;
rS ^ + 2r
dr 2 dr
dr
The solution of (8) is readily found, it is :
where
and
(9)
It is evident then that r m and -^i are particular solutions of
equation (6).
If we choose for a 2 a value which is:
a 2 = m(m + 1),
then equation (9) is satisfied, since
LEGENDRE'S FUNCTION
191
It has been shown that r m is a particular solution of R', thus
using this solution at first, we get
V = r m B\
Substituting this in equation (3) we get,
and,
36
Equation (11) can be solved for 0'.
We have,
d I . a0'\ . a 2 0'
Let
80'
a: = cos 0, there sin = \/l x 2 '.
rift'
In equation (12), is to be determined -- and - ^-'
ou do
80' 86' 8x 86' . 86'
^ = " smd== "
\j ^/v \j \J v/**/
/ a a0'\ ax _ r_a_ / a0_'
\ax a0 / a0 ~ Lax \ ax
a 2 0'
a0 2
(14)
U^l/
r) f) f
Substituting the value of T~T- from equation (13) and the value
ofj
of j from (14) in (12), we get:
a
Thus equation (11) becomes:
r) 2 /?'
m(m + 1) 8' + ~ (1 - x
^-(x + x) = 0,
+ !)' = (16)
192 ELECTRICAL ENGINEERING
This equation, which very important, is called LEGENDRE'S
equation.
It can also be written:
[(1 - x) ~] + m (m + 1) tr - (17)
since>
Assume now that 0' can be expressed in whole powers of x
multiplied by constant coefficients, that is,
9' = 2a n x n = a + aix' + a 2 z 2 + a 3 z 3 + . . . (18)
Referring to equation (17),
~\nf
(1 - x 2 ) - = ai - aiz 2 + 2a 2 x - 2a z x 3 + 3a 3 x z - 3a 3 a: 4
(1 - x 2 ) = - 2aix + 2a 2 -
and,
m (m + 1) 6' = m (m + I)a + m (m + 1) aiz' + m (m + 1) a 2 ^ 2 +
m(m + 1) a 3 z 3 + . . .
Collecting the coefficient for similar powers of x we get:
[2a 2 + m(m + I)a ] is the constant term;
[6a 3 2ai + m(m + l)oj is the coefficient of x 1 ',
[ 6a 2 + 12a 4 + m(m + 1) a 2 ] is the coefficient of x 2 ;
Since, from equation (17), each of these coefficients is zero, we
get:
m(m -f- l)ao m(m -\
2
*** a a i >
D
m (m -f- 1) a 2 + 6a 2 _ m (w + 1) ~ 6
It is seen that if a = 0, all the even terms disappear; if ai =0,
the odd terms disappear.
LEGENDRE'S FUNCTION 193
The coefficients are related in a comparably simple manner, as
follows:
I ON L'\"*
or, (fc+l)(fc
m(m + 1) -
(fc +!)(*-
From (20) it follows, that, if fc = m - 2,
(m - 2+ l)(m - 2 + 2) m(m - 1)
~ (m - m + 2)(m + m - 2 +1) am ~ 2(2m - 1)
m(m - 1) (m - 2) (m - 3)
_
2.4(2m- I) (2m - 3)
m(m l)(m 2)(m 3)(m 4)(m 5)
x i _ . L- _ ft . rff>
2 4 6 (2m - l)(2m - 3) (2m - 5)
It is thus possible to express equation (18) as follows: 6' = 2 a n z n ;
if the highest power of x is x m , then we get:
,
2 (2m - 1)
W ( m - 1) (m - 2) (m - 3) ]
2 4(2m - 1) (2m - 3)
where a m is entirely arbitrary, and it is convenient to choose a
value,
(2m - 1) (2m - 3) (2m - 5) ... 1
a m = 7
because, for this value of a m , 6' 1 when x = 1.
. , = (2m - 1) (2m - 3) (2m - 5) ... 1 r _ m(m - 1) m _ 2
m! 2(2m - 1) ^
m(m-l)(m-2)(m-3) 1
2-4(2m - 1) (2m - 3)
Since 6' is a function of #, and contains no higher power of x than
>x m , it is customary to write, instead of 6', P m (x), or since x was
cos 0, P m (cos 6).
Before enumerating some values of 6', recollect that (factorial 0) = 1,
or 0! = 1, or I? = 1; and since |1 = 1, 0=1 = 1. This is readily seen
In 1
since j n _ l = n] forn = 1, = = 1, .*. |0 = 1.
194
ELECTRICAL ENGINEERING
Example. Find P 3 (cos 0).
m = 3, .'.P 3 (z) =
(6 - 1) (6 - 3) (6 - 5)
1-2-3
r 8 3 3-1 , -i
2 ' 6 - l x .
Note that only three terms can be used in the numerator in
front of the parenthesis, -since the last term must end with 1 as
is shown in equation (22).
The parenthesis contains only two terms, because the next
term would give a negative exponent, and we have assumed that
the powers of x are positive integer numbers.
Thus, for m = 3,
P 3
Similarly, for m = 2,
(4 - 1) (4 - 3)
1-2
For m = 1,
Pi(s) = ^ x = x
or,
V = A rP (cos 0) +
(23)
For m = 0,
But we assumed as a particular solution:
V = r"0' .', V = 2A m r m P m (x), or 2A m r m P m (cos 6) (24)
(cos 0) + A 2 r 2 P 2 (cos 0) +
A 3 r 3 P 3 (cos + . . . (25)
Referring now to equation (10), we see that there is also another
particular solution, namely:
or,
_
A 2 P 2 (cos^)
(26)
Before applying these equations to some practical problems, it
may be of interest to note that the LEGENDRE'S function can be
LEGENDRE'S FUNCTION
195
obtained by expanding ^ where R' is the distance between two
points (Fig. 85).
and
R' = \r 2 + ri 2 - 2rri cos 0.
If n > r, l f ; Ur 5 t __ 2r .T^ A
where A = (1 + h 2 -
where T
p = cos e.
FIG. 85.
Expanding A by the binomial theorem, we get:
hp
Po +
- 3p)
+ /i 2 P 2 + /i 3 P 3 +
. . (27)
The similarity between (23) and (27) is obvious.
Returning now to the problem of a circular wire carrying
current, we have shown that the potential at a point on the axis,
that is, r coincides with i/-axis and 6 = 0, is :
where r and R are shown in Fig. 86.
If R > r, see Fig. 86, then
FIG. 86.
196 ELECTRICAL ENGINEERING
Remembering that when -^ = K is a fraction
Since equation (23) holds for all values of 0, it also holds when
6 = 0. Thus we can readily determine the coefficients A , Ai,
A 2 , etc., which are:
Ao = 2*1,
A -
R '
A, = 0,
_ .,_, 1
A 4 = 0,
A 6 = 0,
AT- +
y = 27r7[l -- ^P! (cos 0) + M ^3^3 (cos 0) -
M ^5 (cos 0) + ^ P 7 (cos 0) + . . .] (28)
If r > R, then,
r ^ !. 3B 4 l-3-5fl 6 -i
2.7 [1 - 1 + K - - - + ^- Q - +
[7? 2 P^ z?6 n
>i^-M^+K 6 |+ ] (29)
From equation (26) we get:
+ P l *e + A l *&
.*. Ao = 0,
R2
1 = 2?r/ 'T'
LEGENDRE'S FUNCTION
197
A 2 = 0,
A 3 = - 2irl
A, = 0,
A 6 =
/. V = 2irl [^^1 (cos 0) - % ^ 4 P S (cos 6) +
7?6
-,
. ;J (30)
As a second application of the use of the LEGENDRE'S function,
the following problem will be considered.
Find the potential at points outside of a thin circular disc, Fig.
87, charged to a certain potential, V.
It will be proven that the distribution of the surface charge is :
where Q is the total charge, that is, the charge on both sides.
FIG. 87.
FIG. 88.
We first calculate the potential at a point PI on the axis
(Fig.
/"Vo = /"*ro =
2Q
Q 1 r 2 -
= 2K cos ^T^
as can be readily found by simple integration.
This expression then must be expanded in a power series.
This can not be readily done, but its derivative with respect
to r becomes a simple expression, which can be expanded,
the resulting series can be readily integrated. Thus,
^. ["-*?- -i r2 - R2 1 = Q
drl2R C r 2 + R 2 1 R 2 + r 2
If R > r, then
Q ^H ,J1 J 6 , 1
fi a + r 2 B 2 L R 2 ~*~ R 4 R & ^ ' ' ' J
198 ELECTRICAL ENGINEERING
Integrating,
R R
-&- -+ c }
OH J
For r = 0, i.e., on the disc, and the potential of the disc will be
proven, to be ^ ' D"'
r - -
" 2
7 -if-- - -4- 4- 1
" R 12 R + 3^ 3 + 5^ 5 "
when r > R, it is found in a similar way that:
'-[?-+-+]
Equation (31) is similar to:
7 = A r P (cos 0) + Air ! Pi(cos 0) + A 2 r 2 P 2 (cos 0)+ . . .
'' Ao = '' Al = ~'' A2 = > As = '^ )A ^ ' etc '
* V = |[I P (COS 6) " i Pl ( cos ^) + 3^^2(cos 0)
Equation (32) is similar to:
_ A P cos0 ' AiPi (cos 0) A 2 P 2 (cos 0)
f l f 2 r 3 - '
/. A = 5- B, A! = 0, A 2 = - ^ y, A 3 = 0, A 4 = ^ y,e
CHAPTER XVII
DISTRIBUTION OF CHARGE ON AN ELLIPSOID
If an ellipsoidal thin shell is formed by two similar, similarly
situated ellipsoids, and the charge per unit volume, p, is constant
in the shell, then the force at any point
inside the ellipsoid is zero, that is the poten-
tial is constant. The outer surface is an
equipotential surface. 1
To prove this, consider the attraction at
o of the two masses at A and B, Fig. 89. FIG. 89.
The volume at A is r z du dr .'. charge, q = pr 2 dudr.
The volume of B is n 2 du dr .'. charge q' = pr^ dudri
.'. The attraction of A at is -^ = pdu dr.
The attraction of B at is 2 = pdu dri.
But from geometry it is known that with two ellipsoids, one of
axes a, b and c, and the other of a (1 + a), b(l -f a) and c(l-f ),
that is, with two similar, similarly situated concentric ellipsoids,
dr must always be equal to dri. Thus the attraction at must
be zero.
In the case of a conducting ellipsoid charged with electricity,
the charge is confined to the surface and the distribution will be
shown to be such as is represented by the thickness of the shell
in Fig. 89. It is greatest where the curvature is greatest and
least on the flat point of the surface.
The problem then is to express the thickness of the shell in
terms of a variable surface charge, cr.
The volume of the shell is evidently = %irabc [(I + a) 3 1];
considering uniform volume charge, the total charge is:
1 NOTE. See "Analytical Statics," vol. II, by ROUTH.
199
200 ELECTRICAL ENGINEERING
But a = pd, where 5 is the variable thickness of the shell,
:.Q-
or
3QS
J s
QS
FIG. 90.
But the thickness of the shell 5 can be ex-
pressed as the distance between two parallel
planes going through any point of the shell.
We have from geometry (see Fig. 90) that
the distance from the center of an ellipsoid to
a tangent plane is :
P = ~ (1)
/ g *
\ a 4 "" fe 4 "*" c 4
Neglecting infinitesimals of higher order than the first,
d = p(l + a) - p = pa.
Qp
. . a =
4irabc
y + a + l)
; or at the limit = 0,
(T =
4:irabc
Consider now a very thin flat elliptic disc in the x y plane
(c is small) we have from (1)
Q
o
when c approaches zero,
47Ttt&
/ x 2
V J - * - v
As a consequence for a circular disc,
Q
a =
- r
DISTRIBUTION OF CHARGE ON AN ELLIPSOID 201
where R is the radius of the disc and r the particu- P
lar distance from the center, where a is the surface
density on the disc.
To find the potential of the circular disc, we calcu-
late the potential at a point on the axis, Fig. 91. FIG. 91.
rdrQ
A = 2Trrdr2<r f
V = - r-- - ~ = 4-7T
Jr-R Vx* + r* J R
4irR\/R 2 -
r*
= __Q T __ rdr
' R J R V(R 2 -r z )(x
In this equation, x is, of course, a constant, being the distance
from the disc at which the potential is to be determined :
On the disc, x = 0,
Q C rdr
Q dr
Q
Incidentally, since the capacity is ^, it follows that the capacity
2
of a disc is - R, which is 2/Tr times that of a sphere of the same
7T
radius.
FIG. 92.
Potentials, Outside and Inside, and in the Body of a Spherical
Shell. Let the uniform charge per unit volume of the mass of
the shell be p, and the inner radius r and the outer radius R,
Fig. 92.
The area of the shaded surface, Fig. 92, is r^<p rA0
= r sin 6 A^> rA0;
202 ELECTRICAL ENGINEERING
the volume of an element of thickness Ar is :
r 2 sin 0A0A0Ar.
If p is the charge per unit volume, then the charge on the
small volume is:
q = pr 2 sin 0A<A0Ar.
Thus the potential function at P due to the charge on the small
volume is:
V = ^ but a = Vri 2 + (c - r cos 0) 2
= Vc 2 -f r 2 sin 2 + r 2 cos 2 - 2cr cos
= V c 2 + r 2 - 2 cr cos ;
or, a 2 = c 2 + r 2 - 2cr cos (1)
pr'sinftfrdfrfr
a r = ro ^ =0 , = c 2 + r 2 = 2cr cos
From (1), a 2 = c 2 + r 2 - 2cr cos 0,
.*. 2ada = 2cr sin 0d0
sin AM = ^ (3)
cr
Substitute (3) and (1) in (2),
r=R
_ f r
Jr
pr'adadrd*
" prdadrd*
^ 2 " f
= Q J
fr = R r? = 2*
r[(c + r) - (c - r)] drdS (4)
C Jr = ro Jv = Q
r = R r
I
= ro Jv
^ I 2irr*dr
= 47T5 /^! 3 - r 3 \ = p(volume of shell) = Q
c \ 3 / = c = c
(5)
If point P had been inside of the shell, then the limits of inte-
gration of a would be r c and r + c.
DISTRIBUTION OF CHARGE ON AN ELLIPSOID 203
.'. Equation (4) would be:
X" Tp X" _. O
F = 2 ( | r[(r + c) - (r - c)]drd<p.
C Jr = ro Jv = Q
2 rr=R r<p=2*
I crdrdp
Jr-
-B
(8)
which is independent upon c, the position of the point P.
Thus the potential is constant inside of a hollow sphere.
FIG. 93.
If the point had been in the body of the shell, Fig. 93, then the
potential would be the sum of the potentials due to the mass
outside and inside of the spherical surface which contains P.
The field intensity or potential gradient is
dv
(The signs should all be reversed for gravitational potentials.)
In the case of the point being outside the sphere,
dV Q 4
dc
3c 2
- r 3 )
and
2Q STTP (R* - r 3 )
c 3 : 3c 3
In the case where the point is inside, it is:
(8)
(9)
204 ELECTRICAL ENGINEERING
where the point is in the shell then :
^Z o f ?cf W-l
" dc = ^ p L" 3 " 3cJ
rV _ 1 _
c 3
2r 2
(10)
dc
31
c 3
Problem. Plot the potential, the potential gradient, and
d 2 V
j-z
dc 2
when V 1 at the center;
ri = 1;
r-o = 0.5
in the case shown in Fig. 94.
For a full discussion see WEBSTER'S "Electricity and Magnetism."
FIG. 94.
FIG. 95.
Potential Outside of a Non-conducting Charged Oblate Ellipsoid,
Let the equation of the oblate ellipsiod, Fig. 95, be:
x 2 y 2 z 2
~o ~I 9 ~T~ ~o =
o 2 a 2 c 2
DISTRIBUTION OF CHARGE ON AN ELLIPSOID 205
Let the total charge of the ellipsoid be Q, and the potential on
the surface be VQ.
The surface intensity at the element ring, generated by ds,
Fig. 96, revolved about the z-axis, has been proven to be:
pQ
(T -
X
FIG. 96.
where p is the distance from the origin to ds, and
5 + * + ~ +
where
Q
,2 ~2
From (1),
a 2 ^ c 2 ~
Differentiating, 2r dr Q 2zdz _
~2 I ^2 ~~ U >
or,
(2)
(3)
206 ELECTRICAL ENGINEERING
The potential at P on the axis due to the ring-shaped element
surface is:
(4)
(r -
Substituting (2) and (3) in (4) we get: the potential at P
due to the whole ellipsoid
Qdz
r,
2cV(r-z) 2
From the equation of the ellipsoid,
substituting in (5),
"=
(5)
f_ c 2V - (a 2 - c 2 ) z 2 - 2rc 2 z + c 2 (a 2 + r 2 )
, (a 2 - c 2 )z + re 2 1 c
, sin" 1 - . = =
2V a 2 c 2 caVa 2 - c 2 + r 2 J _ c
Q f. . a 2 -c 2 + rc . 1 -a 2 + c 2 + rc] /
. sm" 1 x = sin" 1 . : (6)
2V a 2 - c 2 [ aVa 2 - c 2 + r 2 a-ya 2 - c 2 + r 2 J
To find the potential at a j>oint, like PI, which is not on
the z-axis, LEGENDRE'S function may be employed, and the
equation (6) is to be expanded into a series in the terms of r.
In order to obtain an expression which may be easily expanded,
differentiate (6) with respect to r, expand the result into a series,
and then integrate the series. Thus differentiating (6),
dV p Q
(7)
dr (a 2 - c 2 + r 2 )
Expanding (7),
dV P -Q r r 2 r 4 _r^_ l
dr a 2 -c 2 L a 2 -c 2 ~ i (a 2 - c 2 ) 2 " (a 2 - c 2 ) 4 ' 'J 1
when c < r < \/a 2 - c 2 (8)
V Q -[_ c , y( _ c V _
" 2 2 2 2 \Va 2 - C V
DISTRIBUTION OF CHARGE ON AN ELLIPSOID 207
For a point on the surface, i.e., when r = c,
.-. C = V~ + tan-' -7= (10)
^ V a 2 c 2
Since F Pl is a function of V P and /, the solution for V Pl takes
the following form:
V Pl = Ao + AiriPi(cos 5) + A 2 ri 2 P 2 (cos 0) +
A 3 ri 3 P 3 (cos 0) + ....
When
= 0, n = r, Pi = P 2 = P 8 = - - . 1, and F Pl = F P .
^ + - - tan- 1 - 7 J=
Q QVa 2 - c 2 \/a 2 - c 2 J
a 2 - c 2!
= 0;
. U-- -
(a 2 - cV
1 c A P^cosfl)
tan L 7= = I 7^- ^r
(a 2 c 2 )
P 3 (cos e) , P 5 (cos 0) , , P 7 (cos 6) 7 _
I v / ~* 3 * L- M 5 I x y , 7 _i_ I ,
I O / 9 9\ 9*1 F* / *> 9\ *? ' 1 I^ T / 9 9\ A ' 4 I I
3 (a 2 c 2 ) 2 5 (a 2 c 2 ) 3 7(a 2 c 2 ) 4
which is applicable, when
When
expanding (7),
Whenr = c, V P = 0, .'. (7 = 0.
And
_ Q , ii) , A 2 P 2 (cos0) , 3 3 ,
I n" " r! 2 n 3 4 -
208 ELECTRICAL ENGINEERING
When = 0, PI = P 2 = P 3 = . . . = 1, r l = r, and V Pl = V P)
/. ^o = Q;
A! = 0;
4 Q(a-c).
"T"" ;
A* = 0;
= 0;
a2 - c2 ) Ps ( cos ^) . (a 2 - c 2 ) 2 P 4 (cos 0)
~~ ~
(a 2 - c 2 ) 3 P 6 (cos 6>)
7ri 7 -
which is applicable, when
ri >Va 2 -c 2 . (12)
(Two similar series can be derived for an oblong ellipsoid.
For this and the potential at a point inside an ellipsoid, see
W. E. BYERLY'S "Series.")
CHAPTER XVIII
CONCENTRIC SPHERES
Fig. 97 represents a system of concentric spherical shells. It
is desired to find the potential at any point in the medium (which
is assumed free from charge).
Since we are dealing with spherical bodies and since the body
is symmetrical, indeed a sphere, LAPLACE'S equation in spherical
coordinates becomes :
d*V
a PP endlx ) (1)
FIG. 97.
To solve this equation, one first ascertains if the relation
dV = A
dr ~ r 2
is satisfactory. (We may well assume this solution, since it can
be expected that the intensity or force on unit charge varies
inversely as the square of the distance.)
Then,
d^V _2A
dr 2 ~~ r 3 '
Substitute in (1) to see if the solution satisfies the equation
-~ + l = o,Q.E.D.
Thus, 37 _ A
~dr ~ 7*
209
210 ELECTRICAL ENGINEERING
satisfies the equation (1).
o r ,V=- + B (2)
Or we might have solved the equation as follows :
Let _ dV . dW _ dy
y ~~ dr' ' dr 2 dr
.-.*+*.-
- J*-dr -log r2 A A
. . v = Ae = Ae = , 1) -$
e log r* r 2 "
Or again we might have developed the equation directly, without
using LAPLACE'S equation, by assuming a positive charge Q on
the inside sphere.
The intensity of the field at a point in the medium at a distance
r is then by GAUSS'S theorem:
A~n n
R =
.'. V = - \ Rdr = + ^ + B (3)
an equation of the same form as (2).
Referring to equation (2), let Vi be the potential of the inner
sphere of radius r\ and Vz that of the outer, then,
and ' F 2 = - + B.
r 2 r rir 2
r 2
r 2 - ri r
where ri < r < r 2
CONCENTRIC SPHERES 211
To determine the meaning of B assume that the outer shell is
grounded, or which is the same, at zero potential, then
and
from (4),
. v tfi D . K r l
. . V 2 = ~ ~~ - -T jD. . . > = - - V i.
r 2 - ri r 2 - n
From (5), y = Vi n r 2 _ _TI __
r 2 - ri r r 2 - ri *
= _FVi_ rra _ 1 1 == JV
r 2 ri Lr J r 2
- r
7*1 r
If the outside sphere is very far off so that r 2 approaches
infinity and F 2 zero, then,
V z = 0, r 2 = oo ;
7 2 = o = - + 5, /. 5 = 0.
r 2 ? r r
The potential gradient in the space between the conductors is:
7.-F.
It is the greatest at the surface of the inner sphere, where
r = ri.
r 2
The potential gradient at the inner surface of the outer conductor
is evidently :
r 2 ri r 2
Referring to equation (7),
R = 4^ri = ^J equating to (8),
Qi _ 7i-7 2 ra.
r 2 -
14
212 ELECTRICAL ENGINEERING
Example. Calculate the average potential gradient in the
space between two concentric spheres separated by a distance of
2 cm.
Assume that the potential gradient at the surface of the inside
conductor is 100 electro-static units per centimeter, that is, just
about on the point of glowing.
Consider a concentric sphere, Fig. 98, the inner sphere of which
has a charge Q\ and the outer a charge Qo = $2 + Qs-
FIG. 98.
Evidently, Qo = Q 2 + Q 8 .
Since all tubes of force beginning at the surface of the inner
conductor terminate at the inner surface of the outer conductor,
it is evident that the charge Qz = Qi-
.'. Qo = - Qi + Q 8 .
The potential at a point outside of the outer conductor is
thus, from (6),
T7 Qo Qs-Qi ,
y = ^_ = ^ *_, w here r = r 3 .
r T
Since the capacity of an electric field is the ratio between the
charge on the positive boundary and the potential difference
between the boundaries,
c -
r i-F 2
Thus horn (9),
c== Ii^Z_ 2 . rir2 . 1
r 2 ri Fi K 2 / 2 / 1
The capacity of the inside sphere alone is ri.
Capacity of concentric spheres _ r^
Capacity of inner sphere r 2
CONCENTRIC SPHERES 213
If the thickness of the dielectric is small compared with the
radius, then:
C = r , ^^ = T 4' where S = r, - r,.
6 d
4?rri 2 _ area of sphere
as a limiting case, where TI = r^ = we get parallel plates, and,
area on one plate
47r(distance between them)
The capacity is expressed in cm. not in farads. To get the
capacity in farads divide C by 9 X 10. u
The energy input to a condenser is:
W =
Thus the energy stored in the field between two concentric
spheres, is:
Infinite Parallel Planes. LAPLACE'S equation applies in this
case so long as there are no charges between the condenser plates,
_ d*V
dx 2 " dy 2 ~~ dz 2 =
Since the field depends upon the distance between the plates only,
that is, upon one of the coordinates only, we get,
w _ . dv __
"T "^ U, . . j L/o
dx 2 dx
and V = C x + d. ,
If the charge on plate A (Fig. 99) A ifc^T | Ql
is Qi and the potential FI; and the ,
charge on plate B is $2 and the poten- B - - 1 - Q *
tial Vz', and if the distance between
the plates is d]
We have:
Vi = + Cj,
and
F 2 = C d + Ci.
Subtracting,
Vi - V z = - Cod, or, C = - (Fl "T V ^.
214 ELECTRICAL ENGINEERING
:.V -^ -x + Ci; or, since V 1 = d,
d
The potential gradient, that is the potential drop per cm. is:
=-=^
It is constant all through the dielectric.
The total outward flux from A is 47rQi, one-half of this enters
the space between the plates. The inward flux to B is ^irQi,
and one-half of this is added to the flux from A. Thus the total
flux in the space between the plates is:
But the charge on A, Qi, must be numerically the same as that
on B, Q%, since all tubes of force leaving A enter B, thus Qi = $2,
numerically, but of course of opposite sign, which, however, is
taken care of in the above discussion.
Thus the total flux in the gap is 4?rQ, where Q is the charge on
one of the plates.
47T0
.'. R, the intensity of the field, is j- where A is the area of one
side of the plate.
And ' G = R = ^;
or from (1),
4rQ Fi- F 2 . r _ Q A.
A d " ' ' L = Fi - 7 2 "
This could have been calculated in still another way.
Since D
V = - fRdx = - X
For x = 0, V = 7i; /. Ci = 7i.
for , ,, T7 . T7 T7
oj = d, V = 7 2 ; . . V z = Vi -- T d,
CONCENTRIC SPHERES 215
1s . 2 ..
If the plates are separated by uniform insulation of specific
., v ,, ., . KA KA
inductive capacity, A, the capacity is -r, cm., or . , Q 1()ll
farads.
If the dielectric consists of several layers of different specific
inductive capacities then one can consider that the condenser is
made up of a number of condensers in series and the capacity
of each is:
KiA
Ci = -7r> etc.
47rdi
The total capacity is obtained from the well-known relation:
1 1
C ~ Ci '
1
f 7T + . , or,
C 2
1
1
, 1 ,
irdi 4ird 2
Ci
F c 2 +
V * V * '
AI A 2
All these formulae are approximate, however, since no allow-
ance has been made for the effect of the edges, but the plates
were assumed to be infinite.
Concentric Cylinders. LAPLACE'S equation can again be used
if it is assumed that there are no charges between the cylinders.
Moreover since we are dealing with cylinders, it is best to put
LAPLACE'S equation in cylindrical coordinates. Thus we have:
~dr* + r ~fo ~~
let y = -r-> then (1) becomes -j- H y = 0.
The solution of this equation is
-f*L = A = A
y ^^ -. log r " M
I
dV A
216
ELECTRICAL ENGINEERING
To determine the integration constants,
let V = Fi, r = r, (Fig. 100)
and V = F 2 , r = r 2 .
Then, Vi = A log (n) + B,
and T r 2 = A log (r 2 ) + 5.
.'. Fi - F 2 = A (log ri - log 7- 2 ) = A log
and,
><*
FIG. 100.
The potential gradient or the intensity of the electrostatic
field is:
dV V, - V Z /1\ 47TW 2^
where Qi = charge per unit length of conductor, and Z = length
of conductor.
per centimeter length of conductor.
The potential gradient is the greatest at the surface of the inner
conductor, where it is:
1 7, - 7i
Graded insulation between the conductors.
In order that G may be constant at all points of the dielectric
it is evident that the specific inductive capacity must be the high-
est at the inner conductor, and be inversely proportional to the
distance from the inner conductor.
CONCENTRIC SPHERES 217
Let the specific inductive capacity be expressed by the follow-
ing formula:
K = ~, where a is a constant.
With a charge Q on the inner conductor, the flux per centimeter
length is 4irQ, thus the force on unit charge is:
_ 2Q
K2irr ~ Kr
=- C^dr=- (^dr=- f*
J Kr J ar J a
dV 2Q
G = = -- = constant.
dr a
The same result could have been obtained directly from (2),
which, in the general case when Kl, becomes:
R 4,0 2Q
K2irr Kr
a
K = ->
r
Substituting
R =
G)
20
R = - - = constant, Q.E.D.
CHAPTER XIX
CYLINDRICAL CONDUCTORS
Line Charge. Assume that the conductor which is perpendicu-
lar to the page is infinitely long and its diameter so small that
it may be considered as line, and let the charge per unit length
beQ.
The electric field is then represented by radial lines in planes
parallel to the page or, which is the same, at right angles to the
axis of the conductor.
The intensity of the field at a point P, Fig. 101, is obviously:
2Q
And the difference in potential between two points PI and P is:
- C^ dr
Jhi r
- 2Q [log n - log fcj = 2Q log (1)
FIG. 101.
Two equal but opposite line charges separated by a distance 2hi:
Let A and B (Fig. 102) be the locations of the line charges.
The difference in potential between midways between the
charges and P, due to the charge on A alone, is and has been
shown :
V p - V. = 2Q log (2)
The difference of potential between o and P due to the line charge
Q on B is obviously,
V p - V, = - 2Q log J- 1 (3)
TZ
218
CYLINDRICAL CONDUCTORS
219
Thus the difference of potential between and P due to both
line charges is:
V 9 - V = (2Q log - log ) = 2Q log . (4)
Referring to equation (2) or (3), if P lies midway between A
and B, so that r\ = r 2 = hi, then:
V p - V = 0,
thus as long as the charges are equal and opposite, the potential
at is zero, which would, of course, have been concluded without
proof.
V = 2Q log ^ (5)
where V is the potential of P due to the charges on both lines.
From (5), follows
T2 --
= 2Q = a = a constant
(6)
for all surfaces of potential V.
Equation (6) represents a circle, defined by the following relation :
~OA X OB = R 2 (7)
referring to Fig. 103, where is the center of the circle, A and
B Fig. 103) are called the inverse points, and O f the center of
inversion.
FIG. 103. FIG. 104.
To prove that equation (6) represents a circle refer to Fig. 104.
4-
or.
O,
220 ELECTRICAL ENGINEERING
which is the familiar equation of a circle having a radius of
o __ ,
"1-C 2
and its center at a point whose coordinates are:
"1-C 2 from A;
= 0,
/. OA X OB =
from B.
= R 2 ;
(1 - C 2 ) 2
thus, equations (6) and (7) are proved.
The ratio, > can be expressed by a simple equation involving
h, the distance of the center from the neutral plane, and the
radius, R.
FIG. 105.
Referring to Fig. 105.
R 2 = OA X OB = (h - hi)(h +
hi =
or, i = 2 - R 2
But triangles OPB and OP A are similar, since
OP 2 = OA X OB;
- _T?_ Tl ,
'OP ~ OA'
r, = OP = R = B
ri ~ OA ~ h hi ~ a
(8)
(9)
CYLINDRICAL CONDUCTORS
221
Substituting (8) in (9),
rs = R = R(h +
ri ~
h-
We can then determine the potential of a circle, or, which is
equivalent in this case, a cylinder, whose center is h cm. from
the neutral plane and whose radius is R, as
Similarly the potential at a circle around the negative charge
7 2 = - 2Q log
R
(12a)
/. V = V l - 7 2 ,
that is, the potential difference between the two cylinders is:
h +
4Qlog
R
(13a)
For the sake of convenience, will be added other expressions for
Vi, Vz and V, involving hi, and R instead of h and R.
From (8), h 2 = R 2 + /U 2 ,
which, substituted, gives
V l = 2Q log
R
- -2Qlog-
R
and,
= 4Qlog-
4- V hi 2 +
(126)
(136)
It is now evident how we can go from line charges to charges
on actual conductors. It has been proven that the equipotential
surfaces around the line charges are cylinders and hence if circular
cylinders be substituted for the circles, the distribution of the
field is not affected.
The capacity per centimeter length of two such metal cylinders
(that is, of the double conductor) is :
4 log
h +
= cm.
(14)
R
222 ELECTRICAL ENGINEERING
CW--- -f- -/=== farads (15)
9 X 10" 4 log * ^~
or, C m _/. per 1000 ft., of circuit (double conductor)
m-f. (16)
R
'where logic means the ordinary logarithm not the natural
logarithm h is half the distance between conductors, and K
the specific inductive capacity.
If E is the effective value of the alternating-current line
voltage, then the charging current per 1000 ft. of double con-
ductor is readily proven to be:
C m . f .
The capacity to neutral is obtained directly from (lla) and is:
c= 1
n , h + Vh 2 - R 2
2 log - gr-
It is thus seen that the capacity to neutral is twice as great
as that between the lines.
This results, of course, in the same charging current as in the
E
first case, since in this case the voltage is -^- Thus the capacity
of 1000 ft. of one wire to neutral or ground is:
C m -f. = ~ j-- , = m-f. per 1000 ft. of transmission.
logio - ~/j>~
Two Parallel Cylindrical Conductors of Different Diameters
but Equal and Opposite Charges. Since OA X OB = Ri 2 and
(FB X WA = R 2 2 , we have
a(a + 2hi) = Ri 2 , or a hi + \/h\ 2 + R\ 2 (1)
and -'- ' 2hi) = Rz 2 , or = -hi + V/^F^T 2 (2)
and ^ F,= -2Qlog^= -2Qlog| 2
(3)
CYLINDRICAL CONDUCTORS
223
Substituting (1) and (2) in (3),
Vi - V 2 = 2Q log
and
C =
Q
(- hi '+ Vfti 2 + #i 2 ) (- fti + \//ii
1
2 log
RiR:
- hi + Vfti 2 + Ri 2 ) ( - fti + Vfti 2 + #2 2 )
To obtain an expression in terms of h and R, instead of hi and
R, from Fig. 106 we have:
(3 = 2h - 2hi - a (4)
.'. /? + 2hi = 2h - a (5)
FIG. 106.
Substituting (4) and (5) in (2),
(2h -
Solving (1) for 2hi and substituting it in (6),
a
(6)
or,
or,
(2h =
2ha 2 +
a)(2h a) R% ,
- #i 2 - 4/i 2 ) a + 2hRS = 0,
- V(Ri 2 - R* 2 + 4/i) 2 - 16ft 2
4ft
where the sign in front of the radical is minus not plus,
because a = when Ri = 0. Similarly,
4ft
.'. C =
2 log
_ __
(RiR*\
224
ELECTRICAL ENGINEERING
2 log
4/i 2 - (Ri 2 + R 2 2 ) - V 16/i 4 -
which becomes:
C = -
4 log
h + \A 2 - R 2
R
if R is substituted for both RI and R z , a result obtained before.
Construction of Equipotential Surfaces around a Cylindrical
Conductor, Charged to a Certain
Potential, V. Let the distance be-
tween the center of the conductor,
Fig. 107, and ground be h, and the
distance of the equivalent line charge
above ground be hi.
Since the ground is an equipotential
surface, it is evident that the problem
will in no way be affected, if a
second conductor with a charge Q
be placed equidistant below the
ground surface, and the equipotential
surfaces around A be considered as
due to a positive charge, Q at A, and
an equal but opposite ("image")
charge Q, at the inverse point A'.
Suppose that it is desired to draw the equipotential surface
through a point P, distant d from the ground.
The first step is to locate the equivalent line charge in the
original conductor of radius R and distance h from ground.
We have,
/ii 2 = h 2 - R 2 ,
.". hi = Vh 2 - R 2 (1)
from A, the
(2)
(3)
(4)
FIG. 107.
h 2 - R 2
To find the radius of a circle whose center is
location of the equivalent line charge, we have,
But from the figure we have,
hi + ttl = fa + d
.*. i = RI + d hi.
CYLINDRICAL CONDUCTORS
225
Substituting (4) in (2),
(Ri + d- /ii) (2/i! +
-to
- hi)
2d
(5)
The potential of the circle of radius Ri, which goes through the
point, P, is:
Fl = 2 Q log ^ = 2Q log *'
But V, the potential of the conductor, is:
log
Knowing the radius from (5), and the center is Ri + d above
ground, the equipotential surface through P can be drawn, and
the potential of that surface is given by (6).
Potential of a Cylinder due to External Charges. In order
to determine the potential due to a number of charged cylindrical
conductors, it is necessary to calculate the potential of one
cylinder due to charges on other cylinders placed in the vicinity.
FIG. 108.
Consider a line charge Q at B in Fig. 108 and determine the
average potential due to Q on a non-conductive cylinder A. The
potential at P is, as has been shown :
V = 2Q log ->
but from the triangle OPB,
r 2 = d 2 - C 2dr l cos
- -~ cos
ELECTRICAL ENGINEERING
r = d V/c 2 + 1 2k cos <f>,
226
or,
where
Thus the average potential of A is
= TT I
*TrjQ
log
r
Jo
log (1
- cos
- cos
/"2r
where
2Q \oghi-2Q\ogd- I log (a-b cos
a = 1 + fc 2
and, b = 2k.
Evaluating the definite integral (see PIERCE'S "Table of
Integrals") we find that the last term is zero.
Thus,
V A = 2Q(log h - log d) = 2Q log -j 1
(D
Thus, the average potential is independent upon the radius of
the conductor.
But equation (l)has been shown previously to be the potential
at a point distant d from a line charge distant hi above ground.
Thus to determine the potential of a cylindrical conductor A,
due to a line charge at B distant d, the diameter of the conductor
does not enter as long as, with metallic conductors, the field can
be assumed not disturbed by the conductor.
^-^
?> /
>\
I d z
Wa
_X
h*
\
FIG. 109.
Referring to Fig. 109,
The potential of A due to B is:
log -
CYLINDRICAL CONDUCTORS 227
The potential of A due to C is :
F 2 = 2Q 2 log |-
.'. V = Fi -}- F 2 = 2Qi log ^ + 2Q 2 log ^
ai a 2
Lines of Force between Parallel Cylinders. Let s-s (Fig.
110) be a part of a line of force, and N-N a line at right angles
to it. Thus the projection of G\ on the normal is Gi 1 = Gi cos a,
where G\ is the intensity at P due to the line charge at A . Simi-
larly the projection of (7 2 on the normal is G 2 cos 0. The sum
of the projections must be zero, since N-N is perpendicular
to the line of force.
cos a -f G 2 cos |8 =
(1)
But
and
cos
cos a =
Similarly,
L ds
-2Q
TT
as
15
228 ELECTRICAL ENGINEERING
Substituting in (1),
de l + de = o
or 0i + 6 2 = constant.
This equation represents a family of circles through A and B,
with center on the line 0-0.
Construction of Lines of Force. Referring to Fig. Ill, as P
is in the center line,
n
= G l cos
or,
FIG. 111.
Knowing the values of x and the fixed points, A and B, the
lines of force, being circles, can be readily constructed.
Problem. Draw equipotential surfaces around a line charge
placed 10 cm. above the neutral plane, when the charge is 1
electro-static unit per centimeter of conductor.
Find the radius of the conductor containing the line charge
whose potential is 2000 volts. Draw surfaces corresponding
to 400, 800, 1200 and 1600 volts.
Draw lines of force whose intensities at the neutral plane are
120, 110, 100, 90 and 80 volts per centimeter.
Solutions.
First. Radius of conductor: Since 2000 volts corresponds to
6.67 electro-static units, we have:
6.67 = 2Q log
hi + Vhi 2 + R 2 . 10 + \XI66~+ fl 2
- = 2 log -
R
R
.. 10 + VlOO +^ n^Q/ix/QQ 1 AA^
. . logio ^ = 0.434 X 3.3 = 1.445.
R
. 10 + V 100 + R 2
R
= 28.05 .*. R = 0.72 cm.
By a similar process the radii corresponding to 1600, 1200, 800
and 400 volts are found.
CYLINDRICAL CONDUCTORS
229
These being calculated, the corresponding values of A, the
distances from the neutral plane, are found by the relation
h = VV + R 2 -
Second. To find the intersection between the neutral plane
and the line of force of intensity 100 volts per centimeter or 0.333
electro-static units, we have:
- 1 = 10
).333 X 10
-1 = 10X0.447 = 4.47 cm.
Capacity of Two Cylindrical Conductors, when the Effect of
the Proximity of the Earth is Considered. Consider, for the
sake of simplicity, the case of two cylinders of equal radii, and
charges Q and Qi respectively.
FIG. 112.
Referring to Fig. 112, it has been shown that the potential
of A due to its own charge, Q, and the charge on its image, A' is:
2Qlog
R
(1)
It has also been shown that the potential of A due to the Qi, on
conductor B is:
V, = 2Q 1 log ^ (2)
Similarly, the potential of A due to the image of B is :
V, = 2Qi log - 1 (3)
230 ELECTRICAL ENGINEERING
Thus the total effect of conductor B on A is:
V z + V, = 2Q l log ~ (4)
And the resultant potential of A is:
"
V A =- V l + F 2 + V 3 = 2Q log
Similarly,
^-^log^f^-H*]** (6)
Special Cases. Two wires in parallel at same distance from
ground.
Thus h = H, Q = QL .'. V A = F B == 7.
Thus the capacity per centimeter of each wire is:
Q l
2 log
[I"
R J
and the capacity of the two wires taken together, is:
c = - -]> > + -i/M^fiT (8)
\d h + vft 2 # 2
108 U' ~R~ "J
In the case of a transmission line, ft is large compared with
R, and d f is approximately 2ft.
1 1
It has been shown that the capacity of a single wire to neutral
is:
approximately. (10)
2 log " ' - " " - ' 2/i
JLl/ Xl/
Thus the proximity of the other wire has reduced the capacity
of each wire, so that the combined capacity of the two in parallel
is usually not more than 25 to 30 per cent, greater than that of
a single wire.
CYLINDRICAL CONDUCTORS 231
As an instance, let R = 0.5 cm., h = 1,000 cm., and d = 20 cm.
2000
(log -- + log
= 0.0388 cm. per centimeter;
.*. 2Ci = 0.0776 cm. per centimeter
and the capacity of one single wire alone is
C """2000 = 0-0603 cm. per centimeter.
21 six<r
The capacity of the double wire is thus only 28.7 per cent.
greater than that of a single wire.
Second. Assume now that wire A' forms the return for A, so
that the charge on A is Q and that on B is Q.
From equation (5),
2Q log - - - approximately;
and rt ~ . (d' R
V B = 2Q log approximately.
- V =
If the effect of the ground has been neglected, then, as has been
shown, the capacity between the two wires would have been
approximately :
c = "
Comparing equations (11) and (12), it is evident that since -
is always smaller, but usually only very little smaller than
unity, C| is slightly greater than C.
The proximity of the ground has thus slightly increased the
capacity between wires. In transmission lines, the increase
amounts usually to less than 1 or 2 per cent.
CHAPTER XX
MUTUAL AND SELF-INDUCTION OF ELECTRO -STATIC
CHARGES OR FLUXES MAXWELL'S COEFFICIENTS
If among a number of conductors say No. 1, No. 2, etc., a
particular one, say No. 1, is given a charge qi, so that its potential
is Vi, and if all other conductors are connected tpo ground, that
is, are at zero potential, then,
where Ki.\ (with its two indices) is called the coefficient of
self-induction of electrostatic charge, and is, as seen, the capacity
of No. 1 due to its own charge q\, when all other conductors are
at zero potential.
Obviously while the potential of the other conductors is zero,
each has a certain part of the induced negative charge corre-
sponding to qi on No. 1.
The charge on No. 2, for instance, is of course proportional to
the potential of No. 1 and is written:
Similarly,
#3 = Ka.iVi, <?4 = Kt.iVi t etc.
KZ.IJ KS.I, etc., are called the coefficients of mutual induction.
Since Vi is positive, # 2 must be negative, therefore, K 2 .i, or in
general, K with two different indices, is always negative, while K
with same indices is positive.
If instead of grounding all of the conductors except No. 1, we
now ground all but No. 2, and this is given a potential 2, we get,
by a similar reasoning,
q* = K 2 .iV z , qz = #3.2^2, #4 = #4.2^2, and, qi = K^Vz.
Superimposing these conditions, it is readily concluded, that,
if at any time the potential of No. 1 is Vi, that of No. 2 is F 2 ,
etc.
232
MUTUAL AND SELF-INDUCTION
233
The following relation obtains, if Qi, Q 2 , Qs, etc., are the total
charges on No. 1, No. 2, etc.:
Qi =
A little consideration will show that
KLZ = /V2.i, etc.
The applications of these relations will be
illustrated in the case of the two similar
overhead wires (Fig. 113). The immediate
problem being to determine the values of
KI.H Kz.2 and Ki.%.
On account of symmetry, KI.I = Kz.z y
thus we have really only two unknown
quantities, namely, KI.I and KI.Z.
To determine them, give two equal
charges +Q to the conductors, then Vi F 2 .
(1)
20 x?
,,
From (1), & =
i + K^V, =
= = C =
+
d' 2h\
t
I 2
log (d ' R
[See (9) in the previous article.]
Now give one conductor a charge +Q and the other a charge
Q, so that the potential of No. 1 is Vi and that of No. 2 is
-Fi, then from (1), Qi = Vi (K^ - K^),
= c f -
2 log '
(3)
[See (11) in the previous article.]
From these equations it follows that :
lo (}
KI.I = - 5 '
'd 2h\ id' 2h\
^d' R/ \d R/
log "
r - r
234 ELECTRICAL ENGINEERING
and, , 2/i
#1.1 g R C + C'
#1.2 , d' C - C'
Numerical application, Fig. 113:
Let R = 0.5 cm.
h = 1000 cm.
d = 20 cm.
/. d' = 2000 cm.
.' 7^ = 4000,
? = 100 '
and > I =, o.oi.
'* C = 2 log 400,000 = 0< 388 '
C' = s-^ -777 = 0.1352.
2 log 40
.'. K L1 = 0.087,
Ki . 2 = - 0.0482,
f^- 1 = - 1.806.
A 1.2
Discussion. To show the application of these coefficients, the
following problems will be considered.
A. Compare the capacities between a wire and ground,
(a) when the wire is alone; (b) when an adjacent wire is grounded.
B. Compare the charging currents for the same applied voltage
between the two conductors when the two wires are insulated,
and when one is grounded. In the latter case, give the relative
proportions of the current in the grounded wire and in the ground
itself.
The numerical case will be: R = 0.5 cm.;
h = 1000cm.;
and, d = 20 cm.
The problems will be best solved by the use of the MAXWELL'S
equations, viz.
Qi = #i.iFi + Kt.iVt + Ki. t V*,
Q2 = #J.lFi + #2. 2 F 2 + #2.3^3,
and, Q 3 = K Z . 1 V 1 +
MUTUAL AND SELF-INDUCTION 235
In these equations, index 1 refers to conductor No. 1, index
2 to No. 2, and index 3 to the ground.
Since the potential of No. 3 is zero and since we assume two
similar and similarly placed conductors,
V 3 = 0, KM = #2.2 and K lft = K,. 3 .
/. Qi = K 1 . 1 V 1 + #!. 2 F 2 (7)
Q* = #i. 2 F! + #i.iF 2 (8)
and Q 3 = #i. 3 Fi + K^V* (9)
Case A. (a) It has been shown that with a single conductor
suspended above ground, the capacity is:
C = -- ^r = 0.0601 cm. per cm. (10)
r i ""'
21og fl
Thus if V is its potential the charging current is:
ri = 0.0601 ^
(6) since No. 2 is grounded, T 2 = 0.
Thus from (7), Qi = ^i.iFi .'. capacity = KI.I = 0.087, and
ri = 0.087 ^
The capacity of wire No. 1 is increased 45 per cent, by the
proximity of the grounded adjacent wire No. 2.
Case B. Under normal conditions,
Qz = - Qi and KI.I = K 2 .z,
.'. Qi = tfi.iFi + X!. 2 F 2 ,
Thus the capacity between the conductors is:
C = g " ~ g " = 0.0676.
i
If FI F 2 = F, and if z'i is the current in conductor No. 1,
then
dV dV dV
i l = C^ = y 2 (#!.! - #,.,) ^- = 0.0676^-
236 ELECTRICAL ENGINEERING
When No. 2 is grounded, Vz = 0.
.'. Qi = Ki.iVi = Ki.iV, thus the capacity, C" =
= 1.285.
*1 JV.1.1 ~ A- 1.2
The charging current in conductor No. 1 is increased 28.5
per cent, by the proximity of the adjacent grounded wire.
The charge in conductor No. 2 is:
Qz = K 2 .iVi -f #2.2 V z = Ki. 2 V, since 7 2 = 0.
But Xi.t = - 0.135 + 0.087 = -- 0.048
Thus ^2= -0.048 f
The current carried in the ground is obviously
- 2 3 = (0.087 - 0.048) -^
.: ,-,-- 0.039
If the current in No. 1 after grounding No. 2, is taken as 1 amp.,
then wire No. 1 carries 1 amp., No. 2, 0.554 amp. and the
ground, 0.446 amp.
Problem. Assume three similar horizontal conductors of
R = 0.5, h = 1000 and d = 20.
Give the relative values of the charging current between No. 1
and No. 3 if No. 2 is indulated, and if it is grounded. Also give
the charging current if No. 2 is removed entirely. Consider
the current in the last case to be unity.
CHAPTER XXI
TWO-CONDUCTOR CABLE
Since the conductors as well as the lead covering are of metal,
the surfaces of each are equipotential surfaces. In order to
simplify the calculations it is desirable to substitute for the sheath
and each conductor a system of conductors, i.e., the conductor,
and its image, which will give the same distribution of potential.
Consider first the system of Fig. 114 consisting of A, its image
A' and the lead sheath. It is necessary to determine the position
of the line charges at distance hi from the neutral plane, so that
the conductor A and the sheath are equipotential surfaces.
From what has been shown previously, it is evident that the
following relations exist:
and
hi 2 = h 2 r 2 , when considering the conductor;
hi 2 = (h + a) 2 r*i 2 , when considering the sheath.
i 2 - r 2 - a 2
Having determined h from (1),
hi is determined, as hi = \/h 2 r 2
(1)
(2)
FIG. 114.
Referring to Fig. 114, it is evident that the potential of A is
due to its own charge and the charge on its image, and the charges
on B and its image.
TVf)
It is also recollected that the latter potential is: 2Q log =
mp
237
238 ELECTRICAL ENGINEERING
if we neglect the shortening of the lines of force from m to p in
going through conductor B, where np is the distance between the
line charge in B and the center of A, and mp is the distance be-
tween the line charge in B' and the center of A.
.'. np 2a -f- h hi,
and mp = 2a -f- h + hi.
21 r' + VW-r'^a + h-h!
to neutral (3)
i" r* za -f AI Aii \
Approximation. Frequently, in fact almost always, the follow-
ing approximation can be made :
h = hi.
(4)
2 log - "'
r a
If furthermore r 2 is small compared with ri 2 a 2 , and is small
compared with A 2 , then,
h -4- A//? 2 r 2 2h ri 2 a 2
-
/I
.
2a r r ar
thus, c = _^ __ ^ the capac j ty to neutral (5)
2 log ( 1 ,
\r r^ +
Thus, the capacity between the two conductors is approximately
C " /aa'r.. - (6)
4 lo g (7- " ^rf
or, in microfarads per 1000 ft. of cable,
(7)
,
log
l
To determine the capacity of the two conductors in parallel
against the sheath, the two conductors are given positive charges,
+Q, and hence the charges on the images are Q.
TWO-CONDUCTOR CABLE 239
The potential of A due to its own charge and the charge on
its image is:
V' A - 2Q log *X^Ei
The potential of A due to the charges on B and its image is:
or, using the same approximations as before,
(9)
The potential of the sheath, if insulated, due to the charges
in A and its image is:
V. = 2Q log *'
Similarly, due to B and its image is :
y". = 2Qlog^-
r\
.-. F. = 4Q log (10)
Tl
Using the same approximations as before,
V 8 =- 4Q log ^ (11)
Thus the potential difference between the sheath and either
of the conductors (when they are connected in parallel) is
approximately :
V -V A -A. = 2Qlog-~- - 2Qlog
(12)
Thus the total capacity between the two conductors in parallel
and the sheath is:
log
240
ELECTRICAL ENGINEERING
In connection with this it may be of interest to determine the
capacity between the conductor and the sheath in a single con-
ductor, eccentric cable, Fig. 115.
The potential of A due to its own charge and the charge on
its image is:
n + VV + r 2
V A = 2Qlog
FIG. 115.
The potential of the sheath due to the charge on A and its
image is:
7 8 = 2Q log
hi + Vhi 2 +
.'. C =
II
(13)
Denoting the conductor A with 1, B with 2 and the sheath
with 3, the values of KI.I, KI.% and KI. S , are respectively identical
with K 2 .2, K 2 .i and ^2.3. To determine them we have
and, Q 3 = K Z . 1 V 1 + ^3. 2 7 2 + # 3 . 3 7 3 .
If we are concerned with the distribution of currents in the
conductors and lead sheath, it is convenient to consider the sheath
grounded, that is, 7 3 = 0.
and,
TWO-CONDUCTOR CABLE 241
If then FI = F 2 = F, that is, if both conductors are given
the same positive charge, then
Qi V(# 1;1 + #,. 2 ) /. C = #1.1 + #1.2;
but C has been determined in (12) which gives,
(14)
If the two conductors have potentials FI and FI, respectively,
then:
Qi = #1.1^ + # 1>2 F 2 = F^L! - #i. 2 ),
.'. C = #1.1 ~ #1.2-
This capacity has been given in (5), which is:
1
C =
2a
r! 2 + a
From equations (14) and (15) the values of KI.\ and KI.Z are
readily obtained.
Consider finally that when the two conductors are in parallel,
that is, at the same potential and the charging current returns
over the grounded sheath, we have,
Qi + Q 2 + Q 3 = 0, and Vi = F 2 = V.
.'. (Ki.i + Ki.2 + KM + #2.2 + #3.1 + #3.2) F = 0,
or, 2#!.! + 2Ki. 2 + 2#!. 3 = 0.
.*. #1.3 = - (#1.1 + #1.2) (16)
Problems. Find the charging current under the conditions
shown in Figs. 116-120, when r t = 4r; a = 2r; .'. h = 2.75r;
hi = 2.55r; KI.I + #1.2 = 0.4; #1.1 - KI.I = 0.57 .'. #1.1 =
0.485; #1.2 = 0.085 and #1.3 = 0.4 (using no approximations).
(a) (Fig. 116.) F! = F 2 , F 3 = 0.
Q 2 = Fi(#i. 2 -f- #2.2).
.'. ii (#1.1 + #1.2) -jj,
dV
and iz = (#1.1 + #1.2) ~ 9
242 ELECTRICAL ENGINEERING
the total charging current is
1 "*" l '^ ~dt ' ' dt'
(6) (Fig. 117.)
'i.iVi + Xi. 2 F 2 ,
.-.<h -
i.i - K
1.2
dV
: '
(c) (Fig. 118.) 7 2 = - Fi, F = 27!.
o
FIG. 116. FIG. 117.
S
FIG. 118.
FIG. 119.
(d) (Fig. 119.) F 2 = 0.
FIG. 120.
= 0.485 .
dV
TWO-CONDUCTOR CABLE
243
(e) (Fig. 120.) Charging current in the eccentric cable: Since
the shortening of the lines of force in going through a conductor
is neglected, when the formulas were developed; so the solution
of case (6) is a solution of case (e),
dV
0.47
~df
Three-phase Cable. (Fig. 121.) -The location of the inverse
points is determined as in the case of the two-conductor cable.
Thus h 2 = h\ 2 + r 2 , when considering A and A' ';
FIG. 121.
and (h -f- a) 2 = hi 2 + ^i 2 , when considering the sheath.
' h - r ' ~2a~ -'
and At =
Thus /ii is known.
Let, at a given instant, the charges on A, B and C be Q A ,
and Qc respectively.
The potential of A due to the charges on A and A' is:
V = 2Q A log -
16
(1)
244 ELECTRICAL ENGINEERING
The potential of A due to the charges on B and B' is :
V = 2Q B log
y
The potential of A due to the charges on C and C" is:
V = 2Q c \ 0g ~-
\J
/?t -4- -V/) 2 I r 2 ~
.'. V A = 2Q A log - ^ + 2(Q B + <?.) log ? (2)
x is the distance between the inverse point of ' and the
center of A, and ?/ is the distance between the inverse point of
B and the center of A.
With a very slight approximation, the distance y may be
counted between the respective centers, thus,
y = a V3 (3)
and,
x * = (h + hi + a) 2 + a 2 - 2a(h + fci + a) cos 120 (4)
leth + hi + a = D then,
x 2 = D 2 + a 2 + aZ) (5)
It has been shown previously, that since the sheath is an equi-
potential surface,
(a + a )D = n 2 = (a + h - /ii) D = n 2 ,
.'. D = - T-Y^- (6)
a + h hi
Approximations based on the usual conditions, that h is very
nearly the same as hi and r 2 is small compared with h 2 .
Referring to equation (2),
hi + Vhi 2 + r 2 2/i D - a
r r r
2
(7)
Referring to (6), n _ V .
U ~ a + h-h," a
a
Referring to (3), y = a V3 (10)
The potential of A due to charges on A and A' is = 2Q A log -
The potential of B due to charges on A and A' is = 2Q A log -/=
TWO-CONDUCTOR CABLE 245
The potential of C due to charges on A and A' is = 2Q A log -
The potential of A due to charges on B and B' is = 2Q B log
D a
The potential of 5 due to charges on B and 5' is = 2Q B log -
/v
The potential of C due to charges on B and B' is = 2Q# log
The potential of A due to charges on C and C' is = 2Q C log - -r---
The potential of B due to charges on C and C' is = 2Q C log
The potential of C due to charges on C and C' is = 2Q C log - -- -
Since I,V A + 2F B + 27 C = in a three-phase system, we get,
by adding all the equations given above,
2 log ~ =
or, Q A +QB + Qc = (11)
which really needed no proof from our knowledge of the char-
acteristics of the three-phase system.
From (11) follows that Q B + Qc = - Q A -
/. V A = 2Q A log ^ - 2Q A log ^
(12)
.', the capacity of A to ground or neutral is:
2 log
ryVi 4 + a 4 + r! 2 a 2 /
or in microfarads per 1000 ft. of cable, to neutral,
0.00736
(13)
(14)
+ a 4 + r a
2
246 ELECTRICAL ENGINEERING
In order to determine MAXWELL'S coefficients, by symmetry,
we have:
and, KI.Q = K 2 .Q = K 3 . .
where index o represents the sheath. It is necessary to calculate
the capacity between all three conductors and the sheath.
Assume thus that the three conductors are given the same
positive charge Q, and that the images therefore have charges
Q. The potential of A due to the three charges is evidently
V A , 2Q lo, + 2Q log + 20 1.8
The potential of the sheath is due to the charges in the three
conductors and since the sheath is symmetrical with reference
to each conductor and its image, we get:
or, from the illustration, neglecting a,
Fo-3X2Qlog^ = 2
or since aD = rf,
log (16)
.'. C = - i, between a conductor and the sheath.
ri 6 - a 6 '
21og 377W
It is now possible to determine the values of Ki m i, K^ and
KI.Q. Assume that the sheath is grounded, that is, V = 0.
/. Qi = K lml Vi + K^V* + K^V* = K^Vt + ^i. 2 (F 2 + V,).
Since Vi + 7 2 + F 3 = 0, F 2 + F 3 = -- y t .
TWO-CONDUCTOR CABLE
1
It follows then from (13) that,
#1.1 #1 2 = -
2 log
\/3 (r! 2 - a 2 )
247
(19)
+ a 4 + ri 2 a 2 /
Considering next the case when all three conductors have
the same charge, then:
Qi = #i.i7i + #i. 2 7 2 + #i. 8 7 8 = (#1.1 + #1.2 + #1.3) 7i =
From equation (18) it follows that,
Kl Q if
1.1 -f- ^^-1.2 -
1
(6 n 6
'i^r
(20)
From (19) and (20) KI.I and Ki 2 can be solved.
To determine K\ m Q, assume that not only the three conductors
but also the sheath is given a potential V, in which case the
charge is confined to the sheath only. Then:
= (#!.! + Xi.2 + tfi.3 + KwW, .' ^1.1 + 2X 1>2 + ^1.0 = 0;
= (X 2 .i + X 2 . 2 + K 2 . 3 + X 2 . ) 7, .'. #1.1 + 2X 1 . 2 + XLO = 0;
= (X 8 .i + #3.2 + K 3 . 3 + X 8 .o)7, /. #1.1 + 2#!. 2 + Xi. = 0;
any one of these equations gives:
#1.0 = - (#1.1 + 2Xi. 2 ) (21)
Thus XI.Q is determined.
Problem. Verify the equations of the charging current under
the conditions given below (Figs. 122-130) and apply the follow-
ing numerical values:
TI = 4r, a = 2r.
(Fig. 122) i =
(Fig. 123) i = 2
FIG. 123.
; ^ - 0<826 -
Xi.i / dt dt
248
ELECTRICAL ENGINEERING
(Fig. 124) i = 3(/M.i + 2/M.j) ^ = 0.903
Ki.i 2 - #i.2 2 dV
(Fig. 125) t
dt
M* dV
dt
. _ ZKi.z*
to ~Tr~ A. 1.1
-IV 1.1
Si =
FIG. 128.
(Fig. 126) ti =
FIG. 129.
K
dV
l * = K ~di
i K dV
13 ~ Kl -*~di
= l '
St =
dt
(Fig. 127) i = 2(i.i + K lm -
~K" K" A
(Fig. 12J
(Fig. 129) i
dV
i =
0.608
= - 0.418
= 0.
dV
dt'
FIG. 127.
v v
FIG. 130.
= 0.
= 1.488
dV
dt'
~dt'
(Fig. 130) Three-phase: z = (K lfl - K^) - = 0.744
CHAPTER XXII
THE ELECTROSTATIC EFFECT OF A THREE-PHASE
LINE ON AN ADJACENT WIRE OR WIRES
The potential of the wire W, Fig. 131, due to A, B, and C
and their images is obviously:
V = 2Q A log g + 2 Q B g + 2Q C log g
2Q B b, + 2Qcd (1)
where . r 2
01 = log >
61 = log p
and, , r 6
Cl = log -
If C is the average capacity of the three lines against neutral.
then: Q A = Ce\, QB = Ce z , and Q c = Ce s , where e\ t e 2 and e s
are the instantaneous values of the Y voltages.
e 2 6i -f e 8 ci)
2CE[ai sin 6> + bi sin (^ + 120) + d sin (^ + 240)].
- 60 + b^ - cO + Ci( Cl - ai) (2)
where 7 is the maximum value of the Y voltage, that is, of the
voltage to neutral.
To determine the average capacity of the three wires: The
potential of A, Fig. 132, due to its own and the other charges is
evidently,
V A = 2Q A log ^ + 2Q B log g + 2Q C log ||-
If the average value of R 2 , R* and RQ is #!, and the average
249
250
ELECTRICAL ENGINEERING
values of R 3 and R$ is D, then the potential of A can be reason-
ably well expressed as:
V A = 2Q A log y + 2Q B log ^ + 2Q C log ~
(3)
WA log - 1 + 2 (Q* + Q c ) log ^
7 ;
r = Radius of
Conductor
FIG. 131.
But Q B + Q c = -Q A , thus
7 A = 2Q A log (~ j^ =
1
FIG. 132.
/. C =
where D is the average distance between the conductors.
E
'.' Vmax. = T?\/<ll(Q>l 61) + 61(61 Ci) + Ci(Ci -
(4)
(5)
Problem. Prove that the maximum value of the induced
potential on a telegraph wire placed under a three-phase trans-
mission line of 100,000 volts (effective) between the lines is
A THREE-PHASE LINE 251
approximately 3100 volts, when H = average height of trans-
mission wires above ground = 1500 cm., D = 300 cm., and
r = 0.5 cm. The telegraph wire is 800 cm. above the ground,
and 50 cm. to the left of the center line of the pole.
It is seen that when the three-phase line is operating under
normal conditions, the voltage induced in an adjacent wire is
only a few per cent., in this case only 3 per cent, of the line
voltage. If, however, one of the three-phase lines is grounded,
so that the system is unbalanced electrostatically, then very
considerable voltage is induced as will be shown.
If ei = E sin e,
e 2 = E sin (e + 120),
and, 6 3 = E sin (0 + 240)
are the Y voltages or phase voltages,
then it is well known that the line voltages are :
V 1 - F 3 = EV3 sin (6 + 30),
and, V 2 - F 3 = EV3 sin (e + 90).
Therefore, if phase No. 3 is grounded or at zero potential,
then we have the relation between the line voltage as shown in
Fig. 133. The line voltages differ 60 in time phase, when one
phase is grounded.
For the sake of simplicity, let :
Vi - 7 3 = V3 sin e = V a ; '
V 2 - V, = EV3 sin (6 + 60) = 7 6 ;
7 3 = = V c .
or, V a = #o sin 0;
where E G = E\/3>
V b = E sin (6 + 60) ; FIG. 133.
Using MAXWELL'S equation, applying index e for ground,
and remembering that V c = V e = 0, we have,
Qa = Ki.iVa 4~ Ki.zVb,
Q b = # 2 .iF + K*. 2 V b ,
Qc = #3.lF + #3. 2 F 6 ,
and,
Q. = K..iV a + K e . 2 V b .
252 ELECTRICAL ENGINEERING
But KI.I X 2 .2, Xi.2 = Xi.s and K e .i = X e . 2 approximately.
* Qa Xi.iFa H- Ki.zVb,
Q b = Xi. s 7. + Xx.iFi,
Q c = K^V a + Xi. 2 F 6 ,
and, Q e = Xi.,7. + Xi. 6 F 6 .
'. Qa = Xi.i# sin + #1.2 EQ sin (0 + 60)
Xi.i + 0.6X1.2) sin + - # Xi. 2 cos (6)
sin 6> + Xi.i ^ sin (0 + 60) = #0(^1.2 + 0.5Xi.i)
Sin + --KLI cos (7)
Qc = Xi. 2 (F + Vb) = E Ki 2 (1.5 sin 6 H pr cos 6) (8)
t
and, Q e = Ki. e (V a + Vb) = E Ki. e (1.5 sin d + -g cos 0) (9)
Assuming for the present that the values of the MAXWELL'S
coefficients are known, it is then possible to obtain, in a manner
similar to that used for the balanced system, the potential of
the telegraph wire.
While in this case we deal with four charges, the effect of the
charge of the earth is not felt at the telegraph wire, because
the earth may be considered as an infinite cylinder, enclosing
all wires; thus the effect of its charge on any point inside it, re-
sults in no potential. The potential of the wire is now readily
obtained from equation (1). The charging current in the three
wires and the earth is found from equations (6) to (9), remem-
bering that = cot.
. . dQ a r \/3 . I
^a = ~TT = L(\CO\ (/Vi i H~ U.O/Vi 2) COS COl ~ Al 2 Sin 001 ,
at L 2 J
r \/3 i
4 = EQCO\ (Xi.2 + O.SXi.i) cos cot ~ XLI sin cot ',
i e = E OJ\ Xi.2 (1.5 COS Cot ^~ SHI CoZ ',
l e = E Q CO\ Ki. e (1.5 COS Cot ^~ Sm W j
It remains now to determine the values of the MAXWELL'S
coefficients.
Give each of the three conductors the same charge Q, and
assume average values of the distance between the conductor
(10)
A THREE-PHASE LINE 253
and ground as H and the distance between conductors as D.
Then we have approximately the following relation:
OI7 9T/ 977
V. = 2Q log + 2Q log ^ + 2Q log ^ = 2Q log
' ' V
We have also,
.'. #1.1 + 2# lt2 - - -^ (11)
2 log Wr
Give now three-phase charges to the three conductors, then,
QA = ^l.l^a + #1.2^6 + K-l-zVc = Ki.iVa ~ #1.2^6 =
V a (#!.! - #x. 2 ).
Thus #1.1 #1.2 is the capacity of one of the three lines
against the neutral, which has been shown to be:
1
. . #1.1 #1.2 ~ T^- (12)
21ogf
From (11) and (12), the numerical values of Kn and #i. 2
can be determined, as well as #1.3, so that all the coefficients are
known.
It may be of interest to consider the problem from another
point of view.
By grounding one conductor, while the potential difference
between the conductors is not changed, the potential of the
system of three conductors has been changed.
It should be possible, therefore, to calculate the charge Q ,
which should be given to each conductor, in order that the
new potential distribution shall exist. The charge should ob-
viously be such that the potential of C shall be reduced to zero.
Before grounding, the potential was +V C , and hence Q Q should
be such as to give C a potential of V c .
:. -V c = 2Q log y 2 + 2Q log |^ + 2Q log j*
254 ELECTRICAL ENGINEERING
SH 3
= 2Q log -j, using the approximations.
Since V c = E sin (cot + 240), the maximum value of the
TjJ
charge is Q = -
The charges on the conductor A after grounding the conductor
C are therefore,
QA + Qo = E sin cot
Similar expressions are of course readily written for the
charges on the conductors B and C.
The potential of the telegraph line after grounding is thus,
V = 2 [(Q A + Q )ai + (Q B + Qo)6i + (Qc + Qo)CJ.
By applying these equations to the numerical example given
previously, it will be found that the induced potential of the
telegraph line will be 25 per cent, of the phase voltage or 14.5
per cent, of the line voltage. In the case of an insulated balanced
system, it was found about 5 per cent, of the phase voltage or
about 3 per cent, of the line voltage.
The Effects of a Grounded Horizontal Wire on the Distribution
of Electricity in the Atmosphere. It has been observed that
frequently considerable potential difference exists between
successive layers of the atmosphere. A potential gradient of
600 volts per m., or roughly 200 volts per ft., is not unusual.
It is of interest then to see how much the potential at a given
height may be reduced by a grounded overhead line such as
is used in high-potential transmission systems.
Assume that the gradient, not far from the earth, is 2 electro-
static units per m. (600 volts per m.). It is readily seen that
the distribution can be quite closely represented by the effect of
a charged cylindrical conductor, say 300 m. or more above the
surface of the earth. The conductor then represents whatever
cause there was for the potential gradient.
The charge per centimeter length of the fictitious conductor
is determined by the fact that the potential at a certain height
A THREE-PHASE LINE
255
is known. Thus according to the assumption, the potential at
15 m. above the ground is 30 electro-static units. Thus referring
to Fig. 134,
01 C
V = 2Q log = 30 .'. Q Q = 155.
Suppose now that it is desired to find the change in a grounded
overhead wire of radius r = 0.5 cm. placed 15 m. above ground.
Since the potential of A, Fig. 135, is zero, it is evident that
the potential of A due to its own charge and the charge on its
image plus the potential of A due to the charge on the fictitious
conductor and its image must be zero.
nr. TT i TL n~L
Thus 2Q log - : + 2Q log JT~ = = 2Q log - + 30.
.'. Q = -
30
= - 1.72 E.S.U.
T
P Abs
FIG. 134.
FIG. 135.
FIG. 136.
The potential at a point P, Fig. 136, distant hi, from the
ground is then:
V = 2Q log -TT + 2 Q log , but 2Q log -^ is, according to
KI TI III
the first assumption of uniform gradient, 0.02/ii (hi being given
in centimeters).
Thus the potential of P is :
V P = 0.02/i! - 3.44 log -
(1)
The effect of two ground wires A and B on the potential at
a point P in the vicinity of the wires :
256
ELECTRICAL ENGINEERING
The potential of A or B due to the fictitious and the two actual
conductors and their images must be zero.
The potential of A, Fig. 137, is:
2Q log |4i + 2 <3 lo S y + 2Q log ^ = =
or 0.02ft + 2Q log y ^ = 0, /. Q = fj^ (2)
If the wires are 2 m. apart and 15 m. from ground then r 4 =
3010 cm. and r 3 = 200 cm.
:. Q = -1.31.
T
The potential at a point P, Fig. 138, is then:
V p = 0.02^! - 2.63 log T -~
(3)
It will be seen that by means of a single ground wire above a
transmission line the potential is reduced by some 30 per cent.,
and when two ground wires are used by some 40 to 50 per cent.,
and that there is little gain in using ground wires of large diameter.
CHAPTER XXIII
THE CURL OF A VECTOR
In vector representation, the curl of a vector is represented
by the cross-product of the differential operator V and the
vector. It is:
V X R = curl R =
dZ dY
i j k
dx dy dz
X Y Z
dX
dx)
/dY _ dX\
~ dill
\dx dy
iCx + jCy + kC,.
The curl of a vector is thus a vector and its components along
the axes are C x , C v , and C z .
It is important to analyze the meaning of this new vector.
dy
c
^Z,
dz
FIG. 139.
Consider a small rectangle in the y-z plane, Fig. 139. Let
the component of R along the ^/-axis be Y and let it change to
FI, as we move along the z-axis from a to b.
dz
dz
Similarly,
257
258 ELECTRICAL ENGINEERING
The line integral around the rectangle is then :
dL = Ydy + Zidz - Y^dy - Zdz
idZ dY\ , ,
= I - - ) dydz.
\dy dz j
Extending this to all three planes, we get: the line integral
around dS,
-}dzdx +
dz dx
dy
= C cos adS, where a is the angle between curl C and the
normal to the surface dS.
The z-component of the curl C x is then seen to be the limit
of the ratio between the line integral of the vector around a small
element in the y-z plane and the area of the element. Since it
is the ^-component, it is, of course, at right angle to the surface,
dydz.
In general,
r. i r AL dL
Curl = lim -TO = -TO,
AS dS
where surface dS is normal to the vector C.
Stokes's Theorem. STOKES'S theorem states that the line
integral of a vector R around any closed contour is equal to the
surface integral of the curl of the vector over the surface or cap
enclosed by the contour.
The theorem holds always when transforming from the line
integral to the surface integral, but applies in the transformation
from the surface to the line integral only when -r + h
-T = 0, that is, only when the curl has no divergence.
Depending upon the system of notations used, it is written
in either of the following ways:
In vector notation, it is:
fR dr = f f (V X R) ' NdS,
which is to be read: The line integral of the electric field in-
tensity along the circuit is equal to the surface integral of the
curl of the vector over any surface (any cap) bounded by the
circuit, where N is the unit, outward drawn normal to dS.
THE CURL OF A VECTOR 259
Obviously, the theorem may also be written:
fRds cos (Rds) = f(Xdx + Ydy + Zdz)
dZ dY\ dX dZ BY d
where I, m and n are defined below.
The theorem can best be proven by calculus of variations,
but may be understood without mathematics by the following
reasoning. Refer to Fig. 140, which shows the
cap divided up into a number of small elements.
It is evident that the sum of the line integrals
around all these small areas resolves itself into the
line integral around the contour, 'since all lines,
except the contour, are traced in two equal and
opposite directions.
Thus if dL is the line integral around one of the small areas,
then
2dL = fR cos (Rds)dS.
But it has been shown, that
fdZ dY\. dX dZ. /BY
dL = "
\.
- dx) dzdx
cos adS, where a is the angle between the curl C and the normal
to the surface dS. (2)
.*. dL = C cos adS'
but dydz = IdS, where I = cos (Nx)' }
dzdx = mdS, where m = cos (Ny) ;
dxdy = ndS t where n = cos (Nz);
by substituting these values in (2), equation (1) is proved.
17
CHAPTER XXIV
THE EQUATION OF THE ELECTROMOTIVE FORCE
It has been shown that the potential difference between two
points in an electric field is the line integral.
V = f(Xdx + Ydy + Zdz) = fGds (1)
where X, Y and Z are the components of the field intensities or
gradient along the x, y and z axes and V is expressed in electro-
static units.
It will be shown later that the conversion factor between the
electro-static units and electromagnetic units of potential is the
velocity of light, v = 3 X 10 10 cm. per sec.
The e.m.f. in the electromagnetic system of units is v times
that in the electro-static system of units. Equation (1) should
be written:
V = vf(Xdx -f- Ydy + Zdz) in electromagnetic units (2)
Experiments have also shown that the e.m.f. in electromagnetic
units in a circuit is equal and opposite to the product of the turns
enclosing the magnetic flux and the rate of change of the flux.
If L, M and N are the components along the x, y and z axes
of the magnetic field intensity, and if I, m and n are the direction
cosines of the normal to the surface dS, and if /z is the permea-
bility then the flux is:
= ffpQL + mM + nN)dS = ffpH - dS
Then the e.m.f. induced per turn is:
V = - ~ = - ^ [ff(lL + mM + nN)dS] (3)
combining (2) and (3), and assuming M constant,
But from STOKES'S theorem, we can write:
f(Xdx + Ydy + Z*i) =//[Kf - ) + (f - g
260
THE ELECTROMOTIVE FORCE 261
Equating (4) and (5), we get:
dY\ /dX dZ\ /dY dX\-\
m \~d~z - dx) + n (to " a J =
r
~ v dt
(6)
If the circuit be closed, a conduction current will flow, and its
magnitude will depend upon the resistance.
NOTE. If the circuit is inductive, this applies equally well, since in these
equations the total variation in flux is considered.
Let I, with components u, v and w be the current density
along the x, y and z axes, and p be the resistivity of the ma-
terial. Let ds with components dx t dy and dz be an element
of the circuit, and A x , A y and A z be the projected areas of an
elemental surface dS, then the resistance along the z-axis is
- dx and dV = (resistance X current) = -- T uA x = pudx,
A x A x
but
V dV Y
X = ~~dx =pu > - X = pu -
Similarly, Y = pv,
and; Z = pw.
It should be noted that X, Y and Z are expressed in electro-
static units. Thus by transforming the relations to electro-
magnetic units, we get:
pu = vX]
pv = zY',
pw = vZ.
The Equations of the Current. Let the components of the
current density along the three axes be u, v and w, in electromag-
netic units. Let I, m, and n be the direction cosines of the
normal to surface dS' t then the total current is:
+ mo + nw)dS.
262 ELECTRICAL ENGINEERING
It was shown by AMPERE that the work done in carrying
unit pole around an element carrying current i was 4iri.
The work done is J* (Ldx -f Mdy + Ndz), where, as usual,
L, M and N are the components of the magnetic field intensity.
.*. f(Ldx + Mdy + Ndz) = 4x7 = 4*ff(lu + mv + nw)dS.
But by STOKES'S theorem,
f(Ldx + Mdy + Ndz) = ff(lC x + mC y + nCJdS.
+ mv + nw)dS.
dN dM
= C x = -r -- -5
a?/ a^
r aL aAr
= C y = - - ,
and, 4 _ c ._ 5M _ aL
" dz dy
Energy of the Electric Field. Consider a small cube-shaped
volume dxdydz, Fig. 141, in the electric field, and let the po-
tential difference between the two sides dxdy be V.
FIG. 141.
The capacity of the field enclosed by the cube has been shown
to be:
M J M J
4ird 4irdz
The energy stored in the field is J^CF 2 , and the potential
dv
V is Zdz, where Z =
_ Kte*y ^ _ KZVxdydz
4irdz Sir
KZ* , KZ* .
- dv, or the energy per unit volume = -^ , when only
the ^-component of the field is considered.
THE ELECTROMOTIVE FORCE 263
If the components of the electric field intensity R, are X,
TT
Y and Z, then, the total energy per unit volume = Wo = ^
O7T
(X 2 + F 2 + Z 2 ).
Similarly, it is proven that the energy stored per unit volume
in the magnetic field is:
W = ^ (L 2 + M* + TV 2 ).
Thus the total energy per cubic centimeter in space occupied
by magnetic and electric field is:
W = ^ [ M (L 2 + M 2 + JV 2 ) + K(X* + F 2 + Z 2 ).]
There appears to be no limit to the possible intensities of the
magnetic field, but for the electric field in air at atmospheric
pressure, experiments indicate a maximum possible gradient, or
field intensity of 30,000 volts per cm., or 100 electro-static
units of potential per cm.
Thus in the electric field the maximum amount of energy at
normal pressure is:
100 2
W max . = -~ - = 400 ergs per cu. cm. or 0.00004 joules per
cu. cm.
Maxwell's Displacement Current. MAXWELL assumes that
when a potential difference exists in any part of a dielectric, an
electric displacement, or a displacement of electricity has
taken place along the lines of electric intensity (force). The
greater the displacement, the greater the difference in potential.
The displacement, however, is resisted by the electric elasticity
of the medium, which, for the lack of a more satisfactory analogy,
can be thought of as being in a way similar to that existing in
an elastic body, against which a particle is pressed.
For a given potential difference, the displacement is greater
the greater the specific inductive capacity; for example, if the
dielectric be glass, the displacement may be five to six times as
great as would be true with air or vacuum.
A metal may be considered to have zero capacity, in other
words, energy can not be stored into it, but electricity would
continue to pass through it as long as a potential difference
existed.
Dielectrics, on the other hand, would permit electricity to
flow up only to a certain distance, and the flow ceases when the
264 ELECTRICAL ENGINEERING
force causing the electricity to flow is exactly equal to the
opposing force due to the elasticity of the dielectric.
The displacement of electricity is in the direction of the lines
of electric force; since the displacement has magnitude as well
as direction, it is a vector quantity.
According to MAXWELL'S theory an electric current is a time
rate of change of the displacement of electricity.
The charge on a body is a measure of the displaced electricity.
Indeed, MAXWELL states that a charge Q on a body causes a
displacement of Q units of electricity out from the body, and he
has defined the displacement D as the charge per unit area. It
is then numerically equal to a 1 , the charge per unit area, but
while or is a scalar quantity, D is a vector.
D can be expressed as a function of the intensity R and the
specific capacity K.
T ^ j. *. i? xi_ /> i j flux ^ 4?r C T
In air the intensity of the field is - - = j In
area area A
other dielectric of specific capacity K,
P 1 47TQ ARK
R = K ~T " Q ~- ^T
Q ARK RK
The surface charge = T = - A
A 4:
Thus the displacement D is also,
The displacement of electricity is in the direction of the field.
Thus if /, g and h are the components of the displacement, and
X, Y and Z are the components of the electric field intensity,
then,
KX
g
47T '
^^ In these equations, the units
are in the electro-static system,
and,
, KZ
h = -T.
4?r
The amount of electricity displaced is the product of current
and time, or considering current per square centimeter or current
density, the displacement is the product of current density and
time.
THE ELECTROMOTIVE FORCE
265
Let Ud, v d , and w d be the components of the current density,
then:
u d dt = df,
Vd dt = dg,
and,
dt = dh.
It has been shown that the conduction current density in
electro-static units was:
X
and,
Z
w = ,
where p is the specific resistance.
Thus the total current density along the x-axis is:
similarly,
= X ,df = X JtdX
p dt p 4ir dt '
Y , dg Y KdY
v + v d = h -7.- = h T- -7:7
p at p 4?r at
Z dfc Z X dZ
Thus, applying AMPERE'S relation, that in electromagnetic
units the curl of the magnetic field intensity is 4?r times the
current density, we get :
47T ,
- (w
similarly,
and,
1/47TZ
9\ p
dX\ _ lr47T
K dt) := vlp
-
dt
dM
dz dx
dM dL
(16)
L p dtJ dx dy
where v at present is the unknown ratio between the units.
The corresponding equations for the e.m.f. were shown to be
jT *\ rr *\ ~\7
fj, (JLJ O j O I
~ v dt dz ~ dx
266 ELECTRICAL ENGINEERING
v dt = f ~?x ^
_ p cW _ dY _ dX
~ v dt = dx " dy
By combining equations (1) and (2), it is possible to arrive at
equations of the electric and magnetic field intensities in any
medium conductor or non-conductor.
Differentiate (la) with respect to t,
7v 7" ~dt + K ~W = dydt ~ ~dzdt
Differentiate (2c) with respect to y,
IJL d 2 N d 2 Y d 2 X
* ' ~~ ~v ~didy = dxdy ~ 'dy 2
Differentiate (26) with respect to z,
fi d 2 M d 2 X d 2 Z
" ~v Htdz == 'dz 2 ~ dxdz ^
2 ~y
Substitute (4) and (5) in (3), and add and subtract ^ =
T ( -^ } , the following equation results:
47TM dX 3 2 X rd*X .^X.^X d_
p dt ' AM dt 2 '' V Idx 2 " dy 2 " dz 2 ' dx
which is the most general equation.
If there is no divergence, that is if we are interested in
medium having no charges, then the equation becomes:
It is readily seen that exactly similar equations not only
result for the Y and Z components of the electric field intensity,
but also for the components of the magnetic field intensity,
L, M and N.
Special Cases. (a) In a dielectric, p = , thus the equations
become :
(8)
THE ELECTROMOTIVE FORCE 267
or in general,
-Qp = a 2 V 2 ?7, where a 2 = ~, and U stands for either X, Y,
Z, L, M or N.
This is the well-known equation of the propagation of any
disturbance at finite speed.
The velocity of the propagation is a = - . In air. k = 1
and M = 1, thus the velocity of propagation of the electric
and magnetic field is v.
This value has been measured and found to be that of light,
thus the conversion factor is the velocity of light. Thus v =
3 X 10 10 . '
This important fact was deduced by MAXWELL in 1865.
(6) In a conductor, the specific inductive capacity may be
assumed as zero, thus we get:
or,
d*U , d 2 U 47r dU .
d^ + ^ = ^-ar in rectangular
coordinates, and,
, , 1 dU .
^ a0T + -^ + r -^T == > ln cylindrical
coordinates.
;
Assuming, as an application, that it is desired to determine
the current distribution at any time in a cylindrical conductor
at any distance from the origin and any distance from the
center of the conductor. If the practical system of units is
used, v 2 = 1; and on account of circular symmetry, the term
involving disappears. Thus the equation becomes:
d z i d z i 1 di 4?r di , ,
a7 2 + a^ 2 + r dr = 7 Hi
Distribution of current in a cylindrical conductor: If it is of
interest to find the distribution along a radius only, the equation
becomes :
dH 1 di _ 47r di , .
dr 2 + r dr ~ p dt
268 ELECTRICAL ENGINEERING
It will be of interest to verify this equation directly. It has
been shown that the work done in ergs, in taking unit pole once
around a conductor carrying current 7 is 4?r7, where 7 is the
current enclosed in the path.
Consider, for the sake of simplicity, a cylindrical conductor,
Fig. 142. Let the instantaneous values of the current density
at distant r from the center be i, and that at r + dr be i + dr.
FIG. 142.
Let the magnetic field intensity at distant r be H] and at
riff
distant r + dr, be HI = H -f dr.
The work done on unit pole in going from a to b is:
#i(r + dr)6 = (H + ^ dr) (r + dr)8 =
(\TT v
Hr + #dr + r -~- dr\ , neglecting the term which in-
volves (dr) 2 .
The work done in going from b to c, or from d to a, is zero,
because we travel on an equipotential surface.
The work done in going from c to d is Hrd.
:. W = e(
Hdr + rdr = edr
And by definition given above,
W = 4irir0dr, neglecting the term which involves (dr) 2 .
H dH
/1 x
(1)
THE ELECTROMOTIVE FORCE 269
The ohmic drop in voltage along 1 cm. of the conductor at
the outer edge of the segment, that is, at r + dr from the center,
perpendicular to the paper, is:
(i + dr) p, where p is the specific resistance.
The drop along the inner edge is ip; thus the difference in the
e.m.f. at the two edges is:
de =
This must be then the e.m.f. which is consumed by the self-
induction due to the flux in the element.
The flux in the element is = pH(dr X 1 cm.) == Hdr (3)
AA. AH
(4)
(5)
From (2) and (4),
di dH di _ dH
Differentiating (1) with respect to t,
di __ 1 dJH d*H
' dt r dt + drdt
Differentiating (5) with respect to r,
Substitute (7) in (6),
di 1 dH p dH
.*. 4?r = _- + - z
Substitute the value of --- from (5) in (8),
di _ 1 p di p dH
dt r n dr IJL dr 2
or,
M dH . 1 di
_
p dt ~ ar 2 "" r dr
in electromagnetic system of units.
Equation (9) is very important in connection with problems
of heat as well as electricity, it has been studied by great mathe-
maticians, notably, MAXWELL and LORD RAYLEIGH.
It is to be noted that the right-hand member of equation (9)
is LAPLACE'S equation transformed to cylindrical coordinates,
270
ELECTRICAL ENGINEERING
when the cylinder has circular symmetry. Thus we could have
written :
Special Case. Flat bar: Referring to Fig. 69, in the case of
flat bar, r approaches infinity, and (9a) becomes:
Si
FIG. 143.
The distribution of flux in a cylindrical conductor surrounded
by an energized solenoid is determined in a similar way. Fig.
143 shows the path of the current and flux. The dots represent
the current, and the lines around the current, the flux.
The result for a cylinder is identical with equation (9), if
H is substituted for i.
Similarly, for a flat bar equation (10) is applicable with the
same substitutions.
CHAPTER XXV
MATHEMATICAL SOLUTION OF EQUATION 11, PAGE
267, DEALING WITH ALTERNATING CURRENT
DISTRIBUTION IN CIRCULAR CYLIN-
DRICAL CONDUCTOR
The general equation is as has been shown:
dH 1 di 47TM di
dr 2 + r dr ~ p dt
Since we are dealing with sine waves, let:
i = i\ cos cot + z*2 sin cot (2)
where ii and i Zj the current densities, are functions of r but not of
t. Substitute first, i = i\ cos cot,
di dii
- =cosa^->
- 2
and > fU -a* sin erf.
at
dH\ 1 dll 4:TTfJL .
.'. cos ut T-^- H cos (^t T = - iico sin cot (3)
Similarly, for i = i% sin co,
J a 2 l2 , 1 dll +47TM .
sin ut T 2 - + - sm o)t = - izoj cos cot (4)
Adding (3) and (4),
' COS M ^ + - fr- H|
and d 2 i z 1 5i 2
W+r dr =
271
272
ELECTRICAL ENGINEERING
+
6 2 r 2
Assume :
i\ == a<) ~\~ air
and
Then:
v- 1 = 01 + 2a 2 r + 3a 3 r 2
-=-- = 2a 2 + 6a 3 r
Let
+
n(n
b n r n
.n-l +
HI'
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
'. ai = 0;4a 2 = m 2 6 ;9'a 3 = 7^ 2 6 1; in general, n 2 a n = m 2 6 n _ 2 (16)
By similar substitutions in (14), we have:
1 = Q; 46 2 = m 2 a ; 96 3 = m 2 aij in general,
(6) and (7) can be written:
dr
and,
dr
dr
Substituting (9), (10) and (11) in (13),
2a 2 r + 6a 3 r 2 + + n(n l)a n r n ~ l -
+ a! + 2a 2 r + 3a 3 r 2 + + na n r n ~ l
6 n = - m 2 a n _ 2 , or (n - 2) 2 & n _ 2 = - m 2 a n _ 4
Combining the last equations in (16) and (17),
m 4
a n =
From (17),
n 2 (n - 2)
(17)
(18)
(19)
Since ai = 0, and 61 = 0, from (18) and (19) all the a's and 6's
with odd indices separately equal to zero. And those with even
indices are as follows:
SOLUTION OF ALTERNATING CURRENT 273
a c = a
m 4
w 4
as = fi-FToS a
and so forth
m
42 . 6 2 *
m 4 m 8
8 2 .^02 6 = + 42"7g2Tg2~
and so forth
(19)]
60 =+ 7^02 [See (16)]
a
m
10
2 2 -4 2 -6 2 -8 2
7o 2 o
and so forth
m 2
8 2
m 2
m 6
4 2 -6 2 -8 2 -(10) 2 -(12) 2
and so forth
Therefore, i\ = a (l ~ 02742 + "02. 42.^2.02 ~ ' * ' )
H 9
,
" t "
and
52 2 2 -4 2 -6 2 2 2 -4 2 -6 2 -8 2 -(10) 2
~2"2~ ~ 2 2 -4 2 -6 2 + ***] +
m 4 r 4
) (21)
LORD KELVIN has denoted the first series in (20) by ber (mr)
and the second in (20) by bei (mr), thus:
oer(x)
2 2 -4 2 -6 2 -8 2
, and
/v.2 ^.6 /rlO
,.XN *> ^ | t'
= 2*~ 2 2T 4 2T 6 2 "*" 2 2 " T 4 2 6 2 8 2 (lO) 2
And
4
ii = a for (mr) + 2 a 2 6et (mr), and,
4
t' 2 = ao 6ei (mr) + 5 a 2 6er (mr)
(22)
(23)
These functions, 6er and 6^', have been worked out and appear
frequently in books on mathematical physics.
274
ELECTRICAL ENGINEERING
Therefore,
i = a her (mr) -\ - 2 a 2 bei (mr) cos ut +
\ a z ber (mr) a bei (mr) I sin cot (24)
The constants a and a 2 are determined from the fact that the
extreme outside layer is not surrounded by any flux. (We
consider only the flux in the wire in this calculation.) Thus the
sine term is zero at all values of t.
Let Jo = maximum value of the current density at the surface,
then,
and
7 = a ber (mR) -\ ^ bei (mR),
2
= ^ her (mR) a bei (mR)
(25)
Equations (25) are readily solved and give:
4a 2 _ JQ bei (mR)
m? ~~~ ber 2 (mR) -f bei 2 (mR)'
IP ber (mR)
ber 2 (mR) + bei 2 (mR)
{[ber (mR) ber (n
and,
bei
(26)
bei
ber 2 (mR) + bei 2 (mR)
(mr)] cos ut + [bei (mR) ber (mr) ber (mR) bei (mr)] sin co} (27)
Thus the square of the effective current density
I r)P7*^ I ? rr I r>^7*^ ( 'YyiT'i
-j~\\ TO [t/C'/ \^ffvL\j) Ut/l \ifvl )
ber 2 (mR) bei 2 (mr) + bei 2 (mR) bei 2 (mr) + bei 2 (mR) ber 2 (mr)]
7 2 [ber 2 (mr) + bei 2 (mr)].
2[ber 2 (mR) + bei 2 (mR)]
V2
(mR) + bei 2 (mR)
(28)
At the center of the conductor, r = 0,
.'. ber (mr) = 1,
bei (mr) = 0.
1
leff. =
_
A/2 Vber 2 (mR) + bei 2 (mR)
at r = 0.
SOLUTION OF ALTERNATING CURRENT 275
With very low frequency, the current density approaches the
direct current case where it is normal and is:
thus the ratio of the alternating current density at the center.
to that of the direct current is:
_ __ 1 __
\/ber 2 (mR) + bei 2 (mR)
For copper,
fj. = 1, = and p = 1600,
If the radius is 1 cm., and the frequency is 60,
mR = 1.72,
[ber 2 (mR) + bei 2 (mR)]~* = 0.87.
.'. the current density at the center is 87 per cent, of that at
the surface, and also 87 per cent, of what it would be with
direct current.
If the conductor had a diameter of 50 cm., the current density
at the center would only be 25 per cent, of that at the surface.
Actual watts consumed in heat are :
'72
[ber 2 (mr) + bei 2 (mr)]d(r 2 )
-(29)
C
2 I
Jo
2[6er 2 (mR) + bei 2 (mR)]
W = (ohmic resistance) (total eff. current) 2 (30)
Ohmic resistance = - (31)
Total current = irK
i2irrdr =
ber 2 (mR) + bei 2 (mR)
ber (mR) \ ber (mr)d(r 2 ) + bei (mR) \ bei (mr)d(r 2 )
r C R C R i
Cosut-\-\bei(mR) I ber(mr)d(r 2 ) ber(mr) I bei(mr)d(r 2 ) sincoZ
L Jo Jo J
272
.'. (total eff. current) 2 = or , , . PN , , . 9 . ^r\-
2[ber 2 (mR) + bet 2 (mR)]
\ C R I 2 \ C R 1 2 1
I ber (mr)d(r 2 ) + j bei (mr)d(r 2 ) (32)
U Uo
276
ELECTRICAL ENGINEERING
7rp/o 2 j|J o 6er(mr)d(r 2 )J 2 +|J o
bei (mr)d(r 2 ) \
2R 2 [ber 2 (mR) + bei 2
(mR)]
The coefficient of skin effect =
w act .
' C R 2 C R
Jo Jo
2 (mr)d(r 2 ) \R 2
J
W
jf ber (mr)d(r 2 )\ 2 + \ i
I 2
bei (mr)d(r 2 )
(33)
(34)
(mr) = 1 - 312 + 25,500
210,100%
(mr) + bei 1 (mr) == 1 + 313 + 3900
I [6er 2
/mr\ 12
21,700 Q +
(mr)
3100
(35)
C R
\ ber
J
' (mr)d(r 2 ) =
I I
SOLUTION OF ALTERNATING CURRENT 277
Substituting (35) (36) and (37) in (34),
K =
10
(38)
The following tables give the coefficient of skin effect at
various values of mil and the values of m for copper, aluminium
and iron.
mR
K
mR
K
mR
K
1
3.0
1.32
6.0
2.39
0.05
1.0001
3.5
1.49
8.0
3.10
1.0
1.005
4.0
1.68
10.0
3.79
1.5
1.026
4.5
1.86
15.0
5.57
2.0
1.08
5.0
2.04
20.0
7.32
2.5
1.17
5.5
2.22
Material
M
p in. e.m.u.
m
Copper. .
1
1700 at 20 C.
216 A/7
Aluminium
1
3,000
162 A/7
Iron
300 to 1200
10,000
. 09 A/M?
The value of /x for iron is usually taken as 300, but experi-
ments on iron wires used as transmission lines seem to give values
of M as high as 1200.
LORD RAYLEIGH has shown that when the penetration is so
slight that the above table can not be used a close approximation
of the " effective thickness" in centimeters of the surface layer
which causes the current is:
== 7
where K is the specific conductivity.
6.6
This formula becomes 6 =
7= for copper approximately.
8 8
^= for aluminium approximately.
16
Vrf
for steel approximately.
CHAPTER XXVI
ELECTROMAGNETIC RADIATION
Introduction. The laws governing electromagnetic radiation
were stated by MAXWELL fifty years ago. The experimental
verification was presented twenty years later by HERTZ in a
series of most extraordinary papers, which were later published
in book form. The practical application was made by MARCONI.
An extensive literature is now available, notably FLEMING'S
"The Principles of Electric Wave Telegraphy and Telephony,"
and ZENNECK'S "Wireless Telegraphy."
In writing this chapter the author has drawn extensively upon
the information which is given in these books. Since it is likely
that students who have not read what preceded this chapter
will want to understand the principles of wireless transmission
it has seemed wise to built up the theory from the fundamental
laws even though this procedure necessarily involves some
repetition of what has been given in previous chapters.
Fundamental Conceptions. Surrounding any body charged
with electricity is an electric field. The intensity of the field
usually varies from point to point, but, at any point it is propor-
tional to the charge, that is, the amount of electricity on the
charged body.
To charge a body we connect it to a source of potential when a
current momentarily flows from the source to the body, the cur-
rent stopping when the potential of the body is the same as the
potential of the source.
If i is the current flowing during an interval of time dt then
the resulting charge on the body is dq = idt, or,
i - ^
1 ~ dt
For reasons that will appear later, it has been assumed that
the outward field of flux from a body charged with Q units of
electricity is
\l/ = 4wQ lines of electric force.
278
ELECTROMAGNETIC RADIATION 279
If the lines are uniformly distributed over a closed envelope of
area A sq. cm., then the density of the electric field is
By the introduction of the constant 4ir in the flux formula this
density becomes in space the same as the force in dynes per unit
charge which is numerically the same as the intensity R of the
electric field at the particular point considered. This is easily
seen from COULOMB'S law, which states that the repulsive force
between two charges Q and Qi is
f _QQi
J - Kr 2
where r is the distance between them and Vi = 1.
In the ideal case the charge is confined to a point and the flux
is distributed uniformly in every direction.
. R = __ = ^Q = Q
area of sphere 4?rr 2 r 2
where r is the distance from the point to the point charge.
or, Q = Rr*
/. / = j3 rQ! = RQ,.
If, therefore, Qi ==!,/= R.
The potential difference between two points in an electric
field is by definition numerically the same as the work done in
moving unit charge from one point to the other.
Thus, if X represent the intensity of the electric field in a cer-
tain direction, say a direction parallel to the x-axis in a rectangu-
lar coordinate system, then the potential difference across a
short element dx is dV = Xdx = force on unit charge at dis-
tance x, or,
Y _dV
= dx'
Similarly v dV
= dy
and _ dV
J 7 *
dz
Y and Z being, respectively, the electric intensities along, or
parallel to, the s&-a&d-y axes.
* *
280
ELECTRICAL ENGINEERING
If we desire to find the potential difference between the ends
of a wire bent in a small rectangle in the x-y planes and the inten-
sities along the x and y axes are X and X\, Y and FI, then refer-
ring to Fig. 144,
dV = Xdx + Yidy - X,dx - Ydy (3)
dy
dx
FIG. 144.
For
y =
y = dy
X = X
X = X,
The rate of change of X as we travel along the ?/-axis is
thus the total change in distance dy is :
dX,
-r dy
dy y
Similarly,
Substituting (4) in (3) we get
dX
(4)
It is one of the properties of the electric field alone when free
from charges that the above potential difference is zero in a
closed circuit.
If, however, an e.m.f. is induced in the rectangular circuit by
change of flux treading through the circuit, then we get:
dt
dN 1 1
-^dxdy
ELECTROMAGNETIC RADIATION
281
where N is the density of the magnetic field perpendicular to the
plane of the electric circuit.
By a similar reasoning we get then the following three impor-
tant equations:
dX
dy =
ar
dx
dX
dz
dy
_
dx
dY_
dz
dN
dt
dM
dt
dL
dt
(5)
where X, Y, Z, L, M, N, are respectively the electric intensities
and magnetic intensities in the same system of units parallel to
the x, y and z axes.
Note that in air the densities are
'the same as the intensities.
The next consideration is in re-
lation to the magnetic effect of a
current.
Let A, Fig. 145, represent the
end view of a wire A carrying a
certain current dl, perpendicular
to the plane of the paper. Let the
curved line be in the plane of the paper.
The magnetic field intensity at P is then // and this is defined
similarly to R as numerically the same as the force on unit pole.
Let, therefore, a pole of unit strength be carried along the curved
path, Fig. 145. The work done per unit pole in completing the
journey once is evidently
W = fll cos 6ds = ( - - cos 0ds
J r
FIG. 145.
but
rda rda
-3 =. cos 6 , . as =
cos 6
r a -2*
:. w :
Jtx =
2dlda =
The work is independent of the position of the current element
and the path. Thus if there are a number of filament currents
inside the path,
I = Zdl
W = 47rJ (6)
282 ELECTRICAL ENGINEERING
This quantity is by physicists called
^ the magnetomotive force, around the
circuit, whereas, engineers would call
I Ll it 47r X m.m.f .
d Consider now a small rectangular
surface, Fig. 146, in the x-y plane of a
magnetic field, and let L and LI, M
* an d MI be the components of the
FIG. 146. magnetic field intensities, along the x
and y axes respectively.
Then the line integral, or work on unit pole around the element
is
Ldx -\- Midy Lidx Mdy.
The rate of change of L as we travel along the y-axis is ^-
thus the total change is dy, thus
L l= L + - dy.
Similarly
dM dL idM dL\
- - dxdy = ( d - - ~) dxdy.
From (6) it is seen that
dy
where I z is the total current flowing through the rectangle per-
pendicular to dxdy.
Depending upon the medium, this current may be the ordinary
conduction current such as flows in a wire or the charging current
which is incident to a change in the electric field, or indeed, the
sum of the two currents.
In this analysis it will be assumed that the air surrounding the
oscillator is free from ionization, so that its resistance is infinite;
thus the only currents considered are the " displacement, or
charging currents."
MAXWELL assumed that surrounding a charged body is an
electric field, the strength of which is proportional to the charge,
and that the intensity of the field is a measure of what he calls
ELECTROMAGNETIC RADIATION 283
displaced electricity. The displacement of electricity is in the
direction of the field intensity, and is thus a directed quantity.
Numerically a charge Q displaces Q units of electricity outward
from the body. Since dQ = idt, it follows that the displacement
current or, as engineers say, the charging current is proportional
to the time rate of change of the electric field intensity.
Or,
dR
where R is the intensity and a a constant to be determined.
MAXWELL worked out his theory on the basis that the dis-
placement is numerically the same as the charge per unit area.
Thus
area
But the outward normal flux from a charge Q is \j/ = 4-n-Q ; thus
the intensity of the field is
R
area area
r>
.'. R = 4:ird or d = 7- in air.
i = ~
4?r dt
where i is the current per unit area or current density.
If, therefore, u, v and w are the components of the displace-
ment current densities along the x } y and z axes and X } Y and Z,
the components of the electric intensities then :
1 dX 1 dY 1 dZ
u = -7 w = -: and w = - A TT (8)
4-7T d 4;r ^^ 4.w dt
everythir^g being given in electro-static units.
From (7) and (8) it is evident that one can write
/dM dL\ . dZ _
( ^r I dxdy = 4:irwdxdy = dxdy.
\ dx dy] dt
or,
dM _ dL dZ
dx ' dy == dt'
284
ELECTRICAL ENGINEERING
dN
dM
dX
dy
dL
dz
dN
dt
dY
z
dx
dt
dM
dL
dZ
dx
dy
dt 1
By a similar reasoning are obtained the following three
relations.
(9)
everything being given in the same system of units.
The simplest form of oscillator, or rather that form which
lends itself to the simplest mathematical treatment, is that used
by HERTZ.
The oscillator consists of two large spheres separated by a
considerable distance and connected by wires through the spark
gap to the source of energy as shown in Fig. 147.
o
Magnetic Field
FIG. 147.
O
It will be assumed that the electric field is due to the spheres
alone, and the magnetic field to the linear conductor.
It will be assumed that the axis of the oscillator is the 2-axis.
Thus the magnetic field which is in the form of rings around the
conductor has, in the x-y plane, no component in the direction
Z and therefore no e.m.f. can be induced in the x-y plane.
However, e.m.fs. will be induced in the direction of the Z-axis.
Whatever the potential distribution in the x-y plane it
must thus be due to the charges on the spheres alone, that is,
due to the electric field alone.
The distribution of potential around an electric double, that is,
ELECTROMAGNETIC RADIATION
285
around two spheres given equal but opposite charges. Referring
to Fig. 148, since
dV = - Rdr,
V = -fRdr = -J^ r = f.
The potential at P is (Fig. 149) :
FIG. 149.
It is
when r is large compared with dZ (see note).
NOTE. Proof:
(10)
a (q
^r
3/
9 cos 6
and
(2
by the use of the binomial theorem it is easily seen that this becomes:
2qdz cos 6
-^2
Equation (10) may be written
where /i, one-half of the length of the oscillator is substituted
for dz.
NOTE. Equation 11 is not limited to spheres but is quite general as long
as the distances dealt with are long compared with the length of the oscil-
286 ELECTRICAL ENGINEERING
lator. Suppose, for instance, that we are dealing with a linear oscillator.
We assume then that the potential at a point P can be expressed as due to
two point charges located at some points on each rod (not the end of the
oscillator) which will give the same potential as the linear conductor actually
gives at distances far away from the oscillator. While this assumption is
quite justified when dealing with points in space far away from the oscillator,
it is obviously not at all permissible at points near the oscillator, because
it is readily seen that the potential distribution at the surface of the two
halves of the oscillator must be such that the surfaces themselves are equi-
potential surfaces and two point charges, no matter where located, can not
give such equipotential surfaces. Fortunately, we are for practical purposes
interested in only what happens far away from the oscillator, where equation
11 applies. The subsequent equations can indeed be used with such linear
oscillator if instead of letting Q or 7 represent the charge and current respect-
2 2
ively, we use the average value along the oscillator which is - Q and - 7.
7T 7T
The ratio between X the wave length and h the height of the sending antenna
is in such case, theoretically 4, but in reality due to various effect nearer 4.8.
When P is far away from the oscillator the electric condition
is not due to the instantaneous value of the charge q at the
oscillator but due to the value of q which existed somewhat
earlier in time.
Thus the charge causing the electric field at P is not q = Q sin ut
but q = Q sin u(t At) where At is the time required for the
distribution to reach P.
If v is the velocity of the propagation which is that of light,
then
T
vAt = r 'or At = -
v
.'. q = Q sin f ut j
If X is the wave length then
27T
.*. q = Q sin I cot rj = Q sin (oot mr) = Q sin (mr cot)
.'. V = - 2Qh ^
dz r
. _ dF 6 2 sin (mr - ut)
(13)
v dV d 2 sin (mr -
'
ay dydz
ELECTROMAGNETIC RADIATION
287
H M
Z the component of the electric field intensity perpendicular to
the x-y plane cannot be obtained from V alone as discussed
above.
We shall now consider some of
the properties of the magnetic field
intensities.
Consider the x-y plane (Fig. 150).
It is obvious that since the lines
of force are circles, the sum of the
projections of the components of
the magnetic field intensities along
the x and y axes on a radius vector
must be zero. Let L and M be the components of H along the
FIG. 150.
x and y axes,
we have,
but
and
or
Since L itself is negative in the position shown,
L cos a + M sin a. =
cos a = -
P
sin a = -
P
'. Lx + My = 0,
- - but x 2 + y 2 =
x
L
M
.'. xdx + ydy = 0.
Thus L_ dx
M dy
or Ldy Mdx =
This is satisfied as long as
L =
du du
and M =
dy dx
(14)
(15)
where u is any function of x and y
N the component along the 2-axis is obviously zero.
From equations (9) and (15)
dt
~dt
"aT
dy
dL
dz
dz
dz dxdz
dL _ d 2 u
dz ~~~ dydz
_ dM dL _ /d z u d*u\
'' ~dx "~ fry "
dx
(16)
288 ELECTRICAL ENGINEERING
Referring now to (13) and differentiating X with respect to t
- <2Qh - ( d * sin (mr ~ w '
dt " 2Qh dt(dxdz ' ~~r
It is evident by comparing (16) and (17) that,
d sin (mr wQ
Substituting this value in (16) we get:
dX .d 3 (nsM. sin (mr
dt dtdxdz
or v on ,
A = ZQfi
F = 2Q/i
N =
where n = sin (mr - cop
It is now a simple matter to get the different derivatives of
n. An inspection of the several terms will readily show that some
are much larger than others. There is little object in investi-
gating conditions close to the oscillator by these equations even
if all terms are used without considerable caution, because an
approximation was made in the assumption that the electric
field emanated from two point charges.
The derivatives contain trigonometric terms having coefficients
of raV 2 mr and unity.
The terms containing w 2 r 2 are so much larger than terms
involving mr and unity that the latter can be neglected. Making
these approximations and placing P in the x-z plane we get
for distances involving several wave lengths,
ELECTROMAGNETIC RADIATION
289
7 =
Z = - -- sin (wr
7, =
.. rr
M = // =
sin 2 .
.
sin (rar cot) sin
v
(19)
(20)
Everything is given in electro-static units at present since all
terms involve Q the charge which is expressed in such units.
FIG. 151.
R the intensity along the surface of a sphere through r is
(from Fig. 151):
R = Z sin B X cos 6 = - sin (mr at) sin (sin 2 + cos 2 0)
or
R =
sin (mr cot) sin
(21)
It is of interest to compare equations (20) and (21)
47T 2
since
and
We can write
X 2
2ir
ma) = -
47T 2
X2T
H = R
(22)
when the charge is given in electro-static units.
The electric and magnetic intensities perpendicular to each
other in space are in time phase. Thus the product of the two
represents power. (This is the case only at some distance from
the oscillator, near the oscillator the large part of the fields is
in quadrature.)
290
ELECTRICAL ENGINEERING
It is remembered that in the ordinary electric circuit involv-
ing capacity and inductance the magnetic and electric field
intensities are in time quadrature and, therefore, the product
represents " wattless power or better reactive power."
Energy Radiated. Equations (20) and (21) can be trans-
formed to read,
and
2Ihm . ,
H = - sin (mr co/) sin
21 hm 2 . ,
K sin (mr co/) sin 6
co r
(22)
Since
dq
q = Q sin (co/ mr) and i = -37 = Qco cos (co/ mr)
dt
.'. I = QcoorQ
= Qco cos (mr co/)
If 7 is expressed in amperes and R in volts per centimeter,
then
and
But
H = 0.21 sin (mr - co/) sin 6
21 Hm 2
R = 300 X 10 - - sin (mr - co/) sin 6
m = and co = 2irf = 2?r -r- .',
A A
0.47T/ h
H = - - - sin (mr co/) sin 6
r A
1207T/ h .
R = - r- sin (mr co/) sin 6
r \
(23)
(24)
From what has been shown, it is remembered that H and R
are perpendicular to each other in space.
The e.m.f. in a circuit is proportional to R, the current is
proportional to H and the power to HR sin a, where a is the angle
between H and R.
Since these fields are perpendicular to each other in space the
energy radiated in time dt eidt is proportional to HR, or, W =
kHRdt and it remains to determine the value of k.
ELECTROMAGNETIC RADIATION
291
The voltage per centimeter is R\ thus e = R when considering
1 cm. of circuit. The m.m.f. that produces a magnetic in-
4:iri
tensity H is - where I is the length of the magnetic circuit
in air.
. . HI
~ 0.47T
or, if we consider 1 cm. length of magnetic circuit,
i =
H
0.47T*
Thus the energy transmitted through a square centimeter area
is:
RHdt
W = eidt =
0.47T
Thus the energy radiated through the whole sphere of radius r
enclosing the oscillator is (from Fig. 152) :
FIG. 152.
s*e = TT rt = T r>Tj
W = I 7r-r- 2wr
J e = Jr = 0.47T
sin Brdddt
h
240 7T 2 I 2 sin 2 (mr - ut) sin 3 OdOdt
but
- o
sin 2 (mr <*t)dt = -~ approximately
19
(25)
292 ELECTRICAL ENGINEERING
and
sm vav = y%
:. w = 1600^
W h 2
.'. watts = ^ = 1600 ^- 2 I 2 .
If I is given in effective current then,
h 2
watts = 3200 ^I 2 (26)
In the case of wireless transmission the radiated power corre-
sponds to one-half of the area of the sphere thus,
h 2
Watts = 1600 ^- 2 / 2 (27)
where I is the effective value of the current. The " Radiation"
resistance is obviously
R = 1600 ^ (28)
It is noted that in the case of wireless telegraphy the energy
radiated is greatest along the equatorial plane, that is, near the
surface of the earth.
Since the receiving antenna is near the earth this result is,
of course, very desirable.
MARCONI'S improvement upon HERTZ'S oscillator resulted
from his connecting the lower end of his oscillator through a
spark gap to ground, by which he was able not only to obtain
the maximum energy, where it was most useful, but also to make
use of half the length of oscillator for the same distribution of
the magnetic and electric field above ground. This will be evi-
dent at once if it is considered that the earth being a perfect con-
ductor, its surface is an equipotential surface.
It is easily proven from the equations given that the energy
received near the surface of the earth through unit surface is
1.5 times the average value of the energy per unit surface.
It is also of interest to note that with an " ideal" simple
antenna where X = 4h and the current is zero at the top at all
2
times and therefore the average value of the current is - 1 that
7T
the power radiated in watts is 40/ e 2 or the radiation resistance
is R r = 40 ohms.
ELECTROMAGNETIC RADIATION 293
In this connection it is of interest to add that MAXWELL'S
general equation of propagation of electromagnetic waves in
space free from electric charges or magnets has been shown to be :
where u is any of the components of the electric and magnetic
intensities.
In the case of spherical waves it is readily proven by trans-
forming the equation in spherical coordinates that any function
of r vt divided by r satisfies the equation. Thus
r
The function used so far was
_ sin (mr coQ
r
which satisfies the above since mr ut = r = m(r vt).
m
In the case of sustained oscillations the function chosen was
obviously most suitable. In the case of damped oscillations we
would naturally choose
TT = ~ e - a(ut - mr} sm(mr - ut)
r
where A and a depend upon the amplitude and damping of the
circuit.
Of special interest is the magnetic intensity // near the surface
of the ground and the electric intensity R perpendicular to the
surface but near the ground.
Equation (21) gives,
R = sin (mr wt) sin = 2h~ ' - sin (mr ut) sin =
r co r
4?r - / - sin (mr cot) sin
r A
where I is expressed in electro-static units.
If the current be expressed in amperes and the potential
gradient in volts per centimeter
4wlh V
X T7* X 300 sin (mr ojt) sin
10
= 377 - sin (mr - ut) sin (29)
r A
294 ELECTRICAL ENGINEERING
Thus the maximum value of the potential gradient near the
surface of the ground is
Rmao. = 377 - r- volts per cm. (30)
/ A
If, therefore, the height of the receiving antenna is hi cm. the
maximum value of the potential difference between earth and
top is
F 377 h h I
El = T~ x hj
or the impedance of the receiving antenna is
Z Ei 377 h ,
2 ~- 7- ~7~ x * onms - ( 31 )
and the effective value of the voltage is :
E. = I f Z 1 (32)
where E f is the effective value of the voltage across the receiving
antenna and J e is the effective current in the sending antenna.
In a simple antenna the current is a maximum at the gap and is
zero at the top. Thus the current is not uniform as is the case in
the HERTZ oscillator.
2
The average value of the current is - 7. With such simple
7T
antenna the wave length X should be 4/i if there were no disturb-
ing effects.
Substituting these values we get as the impedance of the receiv-
ing antenna in an ideal simple antenna
Z\ = 60 ~ ohms (33)
The magnetic field intensity H (equation 10) is similarly modified
to
H =* 2 -h sin (mr ut) sin B = - r- sin (mr wt) sin 6
0) T T A
where I is in abamperes, or if / is expressed in amperes rather
than abamperes
h . . . . /Q .,
H = - r- sin (mr cot) sin 8 (34)
T A
It is of interest to note that this agrees with the intensity due
to an infinitely long conductor if
h - JL
X ~ 27T*
ELECTROMAGNETIC RADIATION 295
If the sending antenna were a simple rod then the current at
the top would always be zero and the average value of the current
2
would be - 7. In that case the wave length would be 4/i.
7T
Substituting these values we get :
0.27
H = - - sin (mr ut)
near the surface of the earth.
Thus the approximation sometimes made in writing
27
H = - sin(rar to/) sin 6
is not very far from right and is correct in the case of an " ideal
simple antenna."
It should again be emphasized that equations (29) and (34)
give the values of the electric and the magnetic intensities several
wave lengths away from the oscillator.
It can very readily be proven by carrying out the differentia-
tions in equation (18) that near the oscillator the magnetic
intensity decreases inversely as the square of the distance and the
electric intensity inversely as the cube of the distance.
Power Factor and Logarithmic Decrement. Prior to the use
of high-frequency alternators for the production of radiation the
trains of waves were oscillating, with decaying current and e.m.f.
in the antenna and the word decrement had therefore a very
significant meaning.
When alternators or oscillating arcs are used the current and the
e.m.f. at the antenna are sustained, and therefore " decrement"
ceases to have any meaning.
It is, therefore, appropriate to discuss the power factor rather
than to try to treat of the decrement in such circuits.
If RQ is the sum of the radiation resistance and the effective
resistance of the wires and the ground connection, then the power
consumed in the circuit is P -= PR , where 7 is the effective cur-
rent. If E is the effective voltage, then
I = 2irfCE = coCE where C is given in farads,
thuS Pf = a>CR (35)
296 ELECTRICAL ENGINEERING
Numerical application :
Let C = 003 m-f. - ~i farads.
h = 50 m.
X = 3000 m.
.'. Radiation resistance = 1600 (^7.7^;) 2 = 0.5 ohm approxi-
, XoUUU/
mately
W = 27T/ = 27T = 27T10 5 .
A
Let the effective resistance of the wires and ground be 2 ohms,
then
R = 2.5 ohms.
.'. Pf = 27rl0 5 j^ 2.5 - 0.0047
or approximately one-half of 1 per cent.
The radiated energy corresponds in this case of course to only
one-fifth of this amount.
The product of the current and the e.m.f. is 200 times as great
as the power consumed in heat and radiations and 1000 times
as great as the power radiated.
Determination of the "Logarithmic Decrement." If a con-
denser is discharged in a circuit of negligible resistance an alter-
nating current will flow indefinitely, and no energy will be ex-
pended since the energy is transferred alternately between the
magnetic and the electric field.
When the current is a maximum (either positive or negative)
the e.m.f. across the condenser is zero; when on the other hand
the current is zero, the e.m.f. is a maximum.
Thus twice in each cycle the magnetic energy
w m =
is transferred to electric energy
W e =
The total amount of energy in joules surging during a cycle
is then
W = L/ 2 where / is the maximum value of the current,
or,
W = CE 2 where E is the maximum value of the voltage.
ELECTROMAGNETIC RADIATION 297
If, however, the circuit contains resistance, 'the current will
not alternate indefinitely but will die down gradually, the rate
of decay being greater the greater the energy consumption.
During these oscillations energy is also transferred between the
electric and magnetic field but each pulse of energy is smaller,
than the preceding by the loss of energy in the resistance.
Ultimately all energy stored in the condenser becomes dissi-
pated in heat or radiated away.
The energy stored in a condenser is %CE 2 joules where E
is the voltage and C the capacity in farads. Thus if the condenser
is charged and discharged N times per sec. the sum of the energy
N
converted to heat and radiated away is -^CE 2 joules per sec.
Zi
or watts.
FIG. 153.
Thus Wi = ^CE 2 watts (36)
z
In a circuit of concentrated inductance and capacity it is
shown in the elementary theory of alternating current that the
oscillating current can be expressed quite accurately by the
following equation:
i = I 6 - at smut.
IT*
Where a = ^ and R Q and L are assumed constants which how-
iLi
ever is not the case in ironclad inductors and in arcs.
298 ELECTRICAL ENGINEERING
The ratio
g
as is readily proven.
The logarithmic decrement is
(37)
Incidentally 5 is also the ratio between the energy absorbed by
the resistance and the surging energy per cycle.
Thus I 2 RoT R T
-~
which agrees with (37).
In the case of the HERTZ oscillator or the umbrella type of
antenna the inductance is confined largely to the linear con-
ductor and the capacity to the spheres or superstructure; thus
we may consider the inductance and capacity as separated rather
than distributed, thus
(38)
(39)
The resistance in the above formula is the sum of the radiation
resistance, the resistance of the wires (taking into consideration
the skin effect), the ground and the radiation resistance.
When an arc is used the resistance of the arc should also enter.
Unfortunately the latter is not a constant but depends upon the
current carried, and hence the decrement is not logarithmic.
However, for the purpose of this article the arc resistance may
be assumed constant at say 5 ohms. For a very complete dis-
cussion of this whole subject the reader is referred to FLEMMING'S
" Principles of Electric Wave Telegraphy."
Equation 39 contains the inductance and capacity as well as
the resistance. The inductance is usually very difficult to
determine since at different wave lengths more or less inductance
is added to that of the antenna proper. The capacity of the
antenna is however, usually not changed but it depends upon the
construction of the aereal. The complexity of the structure is,
however, such that its value can hardly be calculated except in
the very simplest cases rarely used in practice.
ELECTROMAGNETIC RADIATION 299
FLEMMING expresses the approximate capacity of a vertical
wire of radius, h cm. long as:
C v = - - farads,
2 log - X 9 X 10 11
when, as is the case in wireless stations the lower end of the wire
is near ground, the capacity may, however, be say 10 per cent,
greater.
He also expresses the capacity of a horizontal wire placed h\
above ground as
I
C h =
2 log- 1 X .9 X 10 11
where I is the length in centimeters and h the height above
ground.
Thus the capacity of a T-shaped antenna may be approxi-
mated as:
c = c v + c h
obviously the total capacity is not at all proportional to the
number of wires connected in multiple. It is only slightly
increased as the number of wires is increased.
If the value of the capacity is difficult to calculate accurately
it is measured relatively easily and will therefore be assumed
as known. It ranges according to ZENNECK approximately as
follows :
0.001 m-f. in torpedo boat antenna.
0.002 m-f. in battleship antenna.
0.007 m-f. in BRANTROCK station.
0.18 m-f. in NAUEN high-power station.
The capacity of the antenna of the experimental installation at
Union College is 0.0012 m-f.
When the wave length is considerably more than four times
the height of the antenna the current distribution is fairly uni-
form in the conductor, and, the circuit can be treated as consist-
ing of "bunched" rather than distributed inductance and
capacity when the following relation obtains.
T = 27T/LC /. L = ~
300 ELECTRICAL ENGINEERING
thus .C
-f (40)
\
Numerical application: Union College station with an antenna
having a capacity of 0.0012 m-f. sending out waves of 700 m.
length. Assume R = 10 ohms. (By far the greater part of
this is the ground and spark resistance.)
Then 5 = 2 o 10 _!2_ 3 X _
10 10 70,000
In the case of the simple antenna it has been shown that the
2
radiation resistance assuming X = 4/i and I avo > = - / is 40 ohms.
Thus the radiation decrement is :
3 X 10 10 =
2 log 9 X 10 10 log
In reality due to the proximity of the earth and other causes
the wave length is nearer 4.8 than four times the antenna height,
2
and the average value of the current is nearer 0.7 and -.
7T
Substituting these values we get :
R = 34 ohms instead of 40 ohms
and the radiation decrement for the simple antenna is :
,_*!!* 8 X16*-^ (41)
2 log ^9X10"' log"
Abraham gives s 2.45
o - r*
i
] g ~
General Conclusions. Since the power radiated from an
antenna is:
W = 1600 ^ 7 2
it is evident that at a given voltage as the capacity of the super-
structure is increased the current and the wave length are in-
creased. Since, however, the energy is proportional to the square
ELECTROMAGNETIC RADIATION 301
of the current and the wave length is proportional to -\/C it
follows that by adding capacity to the superstructure and there-
fore increasing the wave length the radiated energy is increased.
Therefore, if the capacity is made four times as great, the cur-
rent 2 is 16 times as great and X 2 is only four times as great, and
hence, the radiated energy for the same antenna height is in-
creased four-fold.
Unfortunately, however, there is hardly a practical way of
increasing the top capacity without decreasing the effective
height so that the gain is not as great as indicated and if the
umbrella is carried to an extreme, the effective height may be so
much decreased that the energy radiated may eventually begin
to decrease.
With a given construction of the antenna the wave length may
be increased by the introduction of inductance. In this case the
energy radiated is, however, reduced.
It is noted that for a given current the radiated power is
greater the higher the frequency. This does, however, not
necessarily mean that the power received is greater, since as will
be shown later the absorption of energy in space is much greater
with short wave length than with long.
At times it is necessary to send at two widely different fre-
quencies. The natural wave length may be say 600 m. and it is
desired to communicate at a wave length of 300 m. In that case
a condenser may be connected in the series with the antenna.
Since two condensers in series have a smaller capacity than each
and thus the frequency is increased.
The relation between the effective value of the antenna current
and the maximum instantaneous value of the current and e.m.f.
If the damping is not excessive the discharge current of a
condenser of voltage E can be represented by the following
equation :
_^o
i = EuCe 2L sin u
= Ie~ at sin co
where
/ = EuC and = g = |-
R Q being the total resistance in the circuit which is assumed
constant, not depending upon the current.
302 ELECTRICAL ENGINEERING
The rate at which energy is being converted to heat and
radiated is then:
Ro P -* a t sin 2 coZ.
The energy developed in one train of waves then is,
R I 2 e~ 2at sin 2 utdt = approximately (43)
If the antenna is charged and discharged N times per sec. then
the power is
W _ A/- L (AA\
43/
If I c is the effective value of the antenna current as read by a
hot-wire instrument,
.:I = I e ^jy (45)
and since /
Substituting JRo
and r 1
L =
we get l2R C
Je , (47)
These equations connect the instantaneous max. values of the
antenna current and e.m.f. with the effective current read by a
hot-wire instrument.
Numerical application : At the Union College station R = 10,
12
X = 700 m., C = -, N = 500. Using the small sending set.
I = 2.5 amp.
3 X 10 10
70,000
0.43 X 10 6
ELECTROMAGNETIC RADIATION
303
I = 2.527T.4310 6
= 46 amp.
E = 14,400 volts.
Relation between E.m.fs. Frequencies and Coupling in In-
ductively Connected Circuits. Let e\ in Fig. 154 be the voltage
across the primary capacity, e 2 in Fig. 154 be the voltage across
the secondary capacity.
Then neglecting resistance we get:
^7 =
but,
dt
" dt
~dt
l ~df
- =
and
(1)
(2)
FIG. 154.
Substituting the current values of equation (2) and equation (1),
and writing
ei = Ei sin ut
e z = EI sin (ut + a).
Thus
we get,
(3)
(4)
304 ELECTRICAL ENGINEERING
From (3)
61 = F^rfc^ 2 (5)
Substitute (5) in (4) and assume that C\Li = C 2 L 2 = CL, that
is, assuming that the circuits when independent are tuned to the
same wave lengths, then,
(6)
- 2LCco 2 - co 4 (L 2 C 2 - lf 2 CiC 2 ) -
1 k
" CL(1 -
where
7 M
k = .
(7)
&nu\j \ j. t\j~ \ a. A/~
or,
(8)
where /o is the frequency of each circuit when alone.
It is seen from these equations that two frequencies exist in
the circuit and that they become nearer and nearer alike as the
coefficient of coupling is decreased, that is the less the value of
the mutual induction as compared with the self-induction.
In the case of transformation by ordinary transformer where
the mutual induction is almost perfect, only one frequency will
appear, namely, / 2 the forced frequency which in that case is
/ 2 = / -7=- 1 other words the radiated frequency has only one
V2
value and that value is 70 per cent, of that of each circuit when
alone.
Since
/I _ \2
/2 Xi'
It follows that two different wave lengths are transmitted
and that
fT^fc
X 2 - ' '-
ELECTROMAGNETIC RADIATION 305
Wave meters are used to show the wave length, and hence, if
the wave length is known the coefficient of coupling can be deter-
mined, it is:
X 2 2 - Xa 1
(10)
k =
(11)
A 2 -f AI*
It is evident from the above that the current and voltage
in two such circuits must be expressed as functions of two
frequencies.
Let 61 = AI cos coi 4~ Bia) 2 t
e 2 = A 2 cos coit -f- B 2 cos ai 2 t
These equations are justified since the resistance is negligible,
and hence, no appreciable phase displacements exist between
the two voltages.
For t = 0,
ei = E lt
and
e 2 = 0.
/. E! = A, + B 1 |
= A 2 + B 2 \
Consider then the two waves separately.
We have from (5)
(12)
1 -
and
where
1
A 2
B l
(13)
a. =
ft =
Thus from (12)
1 jLido? 2 2
Ai = E 1 - B 1
A 2 = - B 2
and
A, = E l - Bi .'. A! = -
aEi
(14)
a.
306
But
and
ELECTRICAL ENGINEERING
E l
jCl 1 A
.'. A<t =
a
a
B= - A* /. B 2 = -
a
(15)
.'. Substituting these values in equation (11)
pi
[COS Uit COS 0> 2
a
Thus
From (13)
from (6)
- a
wi-
C02
2MC
- a
(16)
Equation (16) shows the relation between the maximum voltage
and the capacity when the circuits have negligible resistance.
When damped oscillations are considered the equations become
more involved.
FIG. 155.
Consider the simplest case when the primary of the exciting
transformer, Fig. 155, is supplied with power from an alternator
or other source of sustained oscillations.
Due to the mutual induction between the primary and second-
ELECTROMAGNETIC RADIATION 307
ary circuits an e.m.f. EI sin co^ will be impressed upon the
secondary.
The differential equation of the secondary circuit is thus :
EI sin &it = iRz + L^rrr + ^2
at
where e 2 is the voltage across the secondary condenser.
But . ~ de 2
.-. E 1 sin Wl = C,R + C 2 L 2 2 + e 2 (1)
The sine term can be eliminated by two successive differentia-
tions and the result will be a well-known linear differential
equation of the fourth order the solution of which is:
e 2 = E f sm (wi$ + <p) + E'e ~ at sin (w 2 + t) (2)
The first term shows the value of the permanent voltage of the
secondary circuit of primary frequency, the second that of the
transient which very soon ceases to exist.
Thus, if it is desired to study the constants of the antenna the
transient term may be neglected and the permanent voltage
becomes
e^ = E sin (wi -f- v?).
Substituting this value in the differential equation we get
after some simple transformations the following relation between
the maximum value of the secondary voltage and the induced
voltage.
The secondary frequency
/ =
when the circuit contains no resistance and
when the resistance is R 2
Thus 1
\J2LJ1
or 1
77-7- = ^2 +
U 2 i-/ 2
20
308 ELECTRICAL ENGINEERING
where R 2
az = 2L S
T
"
co! 2 - o> 2 2 - a) 2 + (2eoia 2 ) 2
When the secondary circuit has the same natural period as the
primary impressed frequency the secondary current becomes a
maximum.
Thus for
0> 2 = COi I = I r
since a 2 4 is very small compared with 4coi 2 a 2 2 .
This is readily shown to be
.*. The square of the effective value of the secondary currents is
L 2 - C0 2 2 - 2 2 ) 2
or 2 2 is small compared with co 2 2 , thus
I _ 2a 2 a>i
T r =
tW
//I
\/(l
\ \
but
and
B
2
and
" 2L 2 / 2
' 2ZT 2 = 52j
ELECTROMAGNETIC RADIATION 309
r* (-i-Y
I 2 \ irfi /
.'.(I -o; 2 ) 2
where
*
or
_
-
and
If x is near unity then 1 x 2 = (1 + x) (1 x) = 2(1 x)
and
= 27r(l - x)
rjL
\ir 2 - r
If the secondary current J is read by a hot wire instrument
then since the effective values are proportional to the maximum
values,
or,
er
If the frequency of the secondary is so adjusted that J 2 = --
z
then we get
a formula which is used estensively in connection with wave
meter measurements.
Inductively Coupled Oscillating Circuits Having Considerable
Damping. We have shown (equation 5) the following relation
between the effective secondary current J r at resonance and the
induced e.m.f. EQ when the primary is supplied from a source of
sustained power
310 ELECTRICAL ENGINEERING
The corresponding equation when both the primary and second-
ary circuits are oscillating has been worked out by BJERKNES and
others who found that as long as the decrements are small the
following relation obtains:
16/ 3 L
In this equation J max * is the maximum possible value of cur-
rent read by a hot wire instrument in the antenna circuit.
N is the number of condenser discharges per second.
EQ is the maximum value of the e.m.f. induced in the antenna
circuit. L 2 is the inductance of the antenna circuit in henrys,
/ is the frequency and di and d z the logarithmic decrements in
the primary and antenna circuits per full period.
7i =
E = <**MCiEi and E
2 _ _ =
" 16/ 3 "
Similarly,
In the case of sustained primary power.
The maximum instantaneous value of the antenna current is
from (45) remembering that in these equations the decrements
per full period is used.
T 2 A f S T 2
T 2 " max. ^ J "2 A "max. f?
~w~ i\r ~A^ 52
The maximum instantaneous value of the antenna voltage is
/2
(ID
Numerical Examples. Union College small set.
E l = 5000 volts
120 .
Ci == JQ-TO farads
farads
N = 500
ELECTROMAGNETIC RADIATION
311
and
or
X = 700m.; /. / = 0.43 10 6
>! = 0.05 5 2 = 0.10 k = 0.10 .'. did 2 (d! + d 2 ) = ~- 6
i o v 1 20 1 5
. 2 = 50,000 0.43 10 6 ^f^ 25 10 6 0.01 ~~ = 20
10 75
T _ jr
*J max. ~ Tt.tJ
4 \/ OH
- 0.43 10 6 X 0.10 = 1375
%
7 2 = 37 amp.
10 10
278 X 37 = 11,400 volts.
2x0.43 10 6 12
BJERKNES has shown how with a slight modification equation
(6) can be used to determine the decrement of the secondary
circuit which may, for instance, be the antenna circuit by means
of a third tuned circuit which is called a wave meter:
This expression is:
d + 5i = 27r (l - } (12)
where 5 is the decrement of the circuit being tested and <5i is
the decrement of the meter.
The formula is limited as is the case of equation (6) to the
condition that
J r and J being the effective values of the current in the wave
meter.
It is also limited to the condition
that both d and Si are small and
that di is considerably smaller than
8 and that finally Xi and X 2 do not
differ more than, say, 5 per cent.
Referring to Fig. 156, W is the
wave meter which is a calebrated
closed circuit of known inductance,
capacity and therefore of known
natural period. The resistance is FlG - 156 -
made as low as possible so that the decrement of the meter is small.
The value of the current or the (current) 2 is frequently deter-
mined by means of a low resistance heating element, actually a
thermal couple, which supplies a direct current to a galva-
nometer G.
312 ELECTRICAL ENGINEERING
In that case the galvanometer deflection is obviously pro-
portional to the square of the current value.
The procedure is as follows. The meter is loosely coupled to
the antenna and the capacity of the wave meter is varied until
the largest galvanometer deflection G> is obtained and the
corresponding wave length \o is read.
Then the capacity is changed so that the deflection of the
C 1
galvonometer is -~ when the meter reads shorter wave length.
We have then from (12)
5 + d l = 27T (l - ^]
To determine the decrement of the meter it is desirable to
insert in the meter circuit such non-inductive resistance that at
resonance, that is when the wave meter reads Xo, the galvanom-
C*
eter deflection is ~
The capacity is then varied until the galvanometer deflection
C 1
-^ when the wave length is X 2 .
We have then if 5 2 is the decrement due to the added resistance,
It has been shown in equation (8) that the relation between
the effective values of the resonance current with different
decrements are related as follows:
J/ 2 dd* (d l + da)
In our case
Jr' 2 Gr>
2
d\ = d = decrement of the antenna.
d'z = di + 5 2 = decrement of the wave meter in second
test.
di = 8 = decrement of the antenna.
<2 2 = 5i = decrement of the meter in the first test.
ELECTROMAGNETIC RADIATION 313
2 = ( g * + *)(* + fr + frJ
5i(5 + 5 2 )
1-^
Si + 5 2 Xo Si + 5 2
Si } Xi
Xo
- 8 = 2 I ( Xl ~ Xa ) ( x ~ x ^)
* * \ \ i \ o\
AQ AO "T~ *2 ^Ai
Numerical Example.
X = 500 m.
Xi = 485 m.
\z = 475 m.
.-. + .! = 2ir (l -^) = 0.189.
- ^ ( J - 500) - ' 314
- 0.125.
27T 10 X 25
= 0.
500 5
d = 0.126.
Conditions Affecting the Receiving Station. It has been shown
that at some distance from the sending antenna the maximum
value of the potential gradient in volts per centimeter near the
equatorial plane is
G== l20irh I (1)
where / is the maximum value of the current at the sending
antenna, the current being assumed the same at all points of
the conductor. The dimensions are given in centimeters.
A more general formula would be
* ( .
E 2 = j- (2)
where E 2 is the maximum value of the voltage across the whole
receiving antenna, a is a correction factor for the current dis-
2
tribution which is - for a simple antenna and unity for an antenna
7T
in which the height constitutes only a fraction of a quarter wave,
as is most frequently the case in actual practice.
h\, hz, X and r may be given in any units as long as they are
the same, hi and h 2 are the heights of the sending and receiv-
314 ELECTRICAL ENGINEERING
ing antenna, X the wave length and r the distance between hi
and h z .
In order to be applicable to wireless transmission this formula
needs to be elaborated in several respects.
(a) The voltage is actually greater due to the concentration
of energy as the waves sweep over the surface of the earth.
(b) The voltage is smaller on account of the energy which
strays away from the curvature even if the surface of the earth
is assumed to be perfect of conductivity.
(c) The voltage is reduced on account of the energy absorption
of the earth current which effect is prominent near the sending
conductor where the concentration of current is greatest.
(d) The voltage is sometimes increased, but more often re-
duced, due to reflection, absorption, etc., depending upon the
condition of the atmosphere.
FIG. 157.
Conditions (c) and (d) have not been studied theoretically,
but a considerable amount of data has been given from actual
tests, notably by AUSTIN and FULLER. l
The Effect of the Curvature of the Earth. Assume that the
sending antenna is at A and the receiving antenna at B, Fig.
157.
The distance between A and B is -=-. In the case of a plane
z
wave the receiving antenna for the same distance would then be
at C where,
A -C = \R.
Thus in this latter case the energy would be spread over a
AUSTIN, Bulletin, Bureau Standards, 1914.
FULLER, Proc., A. I. E. E., April, 1915.
ELECTROMAGNETIC RADIATION
315
circumference
wnereas due to the curvature of the earth
the circumference is only 2irR. There is, therefore, a concentra-
tion of energy which can be represented by a coefficient
k' =
and since the intensity of the electric field is proportional to the
Venergy, the concentration coefficient for the electric field at a
distance r under the condition given above is
R0. .'. Energy
Let distance AC, Fig. 157, be equal to AB
per unit length of circumference at C is
E
Energy per unit circumference at B is
_E_
2irR sin 6
2irR6 6
2irR sin sin B
or =
(3)
FIG. 158.
The effect of the straying of power on the potential gradient
due to the curvature of the earth is included in the equation
according to theoretical works done by SUMMERFIELD and ZEN-
NECK by the introduction of a divergence factor.
0.0019r
316 ELECTRICAL ENGINEERING
AUSTIN'S experiments indicate, however, that with continuous
waves this coefficient is:
0.0915r
and FULLER'S experiments show
0.0045r
=
(4)
AUSTIN'S equation gives values which lie between ZENNECK'S
and FULLER'S and has the advantage of being simpler than the
other two.
Thus the general formula for continuous waves becomes :
, ,
#2 = kki -- (5)
Note, however, that in equation (4) the dimensions are ex-
pressed in kilometers.
The maximum value of the antenna current in the case of
Tjl
sustained oscillations is evidently ^2 TT where R% is the total
/l2
resistance of the antenna that is the radiation resistance, the
effective resistance, ground resistance, and resistance of the
receiving device.
The equation 01 the current in the case of damped oscillations
is slightly different.
It has been shown that if an e.m.f., EQ, is impressed on a tuned
circuit the following relations obtain:
T 2 =
~
where EQ is the voltage induced, which in our case is E%. di and
d-2, are the decrements in the two circuits.
Thus di and d% are in this case the decrements of the sending
and receiving circuits respectively.
Equation (7) may be written:
/ di\
,,* ^i + ^
But the decrement of the receiving antenna is
2-L 2 /
ELECTROMAGNETIC RADIATION 317
where / M20 h
=
. , =
4/di
wnere j i is in
Tt 2 N m
g*Ji'
but /i 2 = Ji 2 ^ where Ji is the effective value of the sending
antenna current.
4/fl2 2 (l+JW RS (l+J)
\ 2/ \ 2/
and J2 = _ ^/^ (g)
The effective value of the voltage across the receiving antenna
TT ^ AsaJi (10)
_
where 2 is the effective value of the receiving voltage and Ji
is the effective current at the base of the sending antenna.
It is evident from the above that the ratio between the effect-
ive values of the received e.m.f. with sustained and with damped
oscillations is:
damped
if the decrements of the sending and receiving antennas were the
same then the ratio would be V2.
Method of Determining Power Received. AUSTIN based his
determinations on the fact that if in two circuits in parallel we
know the power in one we can calculate the power in the other
and the total power from the relations of the resistances Rys'
in the circuits.
The total power supplied is
FT F( E + E \ E R + S -
~ E \R + S)~ RS
The power of circuit R is
P -^
Fr ~ R
%l - R + S P - R + S
" P r ~ RS S
D I Cf
or the total power = P r .
318 ELECTRICAL ENGINEERING
The minimum power P r required for distinguishing between dots
and dashes of resistance R is determined experimentally by ob-
serving the current in the receiving antenna under conditions
that can be conveniently controlled.
Knowing P r and R and the resistance S which is shunted across
the telephone receiver enables one to determine the total power
received. In FULLER'S experiments at Honolulu this minimum
power was found to be 3.2 X 10 10 watt, when dealing with sus-
tained oscillations.
APPENDIX I
Partial Differentiation. The complete differential of a func-
tion V of several independent variables r, <p, 6 is recalled to be:
"-*+*+S
In words this equation reads: The total differential of V is
the sum of the partial differentials of V with respect to the
independent variables. meaning the derivative of V with
respect to r when <p and 6 are considered constant.
If the independent variables r, <p, and 6 are some functions of
a single other variables t the derivative of V with respect to t
is obtained by simply dividing equation (1) by dt.
Thus: dV dV dr aV d<p dV dd
dt ~ dr dt + d<p dt + d6 dt
If the independent variables r, <p and 6 are functions of several
other independent variables, for instance x, y, z, then the partial
derivative of V with respect to x is obtained in a similar way by
dividing the equation by dx, remembering, however, that now
-j is the partial derivative and should be written -r
dx dx
Thus ^ = iZ^,^I^_4_dZ^ m
dx dr dx ~*~ dp dx "*" a0 az
Similarly 3F = a7 6r 57 a^ dF a0
dy dr dy ~*~ d<p dy "*" d0 ay
and ^Z _ ?Z ^r ^Z ^ ^Z ^
a2 ~ a/- a^ + a^ dz + a<? a^
The second partial derivative of V with respect to x is obvi-
ously obtained from (3) as follows:
aF av ar _a^ /ev\ _ dv_ av
' '
/e\ _
\dr) ' '
dx 2 dr dx 2 dx dx\dr d<p dx 2 dx dx\ d
,dVWded_/dV\
dO dx 2 T dx dx\dd/ v ;
319
320
ELECTRICAL ENGINEERING
dV dV dV
In equation (6) , -r and r are each functions of r, <p and 6.
or d( ou
and,
d fdV\
d 2 V dr d 2 V d<p d 2 V dd
dx \ dr /
d (dV\
dr 2 dx ' drdtp dx ' drdd dx
d 2 V dr d 2 V dip .d 2 V dd
dx \ d<f>/
d<pdr dx dp 2 dx d<pdB dx
d 2 V dr , d 2 V dtp , d 2 V dS
(7)
Substituting these values in equation (6) we get:
dw dV d*r dV ay dV d 2 e
'
dr
_
~
^ /
dr 2 \dx
2d 2 V dr_d<p 2d 2 V dO dr
d6 dx 2
Q 2 V /dS\
d0 2 \dx)
2d 2 V d0
drd<p dx dx drdO dx dx d<pdO dx
A similar expression is, of course, obtained for
d 2 V . d 2 V
f - n and
(8)
dz 2
A complete discussion of partial differentiation can be found
in any text-book on Calculus, for instance, in volume II of WOODS
AND BAILEY'S "A Course in Mathematics."
As an application of the above is given the transformation of
LAPLACE'S equation from rectangular coordinates to spherical and
cylindrical coordinates.
FIG. 159.
(a) Transformation of LAPLACE'S equation to spherical coordi-
nates. Fig. 159.
" "
dx 2 dy 2 dz 2
APPENDIX
321
F = F(rM, r = f L (xyz), B = f z (xyz), <p = f*(xyz),
dV dV dr dV de dV dtp
~dx ~~ ~dr dx " dO dx dtp dx
dx 2 = ~~ dr ax 2 ~*~ dx dx\dr
dV av
"
dx 2
But
de dx 2 ' ax
dx \d<p.
A/^Z\
~dx \dO/
dr 2 dx "^ *~* *~~
dx \ dip) d<pdr dx
- _L
'
_i .
ax a0a<? ax
dX
= aF av aF
' ' dx 2 ' dr dx 2 4 " de
/dr\
aF
ax 2
d_V_ /dr\ d_V_ /d6\ d_F
ar 2 \dx/ de 2 \dx/ dv 2
2a 2 F ar de 2d 2 v dr_ d<p
dx,
2d 2 v de
drde dx dx drd<p dx dx d6d<p dx dx
Q2y A2V
Similar expressions can be gotten for --r and
' ax 2 dy
a 2 F
ar 2
dy''
~ ~dr (dx 2 dy 2 dz 2
dz 2
dy
aF
" "
av
-. a 2 F
"a^
L 5!Zr /M 2 4. /<M 2 /M 2 i
~ a< 2 L\ax/ " \d) ' \dz) J
dz 2 /
+ (?y
dx + dy dy
)x ax dy dy dz dzJ
but in the spherical coordinate system,
\/x 2 4- i/ 2 if
r = (x 2 + y 2 + 2 2 )^, 6 = arc tan - and ^ = arc tan -
dy
dr 30~i , 2a 2 Frar a<^ ar a^ ar a^
"" ax + d ~d dz a^
322 ELECTRICAL ENGINEERING
From the construction, Fig. 159,
x = \/x 2 + y 2 cos <p = r sin 6 cos <p',y = r sin d sin <f>\ z = r cos 6.
x
2\K = - = sin cos v?
2
dx
y
= ^ = sm sin
dy (x 2 + t/ 2 + z*)* r
dr _ 2 _ z
dz ~ (x 2 + y 2 + )W ~ r =
I^A.( X 2 I ^2)^
/2 ^ r y J
X
dx x 2 + y* x* + y 2 + z* " ( x *
Z Z COS <f> COS ^ COS <p
-r COS V? = - - = --
r 2 r r r
y
dy x 2 + 2/ 2 + 2 2 (a; 2 + y 2 )*
cos sin <p
dz ,
sn
A/i\
y ~dx \xl = y _ y r
f x z + y* ~ Vx 2 + y 2 rVx 2
r
sn
r sn
1
d<p x x x r cos <p
y 2 rx 2 + y 2 r sn
a , . dO d<p
= (sin 6 cos <p ) = cos 6 cos <p sin sin <p =
C/^C o3/ C/2/
. cos cos <p . . . . sin <p
= cos 6 cos v X - - + sin sin
I fjll.1. \J OX A A V^ * f*.
r ^ r sin 6
- [cos 2 cos 2 <p -f sin 2
APPENDIX 323
= - [(1 - sin 2 0)(1 - sin 2 <p) + sinV]
= - [1 - sin 2 cos 2 <f>]
dy* = dy (sin sin ^ = = r [1 ~ si
av d n a0 i
^=^cos0 , -sm0- = + -sin'0
d 2 6 cos 6 cos <p / 1 \ dr
> = - = cos cos " cos
1 .
2 (cos cos <p sm cos <p)
i 1 / . cos cos <p
H I cos <p sm -
r \ r
. 1 , . sin <f>
H cos sm
r r sm
= I -j sin ^ cos ^ cos 2 <^ + sin 9 cos cosV
cos . "I
r r Sin 2 <^>
sm
d 1 1 r
= ^- - cos sm $? = = - -I 2 sin cos sin 2
cos i
- COS <p 1
sm
a/ sin 0\
2 .
92 \ r /
r 2 S1
n cos V
a / sin v? \
1 sin <p c
>r 1 ( sin (p cos
r 2 sin fa r sin 2
cos < d
r sin dx
sin <^g dr 1 sin <p cos <90 cos
r 2 sin do; r sin 2 dx r sin dz
J^ / cos <p l^ cos y? d(p _ 1 cos <p cos 60
~ &y \r sin ~ r 2 sin B dy r sin 2 dy
sn
r sin
21
324 ELECTRICAL ENGINEERING
1 sin (p . 1 sin <p cos
^ STl sin ' cos * + F sin* a
r
r sin r sin r 2 sin
i a sm (p
1 cos <p cos cos sin 9
> sin <p cos <p
r sin 2 r
r sin r sin 6
av a 2
' * az 2 + di
r av _
1 2
-[1 sin 2 cos 2 (p + 1 sin 2 <^ + sin 2 0] = -
cos
ax 2 dy 2 dz 2 r 2 sin
W + \a?// " \dz>
sin 2 cos 2 </> + sin 2 sin 2 <p + cos 2 = 1
/a0\ 2 ,d0\ 2 /a0\ 2 = i
W \dyl \dz' r 2 sin 2
/a<p\ 2 /av?\ 2 /a<p\ 2 _ i
\a^/ " (dy) " Vaz/ ~ r 2 sin 2
ar a0 , ar a0 , ar a0 i x
T-^- + T- T,- + T-^ = -(sin cos cos 2 <p -f- sm cos sm 2 v?
ox ox oil dy dz oz r . .
sin cos 0) =
dx dx dy dy dz dz
d0 d(p d0 dtp dd d<p
a 2 F a 2 F a 2 F = aFrav av a 2 ri aFra 2 a 2 a 2
~ ~"
an 2 /an 2 /an 2-| a 2 Fr /a0\ 2 /a0\ 2 /a0\
dx) (dy) h W J^a^Lvax/ w \dJ
v, J Fr /a^\ 2 /a<^\ 2 /a<p\ 2 "i _2 aF i cos aF
^X^l \^/ ' \dv/ \dz) . ~r ~dr r 2 sin a0
. i_ a 2 F
- +
sin 2
i r / aF , a 2 F\ . i / fl aF,
A r ( 2 a7 +r ai^) + sirT0 ( cos g a0- +s
A B
APPENDIX
325
But
.'. LAPLACE'S equation in spherical coordinates
sin 2 $ d<p*
(C) CYLINDRICAL COORDINATES
Referring in Fig. 160
V = F(rBz) r = (z 2 + y *)K f e = tan' ^z = z
x = r cos e, y = r sin z = z
= x - x
dx ~ (x 2 -f y 2 )^ ~ r
dr
dy
dr
dz
cos 6
= sin 6
=
<
Six*
FIG. 160.
60
dx
+
y
r~
sn
cos
326 ELECTRICAL ENGINEERING
?-
dz
dz_dz__fa_^
dx dy dz
av a ,a0 sin 2 2
= cos = sm = H
ax 2 az dx r
av cos 2
^- r -o
O *-*
az 2
a 2 a sin 1 n d8 . 1 n ar 2
v-s = = sm + sm -r- = sm cos
aar aa; r r ao; r^ dx r 2
a 2 a cos i so i a0 2
^-s = - = sm cos -- = 5 sm cos
a?/ z dy r r dy r 2 - dy r 2
^ =
az 2
a *z = d*z = d% =
dx 2 dy 2 dx 2
Vr av av = i
' ' dx 2 dy 2 dz 2 ~ r
a 2 a 2 a 2
dx 2 8v* 8z 2 ~
(I)' +'+-
dr<Wd^rdOdrd<)_
dx dx dy dy dz dz
dr dz dr dz dr dz
+ "
=
dx dx + a?/ dy dz dz ~
' a^ 2 a?/ 2 a^ 2 ~ r ar 2 ar 2 r 2 a0 2 "" a^ 2 =
which is LAPLACE'S equation in cylindrical coordinates.
APPENDIX II
Elements of Vector Analysis. Physical quantities can be
divided into two large and important classes, namely: scalars
and vectors.
A scalar quantity is one that is absolutely determined by its
magnitude. Thus temperature, work, etc., are scalars.
A vector quantity may be denned as one having magnitude,
sense and direction and it is necessary to specify these three in
order to determine a vector. Velocities and accelerations are
examples of vector quantities; forces are strictly not vectors,
since they are characterized not only by their magnitude, sense
and direction but also by the point of application, while vectors
do not have definite position in space. However, forces can
be treated as vectors when proper account is taken of this
difference.
Addition and Subtraction of Vectors. Vectors are added or
subtracted by the well-known parallelogram law:
Thus
a + o = c
and
c o a.
Vectors follow the associative and com-
mutative laws of algebra, and hence very
little explanation is necessary as to the
addition of vectors. a
The sum of three vectors a, b and c is FlG> 161>
given by the diagonal mn as shown in Fig. 161.
Products of Vectors. There are two kinds of vector products :
I. The dot product which is denned as,
a dot b = a b = ab cos (a, b)
where a and b are the two vectors to be multiplied together,
and a and b are the numerical values of the vectors.
II. The cross product which is denned as:
a cross b = a X b = e ab sin (a, b) .
327
328 ELECTRICAL ENGINEERING
where e denotes that the product is a vector. It is the unit
vector perpendicular to the plane formed by a and 6.
The above names have been introduced by WILLABD GIBBS
and they are used principally by American writers.
The reader is familiar with the resolution of vectors into com-
ponents which can be treated according to the laws of ordinary
algebra. The great advantage of vector analysis is that it deals
with vectors directly. It is found useful, however, to resolve
vectors into their components and in such case a vector a is
defined in terms of its magnitude along any direction, say x,
times a unit vector i along x.
For convenience rectangular coordinates are used and the
unit vector along the z-axis is denoted by i, the unit vector along
the y-axis is denoted by j and the unit vector along the z-axis
byfc.
Thus
a = a x i -f- a v j -f- a x k
a = A/a* 2 + a v 2 + a 2 2
a = a(i cos a -f- j cos /3 -f k cos 7)
where a, and 7 are the direction cosines.
Now it will be easily seen from the definition of the dot product
that:
i - i 1 i - j =
j-j = l t-fc =
k-k = 1 j -k =
a - a = a 2
It is also clear that the condition of perpendicularity of two
vectors is that their dot product shall be zero.
The dot product is also called (by HAMILTON)
the scalar product, because the product is a
scalar. The cross product is called the vector
product, because it is a vector.
. a a X 6 gives a vector c, Fig. 162, whose mag-
nitude is (ab) sin (a,6); its direction is along
the normal to the plane of the vectors a and 6,
and finally the sense of c is taken so that as one
goes from a to 6 he follows a right-hand screw. In other words
from a to b we follow the threads of a corkscrew whose direction
of progress determines the sense of 6. This is, of course, the well-
APPENDIX
329
known rule of MAXWELL for the relation between the direction
of flux, the motion of a conductor, and the e.m.f. thereby
generated.
It is clear from the definition of a cross product that in Fig. 163
i X j = k = - j Xi
j x k = i = - k X j
k X i = j = -- i X k
i X * = j X j = k X k = 0.
FIG. 163.
The cross product of two vectors can also be obtained in
terms of the components and the unit vectors i, j and /b; only
it is evident that care should be taken not to invert the order
of factors, since a Xb = 6 X a.
Exercise. Prove that if a x a v a 2 , b x b v b z are the rectangular
components of a and b.
a X b = (a v b z - aj) v ) i + (a f b x - ajb,) j + (a x b v - a v b x )k
or in determinant form,
* j
a X b = a x a v
b x by b z
Exercise. Prove that the absolute value of a X b which is
written a X b
= (a) (b) sin (a, b)
a X 6=
b v 2 + b f 2 ) - (a x b x + a v b
330 ELECTRICAL ENGINEERING
Now it will be noticed that in the last exercise, a 2 x + a 2 y + a z 2
is simply equal to a a.
Thus:
First term = a a
Second term = b - b
Third term = (a 6) 2
= (ab cos a) 2
where = < a, b
so that a X & = V(a )(& &) - (ab cos a) 2
a X 6 = "a 2 & 2 - a 2 6 2 cos 2 a
= a& \ 1 cos a
= ab sin a
The product of a X 6, must be the normal to the plane of the
vectors a and b is seen as follows : Assume c to be the vector and
find a c = a. (a X b)
also b-c = b. (a X b)
Multiplying these out in the ordinary way we find
a c = 6 c = 0,
i.e., ac cos (a, c) =
be cos (bj c) =
which is satisfied when c is normal to the plane ab.
The above are intended to cover the very small part of vector
analysis used in this book. For further information the reader
should consult special treatises written on the subject.
HEAVISIDES' " Electromagnetic Theory;" ABRAHAM and
FOPPL'S "Theory of Electricity and Magnetism" can be recom-
mended highly.
An excellent short treatise on the subject is "Elements of
Vector Analysis" by BURALLI-FORTI and R. MAREOLONGO, and
a somewhat larger work is that of WILLARD GIBBS, edited by
WILSON. Finally COFFIN'S "Vector Analysis" may be men-
tioned among works of reference, it appears indeed as best suited
for the introduction to vector analysis.
INDEX
Attenuation, 121
Austin, 316
Auxiliary equations, 46
B
Ber and bei function, 273
Bjerknes, 313
Capacity between concentric con-
ductors, 70
between parallel planes, 68
between transmission lines, 70
of antenna, 299
of a single wire, 230
of concentric cable, 96
of isolated spheres, 163
of two cylindrical conductors,
224
of two wires in multiple, 230
Charge distribution on an ellipsoid,
199
Circular symmetry, 188
Complete differential, 164
Complimentary function, 46, 86
Concentric cylinders, 215
spheres, 209
Condenser, capacity of, 68
characteristics of, 68
charged, 71
discharged, 72
energy supplied to, 71
Coulomb's law, 157
Coupling, effect on frequency, 303
Curl of a vector, 257
Current, equation of, 261
Curvature of earth, 314
Cylindrical bars, 150
conductor, current and flux
distribution, 268
conductors, 218
D
Differential equations, higher order,
44
operator, D, 45
Differentials and differences, 61
Direct-current generator, field cir-
cuit, 59
Displacement current, 263, 264
Distortionless line, 142
Divergence of a vector, 185
theorem, 186
E
Electric doublet, 284
field, energy of, 262
intensity, 290
Electromagnetic radiation, 278
Electromotive force, equation of, 260
F
Field intensity, 158
Flat conductors, 152 .
Flux and current distribution, 152
Forces between point charges and
spheres, 175
Froelich's equation, 22
Fuller, 316
G
Gauss' theorem, 160
Graph of function y = <r x , 9
Green's theorem, 186
Grounded horizontal wire, effect on
potential distribution, 254
331
332
INDEX
H
Hertz's oscillator, 284
Hysteresis loop, 56
I
Images, method of, 168
Inductance, 11
of air coil, 36
of concentric cables, 97
Inductances, combined, 37
Inductive circuit containing iron, 62
K
Kelvin, 273
Lame's differential parameter, 185
Laplace's equation, 188, 320
Legendre's coefficient, 188
function, 189
Leyden jars, 72
Linear differential equations, 3
Line charge, 218
integral, 163
Logarithmic decrement, 295
M
Magnetic field, energy of, 263
energy stored in, 8
intensity, 290
potential, 180
and current, 183, 184
shell, 181
Marconi, 292
Maxwell's coefficient, 232, 263
Metallic spheres, 169
Mutual induction, 33
imperfect, 51
perfect, 37
O
Oblate ellipsoid, potential distribu-
tion, 24
Partial differentiation, 319
fractions, 21
Poisson's equation, 187
Potential, 162
distribution between point
charge and plane, 171
between two spheres, 175
gradient, 164, 165
of small magnet, 180
outside of thin circular disc, 197
Power factor, 295
received, 317
R
Radiated energy, 290
Radiation, 278
resistance, 292
Receiving station, 313
S
Short-circuited winding, current in,
54
Short-circuit suddenly opened, 84
Shunt motor self excited, 16
Skin effect, 271
Solenoidal field, 186
Solid angle, 183
Step-by-step method, 66
Stoke's theorem, 258
Surface density, 170
integral of distributed vector,
158
Symbolic factors, 89
Three-phase cable, 243
Three-phase line, 249
Tuned circuit, 80
Two conductor cables, 237
U
Unit charge and unit pole, 157
V
Vector analysis, 327
Velocity of propagation, 121-132
W
Wave lengths, 121-132
Weber's equation, 182
THIS BOOK IS DUE ON THE LAST BATE
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AN INITIAL FINE OI
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THIS BOOK ON THE DATE DUE. THE PENALTY
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OVERDUE.
OCT 23 1932
CT
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8 1933
3CT 5 1934
3 194!
L^St-
dc
INTER-L 3RAR 1
a
f LOAN
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YC 19681
84283
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