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Industrial Power Systems 
Handbook 
D O N A L D BEEMAN, Editor 
Manager, Industriaf P w e r Engineering 
Industrial Engineering Seclwn 
General Electric Company, Schenectady, New Yorlc 
FIRST EDIT ION 
McGRAW-HILL BOOK COMPANY, INC. 
1955 New York Toronto London 
Ch.UPh?r 1 by Donald Beeman, Alan Graeme Darling, 
and R. H. Kaufmann 
Short-circuit-current Calculating 
Procedures 
FUNDAMENTALS OF A-C SHORT-CIRCUIT CURRENTS 
The determination of short-circuit currents in power distribution sys- 
tems is just as basic and important as the determination of load currents 
for the purpose of applying circuit breakers, fuses, and motor starters. 
The magnitude of the shoncircuit current is often easier to determine 
than the magnitude of the load current. 
Calculating procedures have been so greatly simplified compared with 
the very complicated procedures previously used that now only simple 
arithmetic is required to determine the short-circuit currents in even the 
most complicated power systems. 
SHORT-CIRCUIT CURRENTS AND THEIR EFFECTS 
If adequate protection is to he provided for a plant electric system, the 
size of the electric power system must also be considered to determine 
how much short-circuit current i t will deliver. This is done so that cir- 
cuit breakers or fuses may he selected with adequate interrupting capac- 
ity (IC). This interrupting capacity should be high enough to open 
safely the maximum short-circuit current which the power system can 
cause to flow through a circuit breaker if a short circuit occurs in the 
feeder or equipment which it protects. 
The magnitude of the load current is determined by the amount Of 
work that is being done and hears little relation to the size of the system 
supplying the load. However, the magnitude of the short-circuit current 
is somewhat independent of the load and is directly related to the size or 
I 
2 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
capacity of t,he power source. The larger the apparatus which supplies 
electric power to the system, the greater the short-circuit current will be. 
Take a simple case: A 440-volt three-phase lo-lip motor draws about 
13 amp of current a t full load and will draw only this amount whether 
supplied by a 25-kva or a 2500-kva transformer bank. So, if only thc 
load currcnts arc considered when selecting motor branch circuit break- 
ers, a 15- or 20-amp circnit, breaker wnuld he specified. However, the 
size of t,he power system back of the circuit breaker has a real bearing on 
the amount of the short,-circuit, current. which can flow as a result of a 
short circuit on the load side of the circuit breaker. Hence, a much 
larger circuit breaker would be required to handle the short-circuit current 
from a 2500-kva bank than from a 25-kva bank of transformers. 
These numbers A simple mathematical example is shown in Fig. 1.1. 
MUST BE CAPABLE OF INTERRUPTING 1000 AMPERES 
El I O O V 
100 A 
~ ~ 1 0 . 1 OHMS 
MOTOR LOAD 
CURRENT 
5 AMP 
APPARENT 
IMPEDANCE 
20 OHMS 
E I 00 SHORT CIRCUIT CURRENT = - : - = 1000- AMPERES 
Z T 0.1 
MUST BE CAPABLE OF INTERRUPTING 10,000 AMPERES 
w 
I000 A 
2 1 = 0.01 OHMS 
MOTOR LOAD 
CURRENT 
5 AMP 
FIG. 1.1 
circuit-current magnitude than load. 
Illustrotion showing that copocity of power source has more effect on rhort- 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 3 
have been chosen for easy calculation rather than a representation of 
actual system conditions. 
The impedance, limiting the flow of load current, consists mainly of 
the 20 ohms apparent impedance of the motor. If a short circuit occurs 
at F , the only impedance t o limit the flow of short-circuit current is the 
transformer impedance (0.1 ohm compared with 20 ohms for the motor); 
therefore, the short-circuit current is 1000 amp, or 200 times as great as 
the load current. Unless circuit breaker A can open 1000 amp, the 
short-circuit current will continue to flow, doing great damage. 
Suppose the plant grows and a larger transformer, one rated a t 1000 
amp, is substituted for the 100-amp unit. A short circuit a t F , (bottom 
in Fig. 1.1) will now be limited by only 0.01 ohm, the impedance of the 
larger transformer. Although the load current is still 5 amp, the short- 
circuit current will now he 10,000 amp, and circuit breaker A must be 
able t o open that amount. Consequently it is necessary to coiisider the 
size of the system supplying the plant as well as the load current, to be 
sure that circuit breakers or fuses are selected which have adequate 
interrupting rating for stopping the flow of the short-circuit current. 
Short-circuit and load currents are analogous t o the flow of xvater in a 
hydroelectric plant, shoivn in Fig. 1.2. The amount of water that flows 
under normal conditions is determined by the load on the turbines. 
Within limits, i t makes little difference whether the reservoir behiiid the 
dam is large or small. This flow of water is comparable to the flow of load 
current in the distribution system in a factory. 
On the other hand, if the dam breaks, the amount of water that will 
flow will depend upon the capacity of the reservoir and will bear little 
relation to the load on the turbines. Whether the reservoir is large or 
small will make a great difference in this case. This flow of water is 
comparable t o the flow of current through a short circuit in the distribu- 
tion system. The load currents do useful work, like the water that flows 
down the penstock through the turbine water wheel. The short-circuit 
currents produce unwanted effects, like the torrent that rushes madly 
downstream when the dam breaks. 
SOURCES OF SHORT-CIRCUIT CURRENTS 
When determining the magnitude of short-circuit currents, it is 
extremely important that all sources of short-circuit current he considered 
and that the reactance characteristics of these sources be known. 
There are three basic sources of short-circuit current: 
1. Generators 
2. Synchronous motors and synchronous condensers 
3. Induction motors 
4 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
All these can feed shorecircuit current into a short circuit (Fig. 1.3). 
Generators are driven by turbines, diesel engines, water wheels, or 
other types of prime movers. When a short circuit occurs on the circuit 
fed by a generatar, the generator continues to produce voltage because the 
field excitation is maintained and the prime mover drives the generator 
at substantially normal speed. The generated voltage produces a short- 
circuit current of a large magnitude which flows from the generator (or 
generators) to the short circuit. This flow of short-circuit current is 
limited only by the impedance of the generator and of the circuit between 
the generator and the short circuit. For a short circuit a t the terminals 
of the generator, the current from the generator is limited only by its own 
impedance. 
FIG. 1.2 
the hydroelectric plant. 
Normal load and short-circuit currents are analogous to the conditions shown in 
SHORT-CIRCUIT-CURRENT ULCULATlNG PROCEDURES 5 
METAL CLAD SWITCHGEAR 
SHORT CIRCUIT 
CURRENT FROM 
INDUCTION 
MOTOR 
FIG. 1.3 Generators, synchronous motors, and induction motors all produce short-circuit 
current. 
HOW SYNCHRONOUS MOTORS PRODUCE SHORT-CIRCUIT CURRENT 
Synchronous motors are constructed substantially like generators; i.e., 
they have a field excited by direct current and a stator winding in which 
alternating current flows. Normally, synchronous motors draw a-c 
power from the line and convert electric energy to mechanical energy. 
However, the design of a synchronous motor is so much like that of a 
generator that electric energy can be produced just as in a generator, by 
driving the synchronous motorwith a prime mover. Actually, during a 
system short circuit the synchronous motor acts like a generator and 
delivers shortcircuit current to the system instead of drawing load cur- 
rent from it (Fig. 1.4). 
As soon as a short circuit is established, the voltage on the system is 
reduced to a very low value. Consequently, the motor stops delivering 
energy to the mechanical load and starts slowing down. However, the 
inertia of the load and motor rotor tends to prevent the motor from slow- 
ing down. In other words, the rotating energy of the load and rotor 
drives the synchronous motor just as the prime mover drives a generator. 
6 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
The synchronous motor then becomes a generator and delivers short- 
circuit current for many cycles after the short circuit occurs on the system. 
Figure 1.5 shows an oscillogram of the current delivered by a synchronous 
motor during a system short circuit. The amount of current depends 
upon the horsepower, voltage rating, and reactance of the synchronous 
motor and the reactance of the system to the point of short circuit. 
LOAD CURRENT 
SYNCHRONOUS 
MOTOR 
-€t 
FIG. 1.4 Normally motors draw 
load current from the source or 
utility system but produce rhort- 
circuit current when a short cir- 
wit occurs in the d a d . 
U l I L I T Y 
SYSTEM 
,- \ 
SHORT CIRCUIT 
CURRENT FROM 
MOTOR 
. . -. . . 
SYSTEM 
SYNCMOYOUS ' 
FIG 1 5 IBmlowl l roce of 0s- 
Yoroll 
. . - . . ._ ,. _ _ .. ,. . . .. . . .. 
cillogrclm of short-circuit current 
produced by a synchronous 
motor 
SHORT ' . - I 
CIRCUIT 
SHORT CIRCUIT CURRENT DELIVERED BY 
A SYNCHRONOUS MOTOR. 
SHORT.CIRCUIT-CURRENT CALCULATING PROCEDURES 7 
HOW INDUCTION MOTORS PRODUCE SHORT-CIRCUIT CURRENT 
The inertia of the load and rotor of an induction motor has exactly the 
same effect on an induction motor as on a synchronous motor; i.e., it 
drives the motor after the system short circuit occurs. There is one 
major difference. The induction motor has no d-c field winding, but 
there is a flux in the induction motor during normal operation. This flux 
acts like flux produced by the d-c field winding in the synchronous motor. 
The field of the induction motor is produced by induction from the 
stator rather than from the d-c winding. The rotor flux remains normal 
as long as voltage is applied to the stator from an external source. How- 
ever, if the external source of voltage is removed suddenly, as it is when a 
short circuit occurs on the system, the flux in the rotor cannot change 
instantly. Since the rotor flux cannot decay instantly and the inertia 
drives the induction motor, a voltage is generated in the stator winding 
causing a short-circuit current to flow to the short circuit until the rotor 
flux decays to zero. To illustrate the short-circuit current from an 
induction motor in a practical case, oscillograms were taken on a wound- 
rotor induction motor rated 150 hp, 440 volts, 60 cycles, three phase, ten 
poles, 720 rpm. The external rotor resistance was short-circuited in each 
case, in order that the effect might he similar to that which would he 
obtained with a low-resistance squirrel-cage induction motor. 
Figure 1.6 shows the primary current when the machine is initially 
running light and a solid three-phase short circuit is applied a t a point in 
the circuit close to its input (stator) terminals a t time TI. The current 
shown is measured on the motor side of the short circuit; so the short- 
circuit current contribution from the source of power does not appear, but 
only that contributed by the motor. Similar tests made with the machine 
initially running a t full load show that the short-circuit current produced 
T. 
FIG. 1.6 , Tracer of oxillograms of short-circuit currents produced by an induction motor 
running at light load. 
8 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
by the motor when short-circuited is substantially the same, regardless of 
initial loading on the motor. Note that the maximum current occurs in 
the lowest trace on the oscillogram and is about ten times rated full-load 
current. The current vanishes almost completely in four cycles, since 
there is no sustained field current in the rotor to provide flux, as in the 
case of a synchronous machine. 
The flux does last long enough to prodnce enough short-circuit current 
to affect the momentary duty on circuit breakers and the interrupting 
duty on devices which open within one or two cycles after a short circuit. 
Hence, the short-circuit current produced by induction motors must he 
considered in certain calculations. The magnitude of short-circuit cur- 
rent produced by the induction motor depends upon the horsepower, 
voltage rating, reactance of the motor, and the reactance of the system to 
the point of short c. "cuit. The machine impedance, effective a t the time 
of short circuit, cmesponds closely with the impedance a t standstill. 
Consequently, the i iitial symmetrical value of Short-circuit current is 
approximately equnl to the full-voltage starting current of the motor. 
TRANSFORMERS 
Transformers are often spoken of as a source of short-circuit current. 
Strictly speaking, this is not correct, for the transformer merely delivers 
the short-circuit current generated by generators or motors ahead of the 
transformer. Transformers merely change the system voltage and mag; 
nitude of current but generate neither. The short-circuit current deliv- 
ered by a transformer is determined by its secondary voltage rating and 
reactance, the reactance of the generators and system to the terminals of 
the transformer, and the reactance of the circuit from the transformer to 
the short circuit. 
ROTATING-MACHINE REACTANCE 
The reactance of a rotating machine is not one simple value as it is for a 
transformer or a piece of cable, but is complex and variable with time. 
For example, if a short circuit is applied to the terminals of a generator, 
the short-circuit current behaves as shown i n Fig. 1.7. The current starts 
out a t a high value and decays to a steady state after some time has 
elapsed from the inception of the short cirroit. Since the field excitation 
voltage and speed have remained snbstantially constant within the short 
interval of time considered, a change of apparent react,ance of the machine 
may he assumed, to explain the change in the magnitude of short-circuit 
current with time. 
The expression of such variable reactance at any instant after the 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 9 
occurrence of any short circuit requires a complicated formula involving 
time as one of the variables. For the sake of simplification in short-cir- 
cuit calculating procedures for circuit-breaker and relay applications, 
three values of reactance are assigned to generators and motors, viz., 
subtransient reactance, transient reactance, and synrhronous reactance. 
The three reactances can be briefly described as follows: 
1. Subtransient reactance X y is the apparent reactance of the stator 
winding at the instant short circuit occurs, and it determines the current 
Row during the first few cycles of a short circuit. 
2. Transient reactance X i is the apparent initial reactance of the 
stator winding, if the effect of all amortisseur windings is ignored and 
only the field winding considered. This reactance determines the cur- 
rent following the period when subtransient reactance is the controlling 
value. Transient reactance is effective up to 45 see or longer, depending 
upon the design of the machine. 
3. Synchronous reactance X d is the apparent reactance that deter- 
mines the current flow when a steady-state condition is reached. It is not 
effective until several seconds after the short circuit occurs; consequently,it has no value in short-circuit calculations for the application of circuit 
breakers, fuses, and contactors but is useful for relay-setting studies. 
Figure 1.8 shows the variation of current with time and associates the 
various reactances mentioned above with the time and current scale. 
Previous loading has an effect on the total magnitude of short-circuit 
CURRENT DETERMINED 
BY SYNCHRONOUS 
OF TOTAL OSCILLOGRAM 
ONLY TWO ENDS SHOWN 
HERE. THIS REPRESENTS 
THE BREAK BETWEEN 
THE TWO PARTS. 
OCCURS AT 
THIS TIME. 
FIG. 1.7 Trace of orcillograrn of hart-circuit current produced by a generator. 
10 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
MAX, SUBTRANSIENT CURRENT- USE SUBTRANSIENT REACTANCE X"d 
/- 
TIME- 
( 8 ) 
FIG 1.8 Variation of generotor short-circuit current wilh time. 
current delivered by a generator. 
given by the machine designer is the lowest value obtainable. 
use will show maximum short-circuit current. 
before a system analysis can he made. 
The value of X i or X y generally 
Hence, its 
Certain characteristics of short-circuit currents must he understood 
SYMMETRICAL AND ASYMMETRICAL SHORT-CIRCUIT CURRENTS 
These terms are used to describe the symmetry of the a-c waves about 
the zero axis. If the envelopes of the peaks of the current waves are 
symmetrical about the zero axis, the current is called symmetrical current 
(Figs. 1.9 and 1.10). If the envelopes of the peaks of the current waves 
are not symmetrical about the zero axis, the current is called asymmetrical 
ENVEWPES OF PEAKS 
OF SINE WAVE ARE 
SYMMETRIGAL ABOUT 
THE ZERO AXIS. 
ZERO 
A X I S 
FIG. 1.9 Symmelrical a-c wove. 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
THE ENVELOPES OF PEAKS 
ARE SVHHETRICAL ABOUT 
ZERO AXIS 
FIG, 1.10 Symmetrical d t e r n a t i n g current f rom a short-circuited generotor. 
1 1 
ENVELOPES OF PEAKS ARE NOT 
SYMMETRICAL ABOUT ZERO AXIS 
AX1 S 
TOTALLY 0 F F SET 
PARTIALLY OFFSEl 
FIG. 1.11 
for the purpose of illustration only. 
circuits. 
Asymmetrical (I-c waver. The conditions shown here ore theoreticol a n d ore 
D-C component will r a p i d l y d e c a y to zero i n actual 
FIG. 1.12 Trace of orc i l logram of a typ ica l short-circuit current 
12 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
current (Fig. 1.11). The envelope is a line drawn through the peaks of 
the waves, as shown in Figs. 1.9 to 1.12. 
For the sake of explanation, many of the illustrations, such as Figs. 
1.11, 1.15 to 1.19, show sine waves of current uniformly offset for several 
cycles. It should be noted that in practical circuits the amount of asym- 
metry decreases rapidly after the occurrence of the short circuit in the 
system. This decrease of asymmetry is shown qualitatively in illustra- 
tions such as Figs. 1.12, 1.20, 1.23, and 1.24. 
Oscillograms show that short-circuit currents are nearly always asym- 
metrical during the first few cycles after the short circuit occurs. They 
also show that the asymmetry is maximum at the instant the short circuit 
occurs and that the current gradually becomes symmetrical a few cycles 
after the occurrence of the short circuit. The trace of an oscillogram of a 
typical short-circuit current is shown in Fig. 1.12. 
WHY SHORT-CIRCUIT CURRENTS ARE ASYMMETRICAL 
In the usual industrial power systems the applied or generated voltages 
are of sine-wave form. When a short circuit occurs, substantially s ine 
wave short-circuit currents result. For simplicity, the following discus- 
sion assumes sine-wave voltages and currents. 
In ordinary power circuits the resistance of the circuit is negligible com- 
pared with the reactance of the circuit. The short-circuit-current power 
factor is determined by the ratio of resistance and reactance of the circuit 
only (not of the load). Therefore the short-circuit current in most power 
circuits lags the internal generator voltage by approximately 90" (see 
Fig. 1.13). The internal generator voltage is the voltage generated in 
the stator coils by the field flux. 
If in a circuit mainly containing reactance a short circuit occurs at the 
peak of the voltage wave, the short-circuit current would start at zero 
and trace a sine wave which would be symmetrical ahout the zero axis 
(Fig. 1.14). 
If in the same circuit (i.e., one containing a large ratio of reactance to 
resistance) a short circuit occurs at the zero point of the voltage wave, the 
current will start a t zero but cannot follow a sine wave symmetrically 
about the zero axis because such a current would be in phase with the 
voltage. The wave shape must be the same as that of voltage hut 90' 
behind. That can occur only if the current is displaced from the zero 
axis, as shown in Fig. 1.15. In this illustration the current is a sine wave 
and is displaced 90' from the voltage wave and also is displaced from the 
zero axis. The two cases shown in Figs. 1.14 and 1.15 are extremes. 
One shows a symmetrical current and the other a completely asym- 
metricd current. 
This is known as a symmetrical short-circuit current. 
WORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 13 
GENERATOR TRANSFORMER 
INTERNAL VOLTAGE OF GENERATOR APPLIED HERE 
ONE LINE IMPEDANCE 
ioxazx 7x 0.m 
REACTANCE, X = 19% 
RESISTANCE. R = 1.4% 
I 
RESISTANCE I S LESS THAN OF THE REACTANCE HENCE MAY 
BE NEGLECTED WITHOUT AN APPRECIABLE ERROR 
INTERNAL VOLTAGE OF GENERATOR 
NEARLY 90' - SHORT CIRCUIT CURRENT 
DIAGRAM 
SHOWING 
SINE WAVES 
CORRESPONDING 
TO VECTOR 
DIAGRAM 
FOR ABOVE 
CIRCUIT 
FIG. 1.13 Diagrams Illustrating the phase relations of voltage and short-circuit current. 
14 SHORT-CIRCUll-CURRENT CALCULATING PROCEDURES 
GENERATED VOLTAGE 
SHORT CIRCUIT CURRENT 
ZERO 
A X I S 
SHORT CIRCUIT OCCURRED AT THIS POINT 
FIG. 1.14 
cirwit. 
Symmetric01 short-circuit current and generoted voltage for zero-power-factor 
-SHORT 
CIRCUIT 
CURRENT 
FIG. 1.15 Asymmetrical short-circuit current and generated voltage in zero-power-factor 
circuit. Condition i s theoretical and is shown for illustration purposes only. 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES I S 
If, in a circuit containing only reactance, the short circuit occurs a t any 
point except a t the peak of the voltage wave, there will be some offset of 
the current (Fig. 1.16). The amount of offset depends upon the point on 
the voltage wave at which the short circuit occurs. It may vary from 
zero (shown in Fig. 1.14) to a maximum (shown in Fig. 1.15). 
I n circuits containing both 
reactance and resistance, the s~,?&&,R&!~~ 
amount of offset of the short- 
circuit current may vary be- 
tween the same limits as for 
circuits containing only react- 
ance. However, the point on 
the voltage wave a t which the 
short circuit must occur to pro- 
duce maximum asymmetry 
dependsupon the ratioof react- 
ance to resistance of the cir- 
cuit. Maximum asymmetry 
is obtained when the short cir- 
cuit occurs a t a time angle 
equal to 90" + 0 (measured 
forward in degrees from the 
zero point of the voltage wave) 
CURRENT 
where tangent 0 equals there- ASYMMETRICAL 
actance-to-resistance ratio of 
FIG. 1.16 Short-circuit current and generated 
the circuit' The short-circuit voltage in zero-Dower-factor circuit. Short cir- 
current will be symmetrical cuit occurred between the 
when the fault occurs 90" from 
that point onthe voltage wave. 
point and peak of the generated voltctge wove. 
This condition i s theoretical and for illustration 
an example, assumeacir- purporer only. The short-circuit current will 
gradually become symmetrical in practical cuit that has equal resistance 
and reactance, i.e., the react- 
ance-to-resistance ratio is 1. The tangent of 45" is I ; hence, maximum 
offset is obtained whenthe short circuit occurs a t 135' from the zero 
point of the voltage wave (Fig. 1.17). 
CiTCUit., 
D-C COMPONENT OF ASYMMETRICAL SHORT-CIRCUIT CURRENTS 
Asymmetrical alternating currents when treatedas a single current wave 
are difficult to interpret for circuit-breaker application and relay-setting 
purposes. Complicated formulas are also required to calculate their 
magnitude unless resolved into components. The asymmetrical alter- 
nating currents are, for circuit-breaker applications and relay-setting 
16 SHORT-CIRCUIT-CURRENT CALCUUTING PROCEDURES 
MAXIMUM OFFSET 
Short-circuit current and generated voltage in circuit with equal reactance and 
The 
FIG. 1.17 
resistance. 
short-circuit current will gradually become symmetrical in practical circuits. 
purposes, arbitrarily divided into simple components, which makes it 
easy to calculate the short-circuit magnitude a t certain significant times 
after the short circuit occurs. 
The asymmetrical alternating current behaves exactly as if there were 
two component currents flowing simultaneously. One is a symmetrical 
a-c component and the other a d-c component. The sum of those two 
components a t any instant is equal to the magnitude of the total asym- 
metrical a-c wave a t the same instant. 
The d-c component referred to here is generated within the a-c system 
with no external source of direct current being considered. In some cases, 
particularly in the neighborhood of the d-c railways, direct current from 
the railways flows through neighboring a-c systems. This type of d-c 
current is not considered in this discussion or in the calculating procedures 
which follow. 
As an example of the resolution of asymmetrical alternating currents 
into components, refer to Fig. 1.15 which shows an asymmetrical short- 
circuit current which is resolved into a symmetrical a-c and a d-c compo- 
nent in Fig. 1.18. If the instantaneous values of the two components 
(dashed lines) are added a t any instant, the resultant will be that of the 
asymmetrical current wave. 
This condition i s theoretical and is shown for illustration purposes only. 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
F I N S T A N T AT WHICH SHORT CIRCUIT OCCURS 
17 
ASYMMETRICAL 
AC COMPONENT 
FIG. 1.18 
current. 
Theoretical Ihort-circuit-cvrrent wove illustrating components of asymmetrical 
In practical circuits, d-c component would decay to zero in o few cycler. 
 INSTANT OF SHORT CIRCUIT 
T O T A L C U R R E N T 
D C C O M P O N E N T 
A C C O M P O N E N T 
ZERO A X I S 
a = b = D C C O M P O N E N T 
FIG. 1.19 
at some point between the zero point and peak of the generated voltage wave. 
lhsoretical condition similar to that shown in Fig. 1.18. 
Components of asymmetrical short-circuit current in which short circuit occurred 
This is a 
I8 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
As mentioned previously, the examples shown in Figs. 1.13 and 1.18 are 
In practical circuits the d-c component for purposes of illustration only. 
decays very rapidly, as shown in Fig. 1.20. 
INITIAL MAGNITUDE OF D-C C O M P O N E N T 
The magnitude of the d-c component depends upon the iustant, the 
short circuit occurs and may vary from zero, as in Fig. 1.14, to a maximum 
initial value equal to the peak of the a-c symmetrical compoiieiit, as i n 
Figs. 1.15 and 1.18. When the short circuit occurs at any other point, 
such as shown in Fig. 1.19, the initial magnitude of the d-c componciit is 
equal to the value of the a-c symmct,riral component a t thc instant of 
short circuit. The above limit,s hold true for the initial magiiitudc of d-c 
eomporient in a system regardless of the reactance and resistance. Ilow- 
ever, the d-c componeut does not continue to flo~v a t a constant value, as 
shown in Figs. 1.18 and 1.19, unless there is zero resistauce i i i the circuit. 
DECREMENT 
There is uo d-c voltage in the system to sustaiu the flax of direct 
current; therefore the energy represeuted by the dirert. component of 
current will be dissipated as ZZR loss from the direct current flowiug 
through the resistance of the circuit. If the circuit had zero resistance, 
the direct current would flow at a constant value (Figs. 1.18 and 1.19) 
TOTAL ASYMMETRICAL CURRENT 
C COMPONENT 
AC COMPONENT 
FIG. 1.20 
short-circuit currenl gradually becomes symmetrical when d-c component diroppearr. 
Trace of orcillogrom showing decay of d-c component and how orymmetricd 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 19 
until the circuit was interrupted. However, all practical circuits have 
some resistance; so the d-c romponent decays as shown in Fig. 1.20. The 
combination of the decaying of d-c and symmetriral a-(* components gives 
an asymmetrical wave that changes to a symmetriral wave whcti the 
d-c component has disappeared. The rate of decay of the currents is 
called the decrement. 
X / R RATIO 
The X / R ratio is the ratio of the reactance to the resistance of the cir- 
cuit. The decrement or rate of decay of the d-c component is propor- 
tional to the ratio of reactance to resistance of the complete circuit from 
generator to short circuit. The theory is the same as opening the circuit 
of a battery and an inductive coil. 
If the ratio of reactance to resistance is infinite (i.e., zero resistance), 
the d-c component never decays, as shown in Figs. 1.18 and 1.19. On the 
other hand, if the ratio is zero (all resistance, no reartance), it decays 
instantly. FOF any ratio of reactarice to resistance in between these 
limits, the d-c component takes a definite time to decrease to substantially 
zero, as shown in Fig. 1.20. 
! I n generators the ratio of subtransient reactance to resistance may be as 
?much as 70: l ; so i t takes several cycles for the d-c component to dis- 
appear. In circuits remote from generators, the ratio of reactance to 
resistance is lower, and the d-c component decays more rapidly. The 
higher the resistance in proportion to the reactance, the more IaR loss 
from the d-c c.omponent, and the energy of the direct current is dis- 
sipated sooner. 
D-C TIME CONSTANT 
Often i t is said that generators, motors, or circuits have a certain d-c 
This refers again to the rate of decay of the d-c compo- time constant. 
O C COMPONENT 
a = 37Y. OF b (APPROX ) 
C- TIME 
CONSTANT IN OF D C COMPONENT 
SECONDS 
FIG. 1.21 Graphic illustration of time constant. 
20 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
nent. The d-c time constant is the time, in seconds, required by the d-c 
component to reduce to about 37 per cent of its original value a t the 
instant of short circuit. I t is the ratio of the inductance in henrys to the 
resistance in ohms of the machine or circuit. This is merely a guide to 
how fast the d-c component decays. 
Stated in other terms, it is the time in seconds for the d-c component to 
reach zero if it continued to decay a t the same rate it does initially 
(Fig. 1.21). 
RMS VALUE INCLUDING D-C COMPONENT 
The rms values of a-c waves are significant since circuit breakers, fuses, 
and motor starters are rated in terms of rrns current or equivalent kva. 
The maximum rrns value of short-circuit current occurs at a time of about 
one cycle after short circuit, as shown in Fig. 1.20. If there were no 
decay in the d-c component, as in Fig. 1.18, the rrns value of the first 
cycle of current would be j .732 times the rrns value of the a-c component. 
I n practical circuits there is always some d-c decay during the first cycle. 
An approximate rrns value of one cycle of an offset wave whether it is 
partially or totally offset is expressed by the equation 
where C = rrns value of offset or asymmetrical current wave over one 
cycle 
a = rrns value of a-c componentb = value of d-c component at one-half cycle 
MULTIPLYING FACTOR 
Calculation of the precise rrns value of an asymmetrical current a t any 
time after the inception of a short circuit may be very involved. Accu- 
rate decrement factors to account for the d-c component a t any time are 
required, as well as accurate factors for the rate of change of the apparent 
reactance of the generators. This precise method may he used if desired, 
but simplified methods have been evolved whereby the d-c component is 
accounted for by simple multiplying factors. The multiplying factor 
converts the rrns value of the symmetrical a-c wave into rms amperes of 
the asymmetrical wave including a d-c component. 
The magnitude of the d-c component depends upon the point on the 
voltage wave a t which the short circuit occurs. For protective-device 
application, only the maximum d-c component is considered, since the 
circuit breaker must be applied to handle the maximum short-circuit 
current that can occur in a system. 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 21 
In the general case for circuits rated above 600 volts, the multiplying 
factor to account for d-c component is 1.6 times the rms value of the a-c 
symmetrical component at the first half cycle. 
For circuits rated 5000 volts or less where there is no local generation, 
that is, where the supply t,o the bus is through transformers or long lines, 
the multiplying factor to ralculate the total current at the first half cycle 
may be reduced to 1.5. For circuits 600 volts and less, t,he multiplying 
factor to calculate the total current at the first half cycle is 1.25 when the 
circuit breaker is applied on the average current in three phases. Where 
single-phase conditions must be considered in circuits GOO volts and less, 
then to account for the d-c component in one phase of a three-phase cir- 
cuit a multiplying factor to calculate the total current at the first half 
cycle of 1.5 is used. 
For some calculations, rms current evaluations a t longer time intervals 
than the first half cycle, such as three to eight cycles corresponding to the 
interrupting time of circuit breakers, are required. Multiplying factors 
for this purpose may be taken from the curve in Fig. 1.22. 
Table 1.2 gives the multiplying factors commonly used for applying 
e 
FIG. 1.22 
various X / R ratio of circuits. 
Charts showing multiplying factors to account for decoy of d-c component for 
22 SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES 
short-circuit protective devices. These factors range from 1 to 1.6, 
depending upon whether the short-circuit calculation is being made to 
determine the interrupting or momentary duty on the short-circuit pro- 
tective device. 
SHORT-CIRCUIT RATIO OF GENERATORS 
This term is referred to frequently in short-circuit discussions. With 
present AIEE procedures of short-rircuit ralrulations, i t has become an 
accessory with no practical significance from this standpoint. For the 
sake of completeness, a definition is given here. 
Short-circuit ratio 
field current to produce rated voltage a t no load 
field current to produce rated current at sustained short circuit 
-~ - 
No further mention will he made of short-circuit ratio. 
TOTAL SHORT-CIRCUIT CURRENT 
The total symmetrical short-rirruit current is made up of currents from 
several sourves, Fig. 1.23. At the top of the figure is shown the short- 
circuit current from the utility. This act,ually comes from ut,ility gener- 
ators, but generally the industrial system is small and remote electrically 
from the utility generators so that the Symmetrical short-rircuit current 
is substant,ially constant,. If there are generators in the indust,rial plant, 
then they cont,ribute a symmet,rical short-circuit rurreiit which for all 
practical purposes is constant over the first few cycles. There is, how- 
ever, a slight decrement, as indicated in Fig. 1.23. 
The other sources are synchronous motors which act something like 
plant generators, except that t,hey have a higher rate of decay of the sym- 
metriral component, and induction motors whirh have a very rapid rate 
of dccay of the symmetrical component of current. When all these cur- 
rents are added, the total symmetrical short-circuit rurrent is typical of 
that shown a t the bottom of Fig. 1.23. 
The magnitude of the first few cycles of the t,otal symmetrical short- 
circuit, current is further increased by the presence of a d-c compouent, 
Fig. 1.24. The d-c component, offsets the a-c ware and, therefore, makes 
it asymmetrical. The d-c component decays to zero within a few cycles 
in most indust,rial power systems. 
It is this total rms asymmetrical short-circuit current, as shown in Fig. 
1.24, that must he determilied for short-circuit protective-derice applira- 
tion. The problem of doing this has been simplified by standardized 
procedures to a poiut xhere t o determine the rms asymmetriral current 
one need only divide t,he line-to-neutral roltage by the proper reactance 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 23 
RG. 1.23 Tracer of orcillogramr of rym- FIG. 1.24 Arymmelrical short-circuit current 
metrical short-circuit currents from utility, from dl sources illustrated in Fig. 1.23 plus 
panerator, synchronous motors, and induc- d-c component. 
lion motors. The shape of the total com- 
bined currents is illurtmted by the bottom 
hace. 
24 SHORT.CIRCUIT-CURRENT U L C U U l l N G PROCEDURES 
or impedance and then multiply by the proper multiplying factor from 
Table 1.2. 
BASIS OF RATING A-C SHORT-CIRCUIT PROTECTIVE DEVICES 
The background of the circuit-breaker rating structure as well as the 
basic characteristics of short-circuit currents must be understood to 
enable the engineer to select the proper rotating-machine reactances and 
multiplying factors for the d-c component to determine the sbort-circuit- 
current magnitude for checking the duty on a particular circuit breaker, 
such as momentary duty or interrupting duty. 
The rating structure of circuit breakers, fuses, and motor starters is 
designed to tell the application engineer how circuit breakers, fuses, or 
motor starters will perform under conditions where the short-circuit cur- 
rent varies with time. In discussing these rating bases, and for the sake 
of clarity, they will be arbitrarily divided into two sections, i.e., the rating 
basis of high-voltage short-circuit protective devices above 600 volts and 
the rating basis of low-voltage Short-circuit protective devices 600 volts 
and below. 
HIGH-VOLTAGE SHORT-CIRCUIT PROTECTIVE DEVICES (ABOVE 600 VOLTS) 
Power-circuit-breaker Rating Basis. The standard indoor oilless power 
circuit breakers as used in metal-clad switchgear will be used here t o 
explain power circuit-breaker ratings. The same fundamental principles 
apply to all other high-voltage power circuit breakers. 
The circuit-breaker rating structure is complicated because of the time 
of operation of the circuit breakers after a short circuit occurs. 
The few cycles needed for the power circuit breaker to open the circuit 
and stop the flow of short-circuit current consist of the time required for 
(1) the protective relays to close their contacts, (2) the circuit-breaker 
trip coil to move its plunger to release the breaker operating mechanism, 
(3) the circuit-breaker contacts to part, and (4) the circuit breaker to 
interrupt the short-circuit current in its arc chamber. During this time, 
the short-circuit current produces high mechanical stresses in the circuit 
breaker and in other parts of the circuit. These stresses are produced 
almost instantaneously in phase with the current and vary as the square 
of the current. Therefore, they are greatest when maximum current is 
flowing. The foregoing discussion showed that t,he short-circuitcurrent 
is maximum during the first cycle or loop, because of the presence of the 
d-c component and because the motors contribute the most short-circuit 
current a t that time. Thus, the short-circuit stresses on the circuit 
breakers and other parts of the circuit are maximum during the first loop 
of short-circuit current. 
During the time from the inception of the short circuit until the circuit- 
breaker contacts part, the current decreases in magnitude because of the 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 25 
decay of the d-c component and the change in motor reactance, as 
explained previously. Consequently, the current that the circuit breaker 
must interrupt, four or five cycles after the inception of t.he short circuit, 
is generally of less magnitude than the maximum value of the first loop. 
The fact that the current changes in magnitude with time has led to the 
establishment of two bases of short-circuit-current ratings on power cir- 
cuit breakers: (1) the momentary rating or its ability to withstand 
mechanical stresses due to high short-circuit current and (2) the inter- 
rupting rating or its ability t,o interrupt the flow of short-circuit current 
within its interrupting element. 
What Comprises the Circuit-breaker-rating Structure. Circuit- 
breaker-rating structures are revised and changed from time to time. It 
is suggested that where specific problems require the latest information on 
circuit-breaker ratings the applicahlc American Standards Association 
(ASA), National Electrical Manufacturers Association (XEMA), or 
American Instituteof Elect,rical Engineers (AIEE) standards he referred to. 
To illustrate the various factors that comprise the circuit-breaker- 
rating structure, an oilless power circuit breaker for metal-clad switchgear 
rated 4.16 kv 250 mva* has been chosen. The complete rating is shown 
on line 5, Table 1.1. The following will explain the meaning of the several 
columns of Table 1.1, starting at the left. The rircuit-breaker-type 
designation, column 1, varies among manufacturers. For the sake of com- 
pleteness the General Electric Company nomenclature is used in this col- 
umn. The remainder of the items are uniform throughout the industry. 
1. Type of Circuit Breaker (AM-4.16-250) 
AM = magne-blast circuit breaker 
4.16 = for 4.16-kv class of circuits (not applicable to 4800- and 4800- 
volt circuits) 
250 = interrupting rating in mva a t 4.16 kv 
2-4. Voltage Rating 
2. Rated kv (4.16): the nominal voltage class or classes in which the 
circuit breaker is rated. 
3. Maximum design kv (4.76): the maximum voltage a t which the cir- 
cuit breaker is designed to operate. The 4.16-kv circuit breakers, 
for example, are suitable for a 1330-volt system plus 10 per cent for 
voltage regulation or 4.76 kv. 
(Note: 4330 is 4% X 2500.) Some utility syst.ems operate a t 1330 
volts near the substation. 
4. Minimum operating kv a t rated mva (3.85) : the minimum voltage a t 
which the circuit breaker will interrupt its rated mva or in this case 
it is 3.85 kv. At any voltages below this value, the circuit breaker 
* blegavalt-amperes (see Appendix). 
t 
i. 
16 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
I ! 
I ( 
t a 
/ 
I 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 27 
is not designed to interrupt the rated mva but will interrupt some 
value less than rated mva. 
This is very significant in the rating of power circuit breakers for, 
as poiuted out later, the circuit hreaker will interrupt a maximum of 
only so many amperes regardless of voltage. At any voltage less 
than the minimum operating voltage the product of the maximum 
kiloampere interrupting rating times the kv times the square root of 
3 is less than the mva interrupting rating of the circuit breaker. 
5-6. Insulation Level (Withstand Test) 
5 . Low-frequency rrns kv (19): the 60-cycle high-potential test. 
6. Impulse crest kv (60) : a measure of its ability to withstand lightning 
This is applied with an impulse generator as a and other surges. 
design test. 
7-9. Current Ratings in Amperes 
7. Continuous 60 cycles (1200 or 2000): the amount of load current 
which the circuit breaker will carry continuously without exceeding 
the allowable temperature rise. 
8-9. Short-time Rating 
8. Momentary amperes (60,000) : the maximum rms asymmetrical cur- 
rent that a circuit breaker will withstand including short-circuit cnr- 
rents from all sources and motors (induction and synchronous) and 
the d-c component. This rating is independent of operating voltage 
for a given circuit breaker. 
This is just as significant a limitation as mva interrupting rating. 
It defines the ability of the circuit breaker to withstand the mechani- 
cal stresses produced by the very large offset first cycle of the short- 
circuit current. This rating is nnusually significant because the 
mechanical stresses in the circuit hreaker vary as the square of the 
current. It is the only rating that is affected by the square law, and 
therefore is one of the most critical in the application of the circuit 
breakers. The rating schedules of power circuit breakers are so pro- 
portioned that the momentary rating is about 1.6 times the maximum 
interrupting rating amperes. 
9. Four-second (37,500): the maximum current that the circuit breaker 
will withstand in the closed position for a period of 4 sec to allow for 
relaying operating time. This value is the same as the maximum 
interrupting rating amperes. 
10-13. Interrupting Ratings 
10. Three-phase rated mva (250): the three-phase mva which the circuit 
breaker will interrupt over a range of voltages from the maximum 
design kv down to the minimum operating kv. In this case the 
28 SHORT-CIRCUIT-CURREM CALCULATING PROCEDURES 
interrupting rating is 250 rnva between 4.76 and 3.85 kv. The mva 
to be interrupted is obtained by multiplying the kv a t which the cir- 
cuit breaker operates times the symmetrical current in kiloamperes to 
be interrupted times the square root of 3. The product of these must 
not exceed the rnva interrupting rating a t any operating voltage. 
11. Amperes a t rated voltage (35,000): the maximum total rms amperes 
which the circuit breaker will interrupt a t rated voltage, i.e., in the 
case of the example used above 35,000 at 4.16 kv (4.16 X 35.000 x 
fi = 250 mva). These figures are rounded. This figure is given 
for information only and does not have a limiting significance of 
particular interest to the application engineer. 
12. Maximum amperes interrupting rating (37,500) : the maximum total 
rms amperes that the circuit breaker will interrupt regardless of how 
low the voltage is. At 
minimum operating voltage, 3.85 kv, this corresponds to 250 mva, 
and, for example, a t a voltage of 2.3 kv this corresponds to 150mva. 
The circuit breaker will not interrupt this much current a t all volt- 
ages, i.e., i t will not interrupt this much current if the product of 
current, voltage, and the square root of 3 is greater than the mva 
interrupting rating. This current limit determines the minimum kv 
a t which the circuit breaker will interrupt rated mva (column 4). At 
any voltage lower than that given in column 4, this maximum rms 
total interrupting current determines how much the circuit breaker 
will interrupt in mva. Therefore, when the voltage goes below the 
limit of column 4, the mva which the circuit breaker will interrupt is 
lower than the rnva rating given in column 10 by an amount propor- 
tional to the reduction in operating voltage below the value of column 4. 
13. Rated interrupting time (8 cycles on 60-cycle basis): the maximum 
total time of operation from the instant the trip coil is energized until 
the circuit breaker has cleared the short circuit. 
What limits the Application of Power CircuitBreakers an on inter- 
rupting-and Momentary-duty Basis? In so far as applying power cir- 
cuit breakers on an interrupting-duty basis is concerned i t can be seen 
from the foregoing that there are four limits, none of which should be 
exceeded. These must all be checked for any application. 
1. Operating voltage should never at any time exceed the limit of 
column 3, Table 1.1, i.e., the maximum design kv. 
2. Interrupting rnva should never be exceeded a t any voltage. This 
limit is sig’nificant only when the operating voltage is between the limits 
of columns 3 and 4, Table 1.1. It is not significant when the operating 
voltage is below the limit of column 4, Table 1.1, because maximum inter- 
rupting amperes limit the mva to values less than the rnva rating. 
3. Maximum interrupting rating amperes should never be exceeded 
In this example, this current is 37,500 amp. 
SHORT-CIRCUIT.CURRENT CALCUUTING PROCEDURES 29 
even though the product of this current times the voltages times the 
square root of 3 is less than the interrupting rating in mva. This figure 
is the controlling one in so far as interrupting duty is involved when the 
voltage is below that of column 4, Table 1.1 (minimum operating voltage 
a t rated mva). 
4. Momentary current should never be exceeded a t any operating 
voltage. Modern power circuit breakers generally have a momeutary 
rating in rms amperes of 1.6 times the maximum interrupting rating in 
rms amperes. As a result, where there is no short-circuit-current contri- 
bution from motors, a check of the interrupting duty only is necessary. 
If this is within the circuit-breaker interrupting rating then the maximum 
Short-circuit current, including the d-c component, mill be within the 
momentary rating of the circuit breaker. 
Where there is short-circuit contribution from motors, the momentary 
rating of the circuit breaker may be exceeded, before the interrupting 
rating is exceeded in a given cirruit. Whenever there are motors to be 
considered in the short-circuit calculations, the momentary duty and the 
interrupting duty should both be checked. 
Siuce the 
short-circuit current is maximum at the first half cycle, the short-circuit 
current must be determined a t the first half cycle to determine the maxi- 
mum momentary duty on a circuit breaker. 
To determine the short-circuit current a t the first half cycle, it is neces- 
sary to consider all sources of short-circuit current, that is, the generators, 
synchronous motors, induction motors, and utility connections. The 
subtransient reactances of generators, synchronous motors, and inductiou 
motors are employed in the reactance diagram. Since the d-r component 
is present a t this time, it is necessary to account for it by the use of a 
multiplying factor. This multiplying factor is either 1.5 or l.G, as out- 
lined in Table 1.2. Typical circuits where the 1.5 multiplying factor can 
be used are shown in Fig. 1.25. The procedure is the same, regardless of 
the type of power circuit breaker involved. 
To check 
the interrupting duty on a power circuit breaker, the short-circuit current 
should be determined a t the time that the circuit-breaker contacts part. 
The time required for the circuit-breaker contacts to part will vary over a 
considerable range, because of variation in relay time and in circuit- 
breaker operating speed. The fewer cycles required for the circuit- 
breaker contacts to part, the greater will be the curreut to interrupt. 
Therefore, the maximum interrupting duty is imposed upon the circuit 
breaker when the tripping relays operate instantaneously. In all short- 
circuit calculations, for the purpose of determining interrupting duties, 
the relays are assumed to operate instantaneously. To account for 
How to Check Momentary Duty on Power Circuit Breakers. 
How to Check Interrupting Duty on Power Circuit Breakers. 
SEPES-DIVEN 
SEN-RIO-EIELI', tCA 1 
30 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
2400 
4160 
4800 VOLT 
INCOMING 
LINE FROM 
U T I L I T Y 
TO PLANT LOAD 
( 0 ) NO GENERATION 
IN THE PLANT 
u.-L NO GENERATION 
ON THIS BUS 
2400, 4160 OR 
( C ) TO LOAD 
13.6 KV 
HIGH VOLTAGE 
INCOMING LINE 
$ :,oA,,60 4600 V BUS 
TO PLANT LOAD 
NO GENERATION (b) 
IN THE PLANT 
U U 
USE 1.6 
MULTIPLYING 
FACTOR 
NO GENERATION 
FIG. 1.25 
circuits rated less than 5 h. 
One-line diogrom of carer where the multiplying factor 1.5 may be used on 
. . , c .. .: . 
, , ,.. . . 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 31 
variation in the circuit-breaker operating speed, power circuit breakers 
have been grouped into classes, such as eight-cycle, five-cycle, three-cycle 
circuit breakers, etc. It is assumed for short-circuit-calculation purposes 
that circuit breakers of all manufacturers, in any one speed grouping, 
operate substantially the same with regard to contact parting time. 
Instead of specifying a time a t which the short-circuit current is to he 
calculated, it is determined by the simpler approach of specifying the 
generator and motor reactances and using multiplying factors. These 
factors are listed in Table 1.2. 
In industrial plants, eight-cycle circuit breakers are generally used. 
Normally, the induction-motor contribution has disappeared, and that of 
the synchronous motors has changed from the subtransient to the transient 
condition before the contacts of these circuit breakers part. Therefore, 
in calculating the interrupting duty on commonly used power circuit 
breakers, generator subtransient reactance and synchronous-motor 
transient reactance are used and induction motors are neglected. The 
elapsed time is so long that usually all the d-c component has disappeared. 
What d-c component is left is more than offset by the reduction in a-c 
component due to the increase in reactance of the generators. Hence, a 
multiplying factor of one (1) is used. 
In very large power systems, when symmetrical short-circuit interrupt- 
ing duty is 500 mva or greater, there is an exception to this rule. In such 
large power systems, the ratio of reactance to resistance is usually so high 
that there may be considerable d-c component left when the contacts of 
the standard eight-cycle circuit breaker part. To account for this, the 
multiplying factor of 1.1 is used in determining the total rms short-circuit 
mva that a circuit breaker may have to interrupt in these large systems. 
The multiplying factor of 1.1 is not applied until the symmetrical short- 
circuit value reaches 500 mva. 
High-voltage fuses are either of the current- 
limiting type, Fig. 1.26, which open the circuit before the first current 
peak, or of the non-current-limiting type, which open the circuit within 
one or two cycles after the inception of the short circuit. For the sake of 
standardization, all fuse-interrupting ratings are on the basis of maximum 
rms current that will flow in the first cycle after the short circuit occurs. 
This is the current that will flow if the fuse did not open the circuit 
previously, i.e., fuses are rated in terms of “available short-circuit 
current.” 
To determine the available short-circuit current a t the first cycle for 
the application of high-voltage fuses, use the subtransient reactances of 
all generators, induction motors, synchronous motors, and utility sources 
and allow for the maximum d-c component. The multiplying factor for 
allowing for d-c component is 1.6, the same as for allowing for d-c compo- 
High-voltage Fuses. 
TABLE 1.2 Condensed Table of Multiplying Factors and Rotating-machine Reactances 
To Be Used for CaLdatina Swt-dreuit Cunanh for Circuit-breaker, Fuse, and Motor.rtartor Applicdons 
Eight cycle or slower (general case). .......... Above 600 wlh 
Rva cycle.. .............................. Above 600 volt,Any ploee where symmetricmi I .O short-circuit kva i s loss than 1.1 
500 mva 
........................... 1.6 
Lar than 5 k.. .......................... 601 to 5000 volh Remote from generating do- 1.5 Generol GOSO.. Above 600 volt) Near generoting station 
lion (X/R rotio l e u thon I01 
High-voltaqe Fuses 
All typos, including dl wrront-limiting fuses. .... Above 600 wih Anywhere in system I .O 
w u 
Interrupting duty 2 
Subtransient 
Subtransient 
ii s s z Momentary duty 
Subtransient 
Subtransient 
5 
Three-phose Ino interrupting duly 
Subhqndent 1 Transient 1 Neglect 
Maximum rms ampere interrupting duty 
I I 
1 Generators. 1 I 
0 
a 
w 
C 
frequency 1 changers I 
All types, including dl current-limiting fuses.. ... Above 600 volt' 
Non-current-limiting lypes only.. ............. 601 to 15,000 wlh Remote from generoting %to. I Anywhere in system 1 tion ( X / R mtio leu lhm 41 
1.6 Subtronsient Svbtronrient Svbwmrient 
1 .? Subwoniiont Subhmrient Subtransient i i i 
All h e p o w e r ratings.. .................... Anywhere in system 2400 and 4i60Y 
Wlh 
All horsepower rotingr.. .................... 2400 and 4160Y Anywhere in system 
Yolh 
1.0 
I .6 
CIrmit breaker w conladm lype. . ........... 
Cirwit b r w b r or contocto~ lype. ............ 
Clrcvit b r e e b r or contartor type.. ........... 
601 10 5000 volts 
601 to MOO volts 
601 lo 5000 volts 
bywhere in system 
lion lX/R ratio leis than 101 
temote from gener.ting 11.- 
m z Apparatus. 600 Volts and Below 
1.6 
1.5 
Air circuit breakers or breaker-contactor combino. 
lion motor stoners.. .................... 
Low-voltacp furas or fused combination motor 
Slarte" ............................... 
Subtransient 
Subtransient 
600 volts and below 
600 volt* and below 
0 
Subtrmdent Subtransient 
Subtrmdent Subtransient 8 
0 
R 
Interrupting or momentary duty 
Anywhere in system 
Anywhere in system 
I .25 Subtransient Subtianrient Svbtronrienl 
1 .25 Subtransient Subtransient Svbtraniient 
34 SHORT-CIRCUIT.CURRENT CALCULATING PROCEDURES 
nent when determining the momentary duty on a power circuit breaker 
(see Table 1.2). 
The interrupting rating of fuses in amperes is exactly parallel, in so far 
as short-circuit+urent calculations are concerned, to the momentary 
rating of power circuit breakers. 
The ampere interrupting rating of high-voltage fuses is the only rating 
that has any physical significance. For the sake of simplicity of applica- 
tion in systems with power circuit breakers, some fuses are given inter- 
rupting ratings in three-phase mva. The three-phase mva interrupting 
rating has no physical significance, because fuses are single-phase devices, 
each fuse functioning only on the current which passes through it. 
WAVE OF AVAILABLE 
THE FUSE ELEMENTS MELT 
BEFORE PEAK VALUE OF 
AVAILABLE SHORT CIRCUIT 
CURRENT IS REACHED 
1 
FIG. 1.26 
See Fig. 1.27 for method of determining available short-circuit current. 
Grophic sxplonotion of the current-limiting action of current-limiting fuses. 
SHORT-CIRCUIT-CURRENT CAKULATING PROCEDURES 35 
These three-phase mva ratings have been selected so they will line up 
with power-circuit-breaker ratings. For example, a high-voltage fuse 
rated 150 mva and a power circuit breaker rated 150 mva can he applied 
on the basis of the same short-circuit-current calculations. Of course, the 
application voltage must he factored in each case. 
High-voltage motor starters generally 
employ for short-circuit protection either current-limiting fuses or power 
circuit breakers. The short-circuit-current calculations for applying 
these motor starters are the same as those for high-voltage fuses and 
power circuit breakers, respectively. 
High-voltoge Motor Starters. 
LOW-VOLTAGE CIRCUIT PROTECTIVE EQUIPMENT (600 VOLTS AND BELOW) 
low-voltage Air Circuit Breokers. The present designs of low-voltage 
air circuit breakers differ from those of high-voltage power circuit break- 
ers because they are substantially instantaneous in operation a t currents 
near their interrupting rating. The contacts often begin to part during 
the first cycle of current. Therefore, low-voltage air circuit breakers are 
subject to interrupting the current a t the first cycle after short circuit and 
withstanding the mechanical forces of that rurrent. It is necessary to 
calculate the current a t only one time for the application of low-voltage air 
circuit breakers. The current determined should be that of the first halt 
cycle and should be determined on exactly the same hasis as for checking 
the momentary duty of high-voltage power circuit breakers, except for a 
change in the multiplying factor as discussed in the next paragraph. The 
suhtransient reactances of generators, induction motors, and synrhronous 
motors are used, and the d-c component is considered (see Table 1.2). 
The multiplying factor for the d-c component is not so high in low- 
voltage circuits as in some high-voltage circuits. This is due to the gener- 
ally lower level of reactance-to-resistance ( X I R ) ratio in low-voltage 
circLits, which causes the d-c component to decay faster than in some 
high-voltage circuits. 
In rating low-voltage air circuit breakers, the average d-c component of 
the three phases is used, which is somewhat lower than that for the maxi- 
mum phase. 
The generally lower ( X / R ) ratio and the use of an average d-c compo- 
nent for the three phases result in a considerably lower multiplying factor 
in low-voltage circuits. The multiplying factor has been standardized 
at 1.25 for the average for the three phases. This is equivalent to a 
multiplier of about 1.5 to account for the d-c component in the maximum 
phase. 
Application of High-voltage Oil Circuit Breokers to 600-volt Systems. 
In the 192Os, 5-kv oil circuit breakers were used extensively on 600-volt 
36 SHORT-CIRCUIT-CURRENT CALCULAnNG PROCEDURES 
systems. The procedure for determining short-circuit currents in sys- 
tems of 600 volts and below is slightly modified for checking duty on oil 
breakers of the 5-kv class as compared with low-voltage air circuit 
breakers. 
Both the momentary duty and interrupting duty must be checked for 
the oil-circuit-breaker application. To check the momentary duty, use 
the same procedure as for low-voltage air circuit breakers, i.e., generators, 
utility sources, induction motors, and synchronous motors (subtransient 
reactance). However, a multiplying factor of 1.5 is used instead of 1.25 
as for low-voltage air circuit breakers. Oil-circuit-breaker momentary 
ratings are based on the maximum current through any one pole, not on 
the average current in the three phases which is employed in the rating of 
low-voltage circuit breakers. 
To determine the interrupting duty, use the generator subtransient 
reactance and utility-source reactance plus the synchronous-motor 
transient reactance and a multiplying factor of 1.0. 
Low-voltage Fuses. Several low-voltage fuses with published a-c 
interrupting ratings are appearing on the market. There are no industry 
standards to follow, but most of these seem to be following air-circuit- 
breaker standards, i.e., using the same rating base and same method of 
determining short-circuit duty as is used for low-voltage air circuit 
breakers. Hence, the procedure will not be repeated here except to 
point out that the 1.25 multiplying factor is used (see Table 1.2). 
So-called National Electrical Code (NEC) plug and cartridge fuses 
have no established a-c interrupting ratings. Many tests have been made 
to determine their a-c interrupting ability, but to date the industry has 
not applied a-c interrupting ratings. 
Low-voltage motor starters are of two 
types: those using fuses andthose using air circuit breakers for short- 
circuit protection. Those using air circuit breakers for short-circuit 
protection are applied 04 exactly the same basis as low-voltage air circuit 
breakers in so far as short-circuit currents are concerned. 
Motor starters using fuses for short-circuit protection are applied on 
exactly the same basisas fuses in so far as short-circuit current is concerned. 
Low-voltage Motor Starters. 
AVAILABLE SHORT-CIRCUIT CURRENT 
In determining the short-circuit current, the impedance of the circuit 
protective device connected in the faulty feeder is neglected. The short- 
circuit current is determined by’ assuming that the protective device is 
shorted out by a bar of zero impedance (Fig. 1.27). The short-circuit 
/ 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 37 
current which flows in such a circuit is commonly called available short- 
circuit cumat. The procedure for determining the available short-circuit 
current is based on setting up impedance or reactance diagrams. The 
impedance of the short-circuit protective device that is nearest the short 
circuit (electrically) is omitted from the impedance diagram. 
Practically all protective devices are so rated and tested for short- 
circuit interrupting ability; hence this procedure may be followed in 
short-circuit calculations. This greatly simplifies the calculations and 
removes the effect of impedance variations between different types and 
makes of devices having the same interrupting rating. I t means that 
one set of short-circuit-current calculations for a given set of conditions 
is all that is needed for applying any type of protective device, regardless 
of the impedance of the devices themselves. 
GENERATOR 0 
TRANSFORMER 
MOTORS 
CABLE 
C A B L E 
SHORT 
CIRCUIT 
SHORT ClRCUlTED 81 
JUMPER OF ZERO 
IMPEDANCE 
FIG. 1.27 
circuit protective devices. 
Connections for determining available short-circuit current for testing rhort- 
38 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
HOW TO MAKE A SHORT-CIRCUIT STUDY 
FOR DETERMINING SHORT-CIRCUIT CURRENT 
FORMULAS FOR SHORT-CIRCUIT STUDY' 
1. Changing ohms to per cent ohms, etc.: 
(ohms reactance) (kva.base) 
(kvt)*(lO) 
(ohms reactance)(kva base) 
(1.1) Per cent (%) ohms reactance = 
Per-unit (90 ohms reactance = (kv)*( 1000) (1.2) 
[see Eq. (1.34)] 
(1.3) 
(1.4) 
(% reactance)(kv)2(10) 
kva base Ohms reactance = 
Per-unit ohms reactance = per cent ohms reactance 100 
2. Changing per cent or per-unit ohms reactance from one kva base to 
another: 
% ohms reactance on kva base 2 
kva base 2 
kva base 1 - - X (% ohms reactance on base 1) (1.5) 
9f reactance on kva base 2 
kva base 2 
kva base 1 
- - X (% ohms reactance on kva base 1) (1.36) 
3. Converting utility-system reactance to per cent or per-unit ohms 
a. If given in per cent ohms reactance on a kva base different than that 
b. If given in short-circuit kva, convert to per-unit ohms thus: 
reactance on kva base being used in study: 
used in the study, convert according to Eq. (1.5). 
(1.6) 
kva base used in reactance diagram 
short-circuit kva of utility system 9 i reactance = 
c. If given in short-circuit amperes (rms symmetrical), convert t o per- 
unit ohms thus: 
Yi reactance = 
d. If only the kva interrupting rating of the incoming line breaker is 
* See pp. 54 to 57 for more prr-unit formulas 
1 kv = line-to-line kilovolts. 
(1.7) 
kva base used in reactance diagram 
(short-circuit current) (d$) (kv rating of system) 
known, 
SHORTT-CIRCUIT.CURRENT CALCULATING PROCEDURES 39 
9f ohms reactance 
(1.8) - 
kva base used in reactance diagram 
kva interrupting rating of incoming line breaker 
- 
4. Determining kva base of motors: 
The exact kva base of a motor = EI 43 (1.9) 
where E = name-plate voltage rating 
When motor full-load currents are not known, use the following kva bases: 
Induction motors: 
0.8-power factor synchronous motor: 
1.0-power factor synchronous motors: 
I = name-plate full-load current rating 
kva base = horsepower rating 
kva base = 1.0 (horsepower rating) 
kva base = 0.80 (horsepower rating) 
(1.10) 
(1.11) 
(1.12) 
5. Changing voltage base when ohms are used: 
Ohms on basis of voltage 1 
- - ')* X (ohms on basis of voltage 2) (1.13) 
In Eqs. (1.1) to (1.4), ohms impedance or ohms resistance may be sub- 
The final product is then per-unit or per 
(voltage 2)2 
stituted for ohms reactance. 
cent ohms impedance or resistance, respectively. 
6 . Determining the symmetrical short-circuit kva: 
Symmetrical short-circuit kva = ~ (kva base) (1.14) 
% X * 
-~ - y? '& (kva base) (1.15) 
(1.16) 
(1.16a) 
(line-to-neutral voltage)2 
ohms reactance X 1000 
kv2 X lo00 
ohms reactance 
= 3 
- - . 
7. Determining the symmetrical short-circuit current: 
(1.17) (100) (kva base) 
(% X*)(v%(kvt) 
(% X*)(&)(kvt) 
Symmetrical short-circuit current = 
(1.18) 
(1.19) 
- kva base 
- - kvt X lo00 
- 
( d ) ( o h m s reactance) 
* X = reactance or impedanoe. 
t kv = line-&line kilovolts. 
L 0 TABLE 1.3 Factor ( K ) to Convert Ohms to Per Cent or Per-unit Ohms for Three-phase Circuits* 
216Y/125 
240 
480 
600 
2,400 
4.1 60 
4,800 
6.900 
7,200 
l1,OOO 
11.500 
12,000 
12,500 
13.200 
13,800 
23,000 
37.4M) 
46,000 
69,OCU 
loot 
P.r c*nt 
'14 
73 
43.4 
27.7 
1.73 
0.56 
0.435 
0.210 
0.193 
0.0825 
0.0755 
0.0695 
0.064 
0.0574 
0.0525 
0.0187 
0.00711 
0.00471 
0.0021 2 - 
Per-""it 
2.14 
1.73 
0.434 
0.277 
0.0173 
0.00576 
0.00435 
0.0021 
0.001 93 
0.000825 
0.000755 
0.000695 
0.00064 
0.000574 
0.000525 
0,0001 87 
0.000071 I 
0.0000471 
0.0000212 
Per cent 
321.5 
260.4 
65.21 
4.166 
2.604 
0.808 
0.651 
0.315 
0.289 
0.123 
0.113 
0.104 
0.096 
0.086 
0.0787 
0.0283 
0.0107 
0.00708 
0.0031 5 
- 
1 50 
Per-""it 
3.215 
2.604 
0.6521 
0.4166 
0.02604 
0.00808 
0.00651 
0.0031 5 
0.00289 
0.00123 
0.00113 
0.00104 
0.00096 
0.00086 
0.000787 
0.000283 
0.000106 
0.0000708 
0.000031 5 
Base kvo 
200 
Per cant 
128 
147 
86.8 
55.5 
3.47 
1.15 
0.868 
0.42 
0.386 
0.165 
0.151 
0.138 
0.127 
0.114 
0.105 
0.0378 
0.0142 
0.00945 
0.0042 
Por-un1t 
4.28 
3.47 
0.868 
0.555 
0.0347 
0.0115 
0.00868 
0.0042 
0.00386 
0.00165 
0.00151 
0.00138 
0.00127 
0.00114 
0.00105 
0.000378 
0.000142 
0.0000945 
0.000042 
300 
Per cent 
t43 
a1 
30.2 
83.3 
5.21 
1.72 
1.302 
0.63 
0.579 
0.247 
0.226 
0.208 
0.192 
0.172 
0.157 
0.0567 
0.0213 
0.0141 
0.0063 
kva base , kva base 
kv' X 10 
* For per-unit, K = For per cent, K = kv = line-to-line kilovolts 
kv' X 1wO 
- 
Per-""it 
6.43 
5.21 
1.302 
0.833 
0.0511 
0.0172 
0.01302 
0.0063 
0.00579 
0.00247 
0.00226 
0.00208 
0.001 92 
0,00172 
0.001 57 
0.000547 
0.00021 3 
0.0001 41 
0.000063 
500 
__ 
Per cent 
07 1 
868 
217 
I38 
8.68 
2.88 
2.17 
1.05 
0.965 
0.413 
0.377 
0.347 
0.32 
0.286 
0.262 
0.045 
0.0355 
0.0236 
0.0105 
Per-""it v) 
I 
- 2 
2 
2.17 K 
0.71 
8.68 
E 
1.38 2 
0.0868 
0.0288 
0.0105 n 
0.00965 
0.00413 
0.00377 5 
0.00347 f 
0 
0.0032 
0.00286 
0.00262 
0.00045 c 
0.000355 R 
0.000236 v, 
0.000105 
0.0217 B -I 
2 
6 
new base in kva, 
100 
t To determine multiplying factors far any other base use figures under 100-kvs base columns multiplied by 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
8. Determining the asymmetrical short-circuit current: 
INFINITE 
H BUSES-SHORT CIRCUIT CURRENT GOES THROUGH HERE 
41 
(1.20) 
Asymmetrical short-circuit current 
Asymmetrical short-circuit kva 
= (symmetrical current) (multiplying factor) 
= (symmetrical kva) (multiplying factor) 
DIAGRAMS 
One-line Diagram. The first step in making a short-circuit study is to 
prepare a one-line diagram showing all sources of short-circuit current, 
i.e., utility ties, generators, synchronous motors, induction motors, syn- 
chronous condensers, rotary converters, etc., and all significant circuit 
elements, such as transformers, cables, circuit breakers, etc. (Fig. 1.28). 
The second step is to 
make an impedance or reactance diagram showing all significant react- 
ances and resistances (Pig. 1.29). In the following pages this will be 
Make an Impedance or Reactance Diagram. 
C UTILITY SYSTEM I TRANS D GENERATOR GENERATOR 
CABLE E 
SHORT 
CIRCUIT 
LARGE CABLE J 
MOTOR 
FIG. 1.28 e diagram c 
480 VOLT 
MOTORS 
, typical large industrial power system. 
FIG. 1.29 Reactonce diagram of system shown in Fig. 1.28. 
42 SHORT-ClRCUIT.CURRENT CALCULAltNG PROCEDURES 
referred to as an impedance diagram, recognizing of course that only 
reactances will be used in many diagrams. The circuit element,s and 
machines considered in the impedance diagram depend upon many 
factors, i.e., circuit voltage, whether momentary or interrupting duty are 
to be checked, etc. 
The foregoing discussion and Table 1.2 explain when motors are to be 
considered and what motor reactances are to he used for checking 
the dut,y on a given circuit breaker or fuses of a given voltage class. 
There are other problems, i.e., (1) selecting the type and location of the 
short circuit in the system, (2) determining the specific reactance of a 
given circuit element or machine, and (3) deciding whether or not circuit 
resistance should be convidered. 
SELECTION OF TYPE AND LOCATION OF SHORT CIRCUIT 
Three-phase Short Circuits Generally Considered. I n most indus- 
trial systems, the maximum short-circuit current is obtained when a 
three-phase short circuit occurs. Short-rircuit-current magnitudes are 
generally less for line-to-neutral or line-to-line short circuits than for the 
three-phase short circuits. Thus, the simple three-phase short-circuit- 
current calculations will suffice for application of short-circuit protective 
devices in most industrial systems. 
In some very 
large systems where the high-voltage-system neutral is solidly grounded, 
maximum short-circuit current flows for a single phase-to-ground short 
rircuit. Such a system might be served from a large delta-Y trans- 
former bank or directly from the plant generators. 
Hence the only time that single-phase short-circuit-current calculations 
need be made is on large high-voltage systems (2400 volts and above) 
with solidly grounded generator neutrals or where main transformers 
that supply a plant from a utility are ronnected in delta on the high- 
voltage side (incoming line) and in Y with solidly grounded neutrals 
on the low-voltage (load) side. 
The calculations of unbalanced short-circuit currents in large power 
systems can best be done by symmetrical components, see Chap. 2. 
Normally, generator and large delta-Y transformer secondaries are 
grounded through a reactor or resistor to limit the short-circuit current 
for a single line-to-ground short circuit on the system to letis than the 
value of short-circuit current for a three-phase short circuit. 
Several tests have been 
made to evaluate the effect of arc drop at the point of short circuit in 
reducing the short-circuit-current magnitude. It was felt by some 
engineers that the current-limiting effect of the arc was pronounced. 
These tests showed, however, that for circuit voltages as low as 300 volts 
Unbalanced Short Circuits in Large Power Systems. 
Bolted Short Circuits Only Are Considered. 
SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 43 
there may be no substantial difference in the current that flows for a 
bolted short circuit and when there is an arc of several inches of length. 
These test,s also confirmed modern calculating procedure as an accurate 
method of estimating the short-circuit-current magnitude in systems 
of 600 volts and less. 
.4rcs cannot be counted on to limit the flow of short-circuit currents 
even in louvoltage circuits; so short-circuit-current calculations for 
all circuit voltages are made on the basis of zero impedance at the point 
of short circuit, or, in other words, a bolted short circuit. This materially 
simplifies calculation because all other circuit impedances are linear in 
magnitude, whereas arcs have a nonlinear impedance characteristic. 
At What Point in the System Should the Short Circuit Be Considered 
to Occur? The maximum short-circuit current will flow through a cir- 
cuit breaker, fuse, or motor starter when the short circuit occurs at the 
4160V. 
I I I 
$? MAX.SHORT CIRCUIT DUTY ON $- $EW:RS FOR SHORT CIRCUIT BREAKERS ON THIS BUS 
1 
T 
?; 
A&?? 
Y T T - 3 
& + * + r y r-x 
MAX. DUTY FOR 
THESE BREAKERS 
OCCURS FOR 
SHORT CIRCUIT 
HERE 
FIG. 1.30 Location of faults for maximum Short-circuit duty on circuit breakers. 
44 SHORT-CIRCUIT-CURRENT CALCULATING PROCEDURES 
terminals of the circuit breaker, etc. (Fig. 1.30). These devices, if 
properly applied, should be capable of opening the maximum short- 
circuit current that can flow through them. Therefore, only one short- 
circuit location (at the terminal of the device) need be considered for 
checking the duty on a given circuit breaker, fuse, or motor starter. 
DETERMINING REACTANCES AND RESISTANCES OF CIRCUITS AND MACHINES 
Typical reactances of circuit elements and machines are given at the 
end of this chapter. These 
tables may be used as a basis for assigning values to the various elements 
of the impedance diagram. The reactances and resistances are all line- 
to-neutral values for one phase of a three-phase circuit. Where the 
reactances of a specific motor, generator, or transformer are known, 
these values should he used in lieu of the typical reactances in this 
chapter. The following is a guide to general practice in selecting and 
representing reactances. 
Whether or not 
the reactance of a certain circuit element of a system is significant 
depends upon the voltage rating of the system where the short circuit 
occurs. In all cases, generator, motor, and transformer reactances 
are used. In systems rated above 600 volts, the reactances of short 
bus runs, current transformers, disconnecting switches, circuit breakers, 
and other circuit elements of only a few feet in length are so low that 
they may be neglected without significant error. 
In circuits rated 600 volts or less, the reactances of low-voltage current 
transformers, air circuit breakers, disconnecting switches, low-voltage bus 
runs, etc., may have a significant hearing on the magnitude of total short- 
circuit current. 
As a general guide, the reactance of the low-voltage secondary-switch- 
gear section in load-center unit substations with closely coupled trans- 
formers and secondary switchgear is not significant for all voltages of 
600 volts and below. However, where there are several transformers or 
generators paralleled on one bus, or connections several feet long between 
a single transformer and its switchgear, reactances of the bus connections 
will generally be significant and should be considered in short-circuit 
calculations. In systems of more than about 1000 kva on one bus a t 
208Y/120 or 240 volts, reactance of all circuit components such as short 
bus runs, current transformers, circuit breakers, etc., should be included 
in the short-circuit study. 
I n systems of more than about 3000 kva on one bus a t 480 volts or 
600 volts, reactances