Short Circuit Current for the Fire Investigator
by Bernard Béland
Béland, Bernard. Short Circuit Current for the Fire
Investigator.
Fire and Arson Investigator. Vol. 46, No. 4 (June 1996). p 1921.
The computation of the shortcircuit current is often required by the
electrical designer for the proper selection of protection devices for the
circuit and other considerations. Fire investigators sometimes require that
information. The purpose of this article is to give the fire investigator
some basic background in that field. While these computations are simple
in many cases, there are other instances which could be very complex. The
discussion will be limited to simple situations.
Theory
A shortcircuit happens when two conductors at different potentials come
into contact. The contact is such that the current takes a path that is
not intended for the current to flow. The intensity of the current can be
anywhere from a few times the normal current to hundreds of times that value.
In some instances of ground faults or arcing conditions, that fault current
could be only around the normal value.
Both too small or too high a shortcircuit current could cause a dangerous
situation. If the shortcircuit current is only about the normal value for
the circuit, then the fault current could flow indefinitely in an unintended
path, since the protection would not open the circuit. Too high a shortcircuit
current implies that the protection devices (fuses or circuit breakers)
may not be able to open the circuit under fault condition. A protective
device has not only a nominal value for the current it can carry, but also
a limited currentinterrupting capacity. For example, a 15A breaker is
intended to carry 15A on a permanent basis and opens the circuit under
overload or shortcircuit conditions. However, its maximum interrupting
capacity is often limited to 10,000A. This implies that if the shortcircuit
current is higher, the breaker would not perform its duty correctly. It
will fail to open the circuit and the current will continue to flow between
the open contacts for a short time to the complete destruction, arcing and,
possibly, an ensuing fire.
To compute the shortcircuit current, one just divides the source voltage
by the total impedance of the circuit. The problem is to evaluate that impedance.
This is simple in many instances. However, in complex circuits involving
both resistances and inductances, the problem is more complex. In threephase
circuits, the problem is difficult to analyze, except for the simple case
of a threephase shortcircuit, which seldom happens in practice. In this
article, we will restrict ourselves to simple cases. In residences, three
cases need to be considered:
1. Linetoline fault
2. Linetoneutral fault
3. Linetoground fault
The most common of these faults is the last one. The first one is very
rare. Only the case of a solid fault will be considered. In the case of
an arcing fault, the voltage drop in the arc limits the current to a smaller
value than that of a shortcircuit. The arcing current is usually two or
more times smaller than that of a shortcircuit. An arc is quite unpredictable
and no accurate guideline can be given.
Impedance Computation
The total impedance to be considered is that of the transformer, the
line from the transformer to the house, the service drop and any cable up
to the fault point. The impedance of the power company transmission line
is neglected. This is almost always justified, at least under residential
and other low power systems. In this analysis, to be accurate, one should
consider both the resistances and reactances. This will not be done so the
results should be considered as approximative. However, the results are
accurate enough in most cases.
Transformer
The impedance of a transformer is given by:
V2 (%Z)
Z= __________________ (1)
100,000 (KVA)
in which:
Z is the impedance in ohms referred to the secondary (lowvoltage)
side
V is the secondary (lowvoltage) side in volts (for a 120/240V
transformer, the whole winding capacity, which is 240, should be used)
%Z is the percentage impedance of the transformer given on the
name plate; it is usually between 2 and 4% for small distribution transformers.
If that information is not available, 2.5% would be reasonably accurate
in most instances. For large transformers (1,000 KVA), that impedance could
be higher.
KVA is the power of the transformer in kilovoltamperes. It is
usually printed in large letters on the transformer.
Equation (1) gives the equivalent impedance of the transformer referred
to on the secondary (lowvoltage) side. For a 120/240V transformer, 240
should be used for V. The computed impedance applies for a 240V fault or
a linetoline fault. In the case of a 120V fault (linetoground or linetoneutral
fault), the result of equation (1) should be multiplied by 3/8 or 0.375
to obtain the proper impedance.
For a 120/240V transformer, equation (1) becomes:
Z=0.576(%Z)/KVA
for a 240V fault
and
Z=0.216(%Z)/KVA
for a 120V fault. 
(2)
(3) 
Conductor
Then, the impedance of all the conductors to the point of the shortcircuit
should be evaluated. This is quite simple if all the resistances of the
conductors are known in ohm/foot. However, one must not forget to include
the two conductors to the fault and back. Occasionally, the two conductors
could be of different sizes. Table I gives the resistance of different sizes
of copper conductors.
Examples
Let us consider a 2.5%Z, 50KVA, transformer that feeds a house with
50 feet of No. 3 aluminum triplex cable. The service has a length of 20
feet of No. 0 copper conductors. A shortcircuit is produced in a No. 18
copper flexible cord, 5 feet from an outlet. That outlet is feed by 30 feet
of a No. 14 copper cable. Obviously, the fault is from linetoneutral,
at 120V. The computation is as follows:
for the transformer (equation (3)),
Z_{x}
= 0.216 x 2.5/50 = 0.0108;
for the No. 3, aluminum triplex,
Z_{3}= 
0.202 x 1.6 x 50 x 2
________________
1,000 
= 0.032; 
for the No. 0 copper conductors,
Z_{0}= 
0.101 x 20 x 2
________________
1,000 
= 0.004; 
for the No. 14 copper cable,
Z_{14}= 
2.54 x 30 x 2
________________
1,000 
= 0.15; 
and for the No. 18 flexible cord,
Z_{18}= 
6.4 x 5 x 2
________________
1,000 
= 0.064. 
The total resistance is:
Z_{T}
= 0.0108 + 0.032 + 0.004 + 0.15 + 0.064 = 0.2608.
The fault current is then:
I_{SC}= 
V
_____
Z 
120
_____
0.2608 
= 460  A 
In the above equations, the subscripts X and T stand for the transformer
and total impedances, respectively. The number indices stand for the size
of the conductors. I_{SC}
stands for the shortcircuit current.
Remarks
The above example is quite typical of actual situations. It will be noted
that the transformer and the service do not contribute much to the total
impedance. Much of the total impedance is due to the triplex and the cables
inside the house. A 120V fault at the transformer would have resulted 120/0.0108
= 13,000A. A shortcircuit at the main switch or the distribution box would
have produced a current of:
120/(0.0108 + 0.032 + 0.004) = 2564A
At the wall outlet, the fault current would have been:
120/ (0.0108 + 0.032 + 0.004 + 0.15) = 610A
Clearly the fault current decreases quickly as a function of the distance
from the power source.
Household circuit breakers usually have an interrupting capacity of 10,000A.
It is sometimes claimed that shortcircuit current of that level is not
exceptional. This is unfounded. Under industrial and commercial conditions,
this could be the case in some places close to large transformers. Under
household conditions, this is possible only if a house is close to a very
large transformer and fed by a short, large size cable. Even then, the fault
would have to be in the distribution box. Let us consider an infinite power
source that feeds a distribution box. The shortcircuit current will be
infinite. If three feet of No. 14 copper cable is added, the resistance
of which is 3 x 0.0025 x 2 = 0.015 ohm, then, the shortcircuit current
is reduced to 8,000A. That length of conductor (6 feet) is often used inside
the box. Obviously, the shortcircuit current decreases quickly with the
distance. With 10 feet of cable, the shortcircuit current is already reduced
to 2,400A or less. With that last current, the breaker will have to open
in about one cycle if damages to the insulation is to be avoided.
From the above considerations, it should be clear that, under household
conditions, the fault current is normally a few hundred amperes. In a recent
study by UL, on 15A circuits, out of 943 measurements at different outlets,
in numerous households, it was found that the shortcircuit current:^{
1}
1. Was less than 100A in 2% of the cases
2. Exceeded 1,000A in 1% of the cases
3. The highest current was 1,650A.
4. The lowest current was 77A
5. The current was between 100 and 500A in 87% of the cases.
6. The average shortcircuit current was 300A.
Clearly, most shortcircuit currents at outlets are at a few hundredampere
level. Shortcircuit current above 1,000A are rare and 2,000A is extremely
unlikely to happen in practice. In the distribution panel or main switch
of a house, the shortcircuit current is usually between a few hundred to
a few thousand amperes. A fault current of 10,000A is extremely unlikely.
It is possible only in extreme circumstances such as with a 200KVA transformer
with 20 feet of No. 0 aluminum triplex.
Obviously, under arcing conditions, the fault current will be even smaller
than that under a solid shortcircuit. In large buildings or under industrial
conditions, the fault current could be much higher in the 100,000A. Even
under these circumstances, these large currents are possible only close
to large power sources. Most of the lighting and small power distribution
systems, even under industrial conditions, have a fault current comparable
to that of many hours. For example, the school at which this author taught
has about 15,000 students. The highest shortcircuit current in the power
laboratory, at 120/240V, is just under 2,000A. Any test that requires
a higher power has to be conducted elsewhere.
Arching at 120V with 1,000A shortcircuit current produces a voltage
of about 70V with 500A for a power of about 30,000 watts. That is still
an impressive power that can do a lot of damage if it is sustained for some
time. However, in most circumstances, the power will be cut off quickly
because of the blown fuse. However a 200A fuse will take many seconds before
it opens. For comparison purposes, most arc welding is done at about 50V
with a current of around 100 to 200A. The destructive power of arc welding
(useful in this case) is due to the long duration of the arc. Arc welding
can melt a lot of steel; however, if the arc lasts for a second or less,
the amount of melting is very minimal and will not pierce a hole in steel
plates.
Conclusion
In this article, it was seen how to evaluate shortcircuit current in
simple situations that cover most residential cases and many others. It
was seen that, in most cases, the available shortcircuit current is usually
a few hundred amperes and is seldom above a few thousand amperes.
Reference
1. "FactFinding Report on an Evaluation of Branch Circuit, CircuitBreaker
Instantaneous Trip Levels," Underwriters Laboratory, File E87837, Project
92ME51901, October 1993.
Reprinted with permission.
