Short-Circuit Calculations 

A Handbook to Accompany the 

Short-Circuit Calculation Program 

From MSHA’s Approval and Certification Center 

By Wayne L. Carey 

January 4, 2006
Short-Circuit Calculations

Introduction

One of the jobs of the Mine Safety and Health Administration’s (MSHA’s) 

Approval and Certification Center (A&CC) is to approve electrical equipment for 

underground mines. An important part of this job is to determine if circuit 

protective devices are sized and set correctly. Sizing of circuit protective devices 

can only be done when maximum available short-circuit current is known. 

Setting of the devices can only be done when minimum available short-circuit 

current is known. The A&CC has a computer program available through the 

MSHA website (www.msha.gov) to calculate available short-circuit currents in 

mine power systems. This program, other programs, or hand calculation can be 

used to perform short-circuit calculations. Any method used requires specific 

data on the mine power system. All this data is converted to resistance and 

reactance values for each component in the system. Impedance can then be 

calculated and then minimum and maximum available currents. This paper 

describes the data needed for each type of circuit component and how to obtain 

the data. It also describes methods used by the computer program to calculate 

available short-circuit currents. 

Utility

The first step in analyzing a power system is to get the data for the power 

available at the site, the utility data. This data can be obtained from the power 

company. When calling the power company, explain the type of information 

you need and ask for the engineering department. It may be helpful to explain 

why you need the information. 

The power company will be able to supply this information for the point in the 

power system where their responsibility for the power system ends and the 

customer’s responsibility starts. A common location for this point is the 

secondary of a pole or pad mounted transformer. If the customer is responsible 

for the transformer, the transition point would be the primary of the transformer. 

Sometimes a pole mounted disconnect will be the transition point. The power 

company will specify where in the system their responsibility ends. 

The data needed is the line to line voltage (VLL), short-circuit kVA (kVASC), and 

X/R. Obtaining the voltage is simple enough. Short-circuit kVA is the power 

available at a bolted three phase fault. Bolted means all three phases connected 

together with no added impedance. X/R is the ratio of reactance to resistance in 

the supply. Short-circuit kVA and X/R may need to be derived from other data. 

 2
Short-circuit current (ISC) is sometimes supplied by the power company rather 

than short-circuit kVA. This current is the current in one phase of a three phase 

bolted fault. The short-circuit kVA can be calculated from the short-circuit 

current using the following equation. 

1000

3 LLSC

SC

VI kVA∗∗ = (Eq. 1) 

If you’re not doing these calculations every day, it is sometimes hard to 

remember when to include the square root of three factor. Power in a three 

phase circuit is three times the single phase power or three times the current in 

one phase times the line to neutral voltage (VLN). The line to neutral voltage is 

the line to line voltage divided by the square root of three. 

1000

3

1000 3

3

1000

3 SC LN SC LL LLSC

SC

IVI V VI kVA∗∗

=∗

∗ = ∗∗ = (Eq. 2) 

Power factor (PF) is sometimes specified instead of X/R. This must be the shortcircuit power factor. Power factor is defined as the cosine of the angle between 

voltage and current. X/R is the tangent of this same angle. X/R can be found 

from power factor by taking the tangent of the inverse cosine of the power factor. 

RX ( PF 1 costan/ − = ) (Eq. 3) 

When neither X/R nor power factor are specified for the utility, it is usually safe 

to assume the impedance of the utility is all reactance and X/R is infinite. Unless 

there are many miles of transmission line, the impedance of the utility will be 

mainly reactance in the generator. This is all the data needed for the utility. 

From this data, the short-circuit calculation program can be used to calculate 

impedance (Z), resistance (R), and reactance (X). This data can also be calculated 

manually. 

Impedance is calculated from VLL and short-circuit kVA. 

1000

2

∗ =

SC

LL

kVA

V Z (Eq. 4) 

Resistance and reactance are then calculated from the impedance using X/R. 

Since: 22 += RXZ (Eq. 5) 

 3
( )2 /1 RX

Z R

+

= (Eq. 5a) And ∗= ( / RXRX ) (Eq. 5b) 

Resistance and reactance are calculated at the voltages for the points in the circuit 

where the short-circuit currents are calculated. For example, even though the 

utility voltage may be 69 kilovolts (kV), if the short-circuit currents are being 

calculated further down the circuit where the voltage is 2400 volts, the resistance 

and reactance will be calculated at 2400 volts. The computer program initially 

calculates resistance and reactance for the utility at the utility voltage. As the 

program works down through a circuit, encountering transformers, it converts 

the resistance and reactance to the new voltage by multiplying by the ratio of the 

voltages squared. 

2

1

2

2

12 V

V RR ∗= (Eq. 6) 

Where: R2 = resistance at secondary voltage 

 R1 = resistance at primary voltage 

 V2 = secondary voltage 

 V1 = primary voltage 

This formula is the result of conservation of energy. Energy into the transformer 

(V12/R1) equals energy out of the transformer (V22/R2). 

Transformers

Transformers are specified by output voltage (V), kVA rating, percent impedance 

(%Z), and X/R ratio. The X/R ratio is the ratio of reactance to resistance. This 

information, with the exception of X/R, is usually on the transformer nameplate. 

If X/R is not specified on the nameplate, the transformer manufacturer may be 

able to supply this. When contacting the transformer manufacturer, it may be 

helpful to have the transformer serial number. If X/R cannot be obtained, a 

value of 4.9 is typical and can be used in calculations. Impedance (Z) is 

calculated from V, kVA, and %Z. 

000,100

% 2

∗

∗ = kVA

VZ Z Or 

VA

VZ Z

2

*

100

%= (Eq. 7) 

Resistance and reactance are then calculated from Z and X/R as they were for the 

utility. Again, these resistances and reactances are for a short-circuit at the 

secondary of the transformer. If the short-circuit is at a point further down the 

circuit and after another transformer, the voltage at the short-circuit should be 

 4
used in equation 7. Alternatively, the calculated resistances and reactances can 

be converted to the new voltage by multiplying by the ratio of the voltages 

squared. 

Three Winding Transformers

The most common three winding transformers used in mining are located in 

underground power centers. They are usually specified like two separate 

transformers with no interwinding impedance. They can be modeled as two 

separate transformers. When three winding transformers have separate 

interwinding impedance specified, it is usually specified as reactance (X) and 

resistance (R) in ohms. Three sets of X and R must be supplied; primarysecondary (ps), primary-tertiary (pt), and secondary-tertiary (st). Xps and Rps are 

measured in the primary with the secondary short-circuited and the tertiary 

open. Xpt and Rpt are measured in the primary with the tertiary short-circuited 

and the secondary open. Xst and Rst are measured in the secondary with the 

tertiary short-circuited and the primary open. The three winding transformer is 

modeled as follows. 

The impedances are related by the following formulas. 

+= ZZZ spps (Eq. 8a) 

+= ZZZ tppt (Eq. 8b) 

+= ZZZ tsst (Eq. 8c) 

( ) Z p 5.0 −+∗= ZZZ stptps (Eq. 8d) 

( ) Zs = *5.0 −+ ZZZ ptstps (Eq. 8e) 

( ) t = *5.0 −+ ZZZZ psstpt (Eq. 8f) 

In the equations, all impedances must be referred to a common voltage. Note 

that Xst and Rst are measured at the secondary voltage. If all impedances are to 

be at the primary voltage, Xst and Rst must be transferred. 

Cables

 5
The minimum data needed for cables is size, length, voltage rating, and type. 

Size, voltage rating, and type of cables is embossed into the jacket. Size and type 

of aerial cable may be shown on a drawing or listed on an invoice. Lengths may 

be measured, determined from mine maps, or paced off and estimated. Some 

cables such as types SHD and MPF have different constructions for different 

voltage ratings and the voltage rating will need to be specified. It will be 

embossed in the cable jacket. It should be noted if more than one cable is used in 

parallel. The spacing of conductors of aerial cable will affect inductance. The 

computer program uses a default spacing of three feet. If the spacing is different 

than three feet, it should be noted. The computer program has stored values for 

resistance and reactance per thousand feet for most commonly used cable types. 

This data is also available in cable standards and manufacturer’s specifications. 

Resistance is specified at ambient temperature. 

Cable operating temperature has an effect on the resistance of a cable. Most 

cables have a rated operating temperature of 90 °C. Aerial cable is rated 75 °C. If 

different operating temperature ratings are specified, they should be noted. The 

computer program uses fully loaded cables at their rated operating temperature 

to calculate minimum available currents. The cables have higher resistances at 

their rated operating temperature ratings than at ambient temperature. The 

resistance at rated operating temperature can be calculated from the resistance at 

ambient temperature using the following formula. 

[ ( 12 12 1 α −+= TTRR )] (Eq. 9) 

Where: R2 = resistance at operating temperature 

 R1 = resistance at ambient temperature 

 T2 = rated operating temperature 

 T1 = ambient temperature 

α = temperature coefficient of resistivity corresponding to 

 temperature T1 (0.00393 for copper at 20 °C) 

When calculating maximum available current by hand, the resistance of the cable 

at ambient temperature should be used. However, the current that causes the 

cable to reach its highest temperature may not be the maximum available 

current. If the cable is initially at a temperature between ambient and its rated 

operating temperature and a short-circuit occurs, it is possible that the cable will 

reach a higher final temperature at a current lower than the maximum available 

current. The current depends on the cable’s resistance, but the resistance 

depends on the current. Therefore, the computer program calculates current 

that causes maximum temperature with an initial temperature of ambient and 

 6
with an initial temperature at one degree increments up to the cable’s rated 

operating temperature. This is one place where the power of the computer really 

comes in handy. The computer program also uses the subtransient reactance 

(discussed below) to calculate temperature after one cycle of short-circuit. This 

temperature is used as the initial temperature along with transient reactance in 

an additional calculation to calculate temperature at the end of the short-circuit 

(when the circuit protective device opens). The maximum available current 

calculated in the computer program is actually the current that causes maximum 

temperature. 

Reactance of most cables is published by the manufacturer. The reactance in 

ohms per 1000 feet of aerial cables with one foot spacing can be found with the 

following formula. 

⎟

⎠

⎞ ⎜

⎝

⎛ ∗=

GMR

X L

1 ln02298.0 (Eq. 10) 

GMR is the geometric mean radius in feet. It can be calculated by multiplying 

the wire O.D. in inches by .03245. 

eRGMR D ∗≅∗≅∗≅ d − 3894.0 03245.0 25.0 (Eq. 11) 

Where: R = wire radius in feet 

 D = wire diameter in feet 

 d = wire diameter in inches 

The reactance of aerial cable depends on the spacing between wires. Reactance 

at spacings other than one foot can be calculated with the following formula. 

( )

⎥

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎢

⎣

⎡

⎟

⎠

⎞ ⎜

⎝

⎛ +=

GMR

spacing new XX old 1 ln

ln 1 (Eq. 12) 

 7
Motors

Motors are a source of short-circuit current. They will act like generators when a 

short-circuit occurs. At a minimum, motors must be specified by horsepower 

which will be listed on the motor nameplate. Their reactance may also be 

specified. Motors have two values of reactance, subtransient and transient. 

Subtransient reactance is the reactance of the motor during the first cycle of the 

short-circuit. Transient reactance is the reactance of the motor during the 

remainder of the short-circuit. The computer program supplies typical values of 

subtransient and transient reactance for motors. Resistance is considered to be 

negligible. AC motors may be further identified as induction or synchronous. 

Synchronous motors can be 6-pole or 8- to 14-pole. The subtransient and 

transient reactance will vary depending on whether the motor is induction, 6-

pole synchronous, or 8- to 14-pole synchronous. It also varies between induction 

motors rated less than or equal to 600 volts and induction motors rated greater 

than 600 volts. Induction motors will not contribute to a short-circuit after the 

first cycle. Typical values of reactance are listed in Table 11. 

Type of Motor Subtransient Reactance 

(per unit) 

Transient Reactance 

(per unit) 

DC 0.15 0.30 

6-pole Synchronous 0.15 0.23 

8- to 14-pole Synchronous 0.20 0.30 

Induction ≤600 Volts 0.25 - 

Induction >600 Volts 0.17 - 

Table 1 

 Subtransient and transient reactances are usually given in per unit values. The 

actual value of reactance can be found with the following formula. 

HP

VX X PU

∗

∗ = 746

2

 (Eq. 13) 

Where: XPU = per unit reactance 

 V = voltage 

 HP = horsepower 

Subtransient reactance should be used to calculate maximum available shortcircuit current. Minimum available short-circuit current does not usually include 

contributions from motors, but when it does, transient reactance should be used 

in the calculation. A&CC does not include motor contribution in minimum 

available short-circuit current. When adding impedances working down 

 8
through a circuit, the impedance of a motor should be added in parallel with the 

total impedance up to the point in the circuit where the motor contributes. This 

reduces the impedance at that point. Adding impedances in parallel is most 

easily done by first converting resistance and reactance to conductance and 

susceptance, adding the conductances and susceptances, and converting the 

conductance and susceptance back to resistance and reactance. Conductance, 

susceptance, resistance, and reactance are related by the following formulas. 

22 R X

R G

+ = (Eq. 14a) 22 R X

X B

+ =− (Eq. 14b) 

22 G B

G R

+ = (Eq. 14c) 22 G B

B X

+

− = (Eq.14d) 

Where: R = resistance 

 X = reactance 

 G = conductance 

 B = susceptance 

Generators

Generators are treated just like motors. They are specified by kVA, subtransient 

reactance, and transient reactance. The kVA will be listed on the generator 

nameplate. Table 2 lists typical values of reactance for dc and four types of ac 

generators2. The four types are two-pole turbine, four-pole turbine, salient pole 

with dampers and salient pole without dampers. 

Type of Generator Subtransient 

Reactance (per unit) 

Transient Reactance 

(per unit) 

DC 0.15 0.30 

2-Pole Turbine 0.09 0.15 

4-Pole Turbine 0.14 0.23 

Salient Pole with Dampers 0.20 0.30 

Salient Pole without Dampers 0.30 0.30 

Table 2 

Reactance is calculated with the following formula. 

kVA

VX X PU

∗

∗ = 1000

2

 (Eq. 15) 

 9
Where: XPU = per unit reactance 

 V = voltage 

 kVA = kilovolt-ampere rating 

Capacitors

Capacitors are specified by kVAR on their nameplates. This is kVA reactive. If 

no tolerance is specified, 15% may be used. Capacitors will feed a short-circuit 

just like motors and generators. The following formula calculates reactance for a 

capacitor using a tolerance of 15%. 

1000

15.1 2

∗

∗ = kVARV X (Eq. 16) 

Where: V = voltage 

 kVAR = kilovolt-amperes reactive 

Rectifiers

Efficiency is the only rating needed for rectifiers. If no efficiency rating is 

specified, 99% can be assumed. Current through the rectifier is reduced by the 

factor Efficiency (%) / 100. 

Minimum Available Current

Minimum available current is calculated for a line to line arcing fault using the 

following formula. 

MAX

LL

MIN Z

V I

∗

∗ = 2

95.0 (Eq. 17) 

Where: VLL = line to line voltage 

 ZMAX = maximum impedance 

This formula is used for AC and the DC output from a three phase rectifier. The 

factor of 0.95 accounts for voltage fluctuations. The maximum impedance is 

calculated from the maximum resistances and reactances for all the elements in 

the circuit. For AC circuits, the current is further reduced by multiplying by an 

arcing fault factor, KA. This factor is listed in Table 3 for various voltages3. 

 10
Voltage (V) KA

V ≤ 480 0.85 

480 < V ≤ 600 0.90 

600 < V ≤ 1040 0.95 

1040 < V 1.0 

Table 3 

For DC circuits the arc voltage depends on the current. First, the current is 

calculated with Equation 17. If the calculated current is greater than or equal to 

600 amperes, the arc voltage is 60 volts. If the calculated current is less than 600 

amperes, the arc voltage is calculated with the following formula4. 

( )⎥

⎦

⎤ ⎢

⎣

⎡ −

= 303

1842 I

ARC eV (Eq. 18) 

Where: I = initially calculated current 

The voltage is then reduced by the arc voltage and the available current is 

recalculated. 

Maximum Available Current

Maximum available current for AC circuits is calculated for a three-phase bolted 

fault using the following equation. 

MIN

LL

ACMAX

Z

V I

∗ − = 3 (Eq. 19) 

Where: VLL = line to line voltage 

 ZMIN = minimum impedance 

Maximum available current for DC circuits is calculated for a line to line bolted 

fault using the following equation. 

MIN

LL

DCMAX Z

V I

∗ − = 2 (Eq. 20) 

Where: VLL = line to line voltage 

 ZMIN = minimum impedance 

 11
Example

A mine is supplied 95 MVA of power at 34.5 kV with an X/R ratio of 5.23. The 

power travels 1200 feet through number 2, ACSR aerial cable with three-foot 

spacing to a substation. The substation has a secondary voltage of 12.47 kV and 

is rated 10 MVA with 6.08% impedance. The power then travels 6000 feet 

through number 2/0, 15 kV, mine power feeder cable to an underground power 

center. The power center has a secondary voltage of 1040 volts and is rated 1350 

kVA with 5.0% impedance. Power then travels 850 feet through number 2/0, 2 

kV, shielded trailing cable to a continuous miner. If a short-circuit occurs on the 

continuous miner at the point where the trailing cable ends, what are the 

minimum and maximum available short-circuit currents? 

Figure 1 

 12
1. Starting at the utility, calculate the impedance using equation 4. 

Remember to use the voltage at the short-circuit, 1040 volts. 

Ω=

∗ = ∗ = 0114.01000000,95

1040

1000

2 2

kVA

V Z LL

From the impedance and X/R ratio, find X and R. 

( )

Ω=

+

=

+

= 0021.0

23.51

0114.

/1 2 2 RX

Z R

( ) RXRX 0112.023.50021.0/ Ω=∗=∗=

2. Now, find the reactance and minimum and maximum resistance for the 

aerial cable. The GMR must first be found with Equation 11. From 

manufacturer’s specifications, the O.D. of number 2, ACSR wire is 0.316 

inch. 

GMR 03245.0 d =∗=∗≅ 0103.0"316.003245.0 '

The reactance is then found with Equation 10. 

⎟ Ω=⎠

⎞ ⎜

⎝

⎛ ⎟ ∗=

⎠

⎞ ⎜

⎝

⎛ ∗= 105.0

0103.0

1 ln02298.0 1 ln02298.0

GMR

X L

Since this is the impedance for one-foot spacing, you must correct for 

three-foot spacing using Equation 12. 

( ) ( ) Ω=

⎥

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎢

⎣

⎡

⎟

⎠

⎞ ⎜

⎝

⎛ +∗=

⎥

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎢

⎣

⎡

⎟

⎠

⎞ ⎜

⎝

⎛ += 1304.0

0103.0

1 ln

3ln 1105.0

1 ln

ln 1

GMR

spacing new XX old

The reactance of 1000 feet of number 2, ACSR aerial cable is 0.1304Ω. The 

reactance for 1200 feet is 1.2 * 0.1304 = 0.1565Ω. This needs to be 

converted from the overhead line voltage to the voltage at the shortcircuit. 

0001.0 Ω=∗=

5.34

04.1 1565.0 2

2

X

 13
From the manufacturer’s specifications, the resistance of number 2, ACSR 

aerial cable is 1.753 Ω per mile at 75 °C. The resistance of 1200 feet is 

0.3984 Ω at 75 °C. Multiplying by the ratio of the voltages squared gives a 

maximum resistance of 0.0004 Ω. To get the minimum resistance, the 

resistance at 75°C must be converted to the resistance at 20 °C using 

Equation 9. 

( ) Ω=

+ = −+ = 0003.0

55*00393.01

0004.0

20751

75

20

α

R

R

3. The next component in the system is the substation. First, find the 

impedance using Equation 7. 

= Ω=

∗

∗ = 0066.0

000,100*000,10

1040*08.6

000,100

% 2 2

kVA

VZ Z

The X/R ratio is not specified for the substation. A value of 4.9 can be 

assumed. Using this value and the impedance, find X and R. 

( )

Ω=

+

=

+

= 0013.0

9.41

0066.

/1 2 2 RX

Z R

( ) RXRX 0065.09.40013.0/ Ω=∗=∗=

4. Next is the 6000 feet of number 2/0, 15 kV, mine power feeder cable. 

From manufacturer’s specifications, reactance of 1000 feet of this cable is 

0.038 Ω. The reactance of 6000 feet at 1040 volts is calculated as follows: 

= 0016.0 Ω=

470,12

1040 *038.*6 2

2

X

From manufacturer’s specifications, resistance of this cable at 20°C is 

0.0792 Ω per 1000 feet. Resistance of 6000 feet at 1040 volts would be: 

= 0033.0 Ω=

470,12

1040 *0792.0*6 2

2

RMIN

The maximum resistance will be at 90 °C, the rated operating temperature 

of mine power feeder cable. 

MAX RR MIN [ ] α( ) =−+= ( ) + 0042.070*00393.01*0033.020901* Ω=

 14
5. The power center is the next component in the circuit. First find the 

impedance. 

= Ω=

∗

∗ = 0401.0

000,100*1350

1040*5

000,100

% 2 2

kVA

VZ Z

The X/R ratio is not specified for the substation. A value of 4.9 can be 

assumed. Using this value and the impedance, find X and R. 

( )

Ω=

+

=

+

= 0080.0

9.41

0401.0

/1 2 2 RX

Z R

( ) RXRX 0393.09.40080.0/ Ω=∗=∗=

6. Last is the 850 feet of number 2/0, 2 kV, shielded cable. From 

manufacturer’s specifications, reactance of 1000 feet of this cable is 0.031 

Ω. The reactance of 850 feet is 0.85 * 0.031 = 0.0264 Ω. From 

manufacturer’s specifications, resistance of this cable at 20°C is 0.0839 Ω

per 1000 feet. The minimum resistance (RMIN) of 850 feet is 0.85 * 0.0839 = 

0.0713 Ω. The maximum resistance is the resistance at the 90 °C rated 

operating temperature of mine power feeder cable. 

MAX RR MIN [ ] α( ) =−+= ( ) + 0909.070*00393.01*0713.020901* Ω=

7. Now that all the resistances and reactances are calculated, we total them 

and calculate impedances and available currents. 

Component Minimum 

Resistance (Ω) 

Maximum 

Resistance (Ω) 

Reactance (Ω) 

Utility 0.0021 0.0021 0.0112 

Aerial Cable 0.0003 0.0004 0.0001 

Substation 0.0013 0.0013 0.0065 

Mine Power Feeder Cable 0.0033 0.0042 0.0016 

Power Center 0.0080 0.0080 0.0393 

Trailing Cable 0.0713 0.0909 0.0264 

Total 0.0863 0.1069 0.0851 

Table 4 

=+= 1212.00851.00863.0 Ω=+ 2 2 2 2

MIN MIN XRZ

 15
=+= 1366.00851.01069.0 Ω=+ 2 2 2 2

MAX MAX XRZ

3436

1366.0*2

1040*95.0*95.0

2

95.0 = = ∗∗ =

MAX

LLA

MIN Z

VK I Amperes 

4954

1212.0*732.1

1040

3 = = ∗ =

MIN

LL

MAX Z

V I Amperes 

Conclusion

Accurate calculations of available short-circuit currents can be made without a 

power systems specialization in electrical engineering. Symmetrical components 

and the per-unit system can be left for more advanced analyses. Hand 

calculations can still be tedious though. A computer program can make short 

work of the calculations. 

 16
 

References

1. Recommended Practice for Electric Power Distribution for Industrial 

Plants, Std. 141-1976, IEEE, 1976. 

2. Electrical Transmission and Distribution Reference Book, East Pittsburgh, 

PA, Westinghouse Electric Corporation, 1964. 

3. William S. Vilcheck, George Fesak, and William J. Helfrich, 

“Instantaneous Circuit Breaker Settings for the Short-Circuit Protection of 

Three-Phase 480-, 600-, and 1040-V Trailing Cables,” IEEE Transactions on 

Industry Applications Volume IE-17 No. 4, pp. 362-368, IEEE, 

July/August, 1981. 

4. George Fesak, William S. Vilcheck, William J. Helfrich, and David C. 

Deutsch, “Instantaneous Circuit Breaker Settings for the Short-Circuit 

Protection of Direct Current 300- and 600-V Trailing Cables,” IEEE 

Transactions on Industry Applications Volume IE-17 No. 4, pp. 369-375, 

IEEE, July/August, 1981. 

 17