Tool · Short circuit

Fault current calculator

Find the three-phase short circuit current at a transformer secondary, with the upstream grid and an optional cable. Uses the IEC 60909 impedance method, and shows the peak current and every step.

Fault current I_k″

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Peak current i_p

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Fault level

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Step-by-step working

How to use this calculator

Enter the upstream fault level. Your utility can give it, usually in MVA or kA. If you do not know it, tick Infinite grid for a safe upper value.
Enter the transformer rating, secondary voltage, impedance and X/R from its nameplate or test report.
If the fault is further away, add the cable length and its resistance and reactance per km from the cable data sheet.

The result is the initial symmetrical fault current I_k″ and the peak current i_p. Use I_k″ to check breaker breaking capacity and relay settings. Use i_p to check the making capacity and the busbar strength.

The formulas

Each part of the circuit becomes an impedance at the secondary voltage U. They add up, and the fault current follows from Ohm’s law:

Z_Q = c × U² / S_kQ (grid)
Z_T = (u_k / 100) × U² / S_T (transformer)
Z_L = (R′ + jX′) × length (cable)
I_k″ = c × U / (√3 × |Z_Q + K_TZ_T + Z_L|)

The peak current depends on the R/X ratio of the whole circuit:

κ = 1.02 + 0.98 × e^−3R/X
i_p = κ × √2 × I_k″

The voltage factor c covers the fact that the real voltage before the fault can be higher than nominal. IEC 60909-0 gives these values:

Nominal voltage	c for maximum current	c for minimum current
Low voltage, up to 1 kV, +6% tolerance	1.05	0.95
Low voltage, up to 1 kV, +10% tolerance	1.10	0.95
Medium and high voltage, above 1 kV	1.10	1.00

For network transformers, IEC 60909-0 also corrects the transformer impedance with K_T = 0.95 × c_max / (1 + 0.6 × x_T), where x_T is the transformer reactance in per unit. It slightly lowers the impedance, so the current comes out a little higher.

Worked example

A 1000 kVA, 400 V transformer has 6% impedance and X/R of 6. The grid fault level is 250 MVA with X/R of 10. The fault is at the transformer terminals. c = 1.05.

Grid: Z_Q = 1.05 × 400² / 250,000,000 = 0.672 mΩ.
Transformer: Z_T = 0.06 × 400² / 1,000,000 = 9.600 mΩ. With X/R = 6, R_T = 1.578 mΩ and X_T = 9.469 mΩ.
Correction: x_T = 0.0592, so K_T = 0.95 × 1.05 / (1 + 0.6 × 0.0592) = 0.963.
Total: R = 1.587 mΩ, X = 9.790 mΩ, |Z| = 9.918 mΩ.
I_k″ = 1.05 × 400 / (√3 × 9.918 mΩ) = 24.45 kA.
κ = 1.02 + 0.98 × e^{−3 × 1.587 / 9.791} = 1.623, so i_p = 1.623 × √2 × 24.45 = 56.1 kA.

The quick rule I = S / (√3 × U × %Z) gives 24.06 kA. It ignores the grid and the c factor, so it can be off in either direction. This is the example loaded in the calculator above.

Typical transformer impedance

If you do not have the nameplate, IEC 60076-5 gives the minimum short-circuit impedance for distribution and power transformers:

Rating	Minimum impedance
Up to 630 kVA	4.0%
631 to 1250 kVA	5.0%
1251 to 2500 kVA	6.0%
2501 to 6300 kVA	7.0%
6301 to 25,000 kVA	8.0%

Real transformers are often a little above these values. The nameplate value is always better.

Common mistake

Using the maximum fault current to check relay sensitivity. The maximum current is right for breaker ratings. For relay pickup, check the minimum current too: use c = 0.95 (LV) or 1.00 (MV), the highest grid impedance and the longest cable.

Questions people ask

What is the difference between I_k″ and i_p?

I_k″ is the RMS value of the AC fault current just after the fault starts. i_p is the highest instant value in the first cycle. It is higher because of the DC offset, and it depends on the X/R ratio.

Why does X/R matter?

A high X/R means the DC part of the fault current dies away slowly. The first peak is then bigger. X/R also affects the asymmetric current that a breaker must interrupt.

What about a line-to-ground fault?

A line-to-ground fault needs the zero-sequence impedance, which depends on the transformer winding and how the neutral is earthed. This tool covers three-phase and line-to-line faults. A line-to-line fault is about 0.87 times the three-phase value.

Do motors add to the fault current?

Yes. Running motors feed current into a fault for a few cycles. IEC 60909 includes them. This tool does not, so add motor contribution separately for large motor loads.

How do I convert fault level in kA to MVA?

Fault level in MVA = √3 × voltage in kV × current in kA. For example, 25 kA at 11 kV is √3 × 11 × 25 = 476 MVA.

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Fault current calculator: short circuit current at a transformer (IEC 60909)