Transformer short circuit current calculator
Enter the transformer rating in kVA, the secondary voltage and the impedance from the nameplate in percent. Leave the supply fault level at 0 for an infinite source, or enter it in MVA for a less conservative figure. The result is the maximum current a bolted fault at the terminals would draw.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Enter the transformer rating in kVA and its secondary voltage (line-to-line for three phase).
- Enter the impedance Z% from the nameplate, often around 4 to 6 % for distribution transformers.
- Choose three phase or single phase.
- Leave the supply fault level at 0 for a worst-case infinite source, or enter it in MVA if the utility has given it.
- Full-load current: I_FL = S / (√3 × V) three phase, S / V single phase
- Transformer impedance in per unit: z = Z% / 100
- Supply impedance on the transformer base: z_s = S_T / S_sc (0 for an infinite source)
- Short circuit current: I_sc = I_FL / (z + z_s)
- Fault level: S_T / (z + z_s)
Cable impedance and motor contribution are not included.
Worked example
A 1000 kVA, 400 V, 5 % transformer on an infinite source: I_FL = 1,000,000 / (1.7321 × 400) = 1443.4 A. The short circuit current is 1443.4 / 0.05 = 28,868 A = 28.87 kA, 20 times full-load current, a fault level of 20 MVA.
With a 500 MVA supply, z_s = 1/500 = 0.002 and the total is 0.052: I_sc = 1443.4 / 0.052 = 27.76 kA (19.23 MVA).
Tips
- Breaking capacity: the breakers and fuses on the secondary side must have a rated breaking capacity above the prospective fault current at their location.
- Lower Z%, higher fault current. A low-impedance transformer regulates voltage better but raises the fault level.
- Check the voltage you enter. Line-to-neutral instead of line-to-line voltage in the three-phase formula overstates the current by √3.
Frequently asked questions
How do I calculate transformer short circuit current?
First the full-load current: I_FL = kVA × 1000 / (√3 × V) for three phase. Then I_sc = I_FL × 100 / Z%. A 1000 kVA, 400 V, 5 % transformer: I_FL = 1443.4 A, I_sc = 1443.4 / 0.05 = 28,868 A, about 28.9 kA.
What does the impedance percentage mean?
Z% is the voltage, as a percentage of rated voltage, that drives full-load current through the shorted secondary. At 5 %, full voltage drives 100/5 = 20 times full-load current.
Why include the supply fault level?
The infinite-source assumption ignores the impedance of the network upstream, so it slightly overstates the fault current. Adding the utility's fault level in MVA adds its impedance in series: 500 MVA ahead of a 1000 kVA, 5 % transformer lowers the fault from 28.87 to 27.76 kA.
Is this the value to choose breakers by?
It is the maximum at the transformer terminals and a starting point. Cable impedance lowers the current further along, and running motors can add to it during the first cycles; a full study per your local standard (for example IEC 60909) accounts for those.