A short-circuit current calculator for transformers
When a fault occurs at the terminals of a transformer, the only thing limiting the current is the transformer’s own impedance and the strength of the network behind it. This short-circuit current calculator gives the symmetrical fault current at the secondary from the rating in kVA, the secondary voltage and the percentage impedance on the nameplate, and shows the working.
You can leave the source as infinitely strong, which gives the highest value and is the usual first check, or enter the utility fault level in MVA to include the source impedance. The page also reports the fault level in MVA, the multiple of full-load current, and the peak (making) current for the X/R ratio you enter, and a table shows how the answer changes across common transformer sizes.
This is a quick estimate for choosing breaker and busbar ratings. It leaves out motor contribution and cable impedance, so use a full short-circuit study for protection coordination, and have the design checked by a qualified electrical engineer.
How to use the Short-Circuit Current Calculator
- Enter the transformer rating in kVA, the secondary line voltage and the %Z from the nameplate.
- Choose three-phase or single-phase.
- Optionally enter the utility fault level in MVA, and set the X/R ratio for the peak current.
- Read the fault current in kA, the full-load current and the multiple, and compare other sizes in the table.
Short-Circuit Current Calculator features
Infinite-bus or real source
Add the utility fault level to include the source impedance.
Peak current
IEC 60909 peak factor from your X/R ratio.
Working shown
Full-load current, total %Z, Isc and peak current step by step.
Size table
Fault currents for 100 to 2,500 kVA at the same voltage and %Z.
The fault current formula
The full-load current of a three-phase transformer is IFL = kVA × 1,000 ÷ (√3 × V). If the source is an infinite bus, the fault current is the full-load current divided by the per-unit impedance: Isc = IFL × 100 ÷ %Z = kVA × 100 ÷ (√3 × V × %Z). For a single-phase transformer drop the √3. This is the method in IEEE Std 141 and in manufacturers’ guides.
Example: a 1,000 kVA, 415 V transformer with 5% impedance has a full-load current of 1,391 A and a fault current of 1,391 × 100 ÷ 5 = 27.8 kA, twenty times full load. If the utility can supply only 250 MVA, its impedance on the transformer base is 1,000 ÷ 250,000 × 100 = 0.4%, the total is 5.4% and the fault current falls to 25.8 kA. The peak current for X/R of 5 is κ √2 Isc with κ = 1.02 + 0.98 e−3R/X = 1.56, about 61 kA.
Using the result
- Breakers and fuses need a breaking capacity at least equal to the prospective fault current at their location; switchgear also needs a making capacity above the peak current.
- Busbars and cables must withstand the fault for the clearing time: compare with the minimum area from the busbar size calculator.
- Connected motors feed current into a fault for a few cycles and raise the first-cycle value, and cable impedance lowers it away from the terminals; neither is included here.
- Lower impedance means a higher fault current: a 4% transformer gives 25% more than a 5% one.
Short-Circuit Current Calculator FAQ
How do I calculate the short-circuit current of a transformer?
Isc = kVA × 100 ÷ (√3 × V × %Z) with V in volts. For 1,000 kVA at 415 V and 5% that is 27.8 kA.
What is the %Z of a transformer?
The percentage impedance is the voltage that must be applied to the primary to circulate full-load current in a shorted secondary, as a percentage of the rated voltage. It is on the nameplate; typical distribution values are 4 to 6%.
Why does the utility fault level matter?
A real network has impedance too, so the fault current is lower than the infinite-bus value. Enter its MVA to include it.
What is the peak current for?
It is the maximum instantaneous value in the first half cycle, used to check the making capacity of switchgear and the mechanical strength of busbars.
Does this include motors?
No. Motors can add several times their full-load current to the first cycles of a fault, so a complete study includes them.
Is this valid for the primary side?
The current on the primary is the secondary value multiplied by the voltage ratio, but the source impedance matters more there. Use a network study for medium-voltage work.