Available Fault Current Calculator (Transformer)

Estimate the highest short-circuit current a transformer can deliver at its secondary terminals, so you can check that breakers, fuses and panels are rated for it.

Transformer nameplate

Find the kVA, secondary voltage and percent impedance on the transformer nameplate. For a single-phase transformer, use the voltage across the secondary winding.

This is a maximum at the transformer terminals. It assumes an unlimited source behind the transformer and no cable impedance. Real fault current at equipment farther along the circuit is lower, and motors can add current for the first moments of a fault. The utility or an engineering study gives the exact figure. Use this to see the order of magnitude and to catch equipment that is clearly under-rated.

How it works

A transformer's impedance limits how much current it can deliver into a short circuit. A transformer with a low impedance can deliver many times its full-load current. The full-load current is found from the kVA rating and voltage, and the fault current is that value divided by the impedance expressed as a fraction.

Full-load current = kVA × 1000 ÷ (√3 × V)   (three-phase)
Full-load current = kVA × 1000 ÷ V   (single-phase)
Fault current = full-load current ÷ (%Z ÷ 100)

Worked example

A 112.5 kVA, 480 V three-phase transformer with 3.5% impedance has a full-load current of 112,500 ÷ (1.732 × 480) = 135.3 A. The available fault current is 135.3 ÷ 0.035 = 3,866 A, about 3.9 kA. Any breaker, fuse or panel connected there needs an interrupting or short-circuit current rating of at least that.

Common questions

Where do I compare the result?

With the interrupting rating (often called AIC) on each breaker or fuse, and the short-circuit current rating (SCCR) marked on an industrial control panel. Both must be at least the available fault current at that point. The fault current guide explains the terms.

Why is a bigger transformer a bigger problem?

For the same impedance, a larger kVA rating delivers proportionally more fault current. Lower impedance does the same.

Calculations are provided for convenience and education. Obtain the available fault current from the utility or an engineering study for design and labeling.