Transformer Short-Circuit (Fault) Current Calculator

How it works

I_FLA = kVA × 1,000 ÷ (√3 × V); I_sc (infinite bus) = I_FLA ÷ %Z; with a finite source I_sc = I_FLA ÷ (%Z + kVA ÷ source kVA); motor contribution ≈ 4 × motor FLA; downstream I = V ÷ (√3 × (Z_source + Z_feeder))

When a bolted short circuit occurs at a transformer’s secondary, the only thing limiting the current is the transformer’s own impedance plus whatever the utility source and cables add. Dividing full-load current by the per-unit impedance gives the infinite-bus fault current — a 1,000 kVA, 5.75% transformer at 480 V can deliver about 21 kA, roughly 17 times its rated current. A real utility source with finite fault capacity adds impedance and trims that a little; running motors add to it, since they briefly act as generators feeding the fault. Every breaker and fuse must be able to interrupt the fault current available at its location, and downstream panels see less because cable impedance intervenes. These figures are the starting point of a short-circuit and arc-flash study, not a substitute for one.

Worked example

A 1,000 kVA, 480 V, 5.75%Z three-phase transformer: FLA 1,203 A, infinite-bus fault 20.9 kA, 19.6 kA with a 250 MVA primary source. Add 300 hp of running motors (≈ 1.4 kA) and the main sees about 21 kA — breakers rated 22 kAIC, though 25 kA is the prudent choice. Sixty metres of feeder drops the bolted value to about 12 kA at the next panel.

Frequently asked questions