Available fault current (prospective short-circuit current) is the maximum current that would flow if a bolted short circuit occurred at a point in the system. It sets the minimum interrupting rating for breakers and fuses and the short-circuit withstand for busbars, cables and switchgear. At a transformer secondary the dominant limit is the transformer's own impedance, so a quick and widely used estimate divides the full-load current by the per-unit impedance.
Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: Prospective short-circuit current, IEC 60909, IEEE 1584.
Safety notice. This is a planning estimate using the infinite-source assumption. A code-compliant short-circuit and arc-flash study (IEC 60909 / IEEE 1584) must account for the utility source impedance, conductor impedance and motor contribution, and be performed by a qualified engineer. See our disclaimer.
The fault-current formula
Two steps give the transformer-secondary fault current:
The %Z term is the transformer's per-unit impedance. Because a short circuit is limited only by that impedance, a 5% transformer can deliver about 100/5 = 20 times its full-load current into a bolted fault. For single-phase, drop the √3 and use the single-phase voltage.
Worked example — 1000 kVA substation
Scenario: A 1000 kVA, 400 V, 3-phase transformer with nameplate impedance 5.75%.
All downstream breakers must have an interrupting rating (kAIC) above 25.1 kA — commonly a 36 kA-rated device is specified for margin. Busbars and switchgear need a matching short-circuit withstand. Size the incomer with the busbar sizing calculator for the same fault level.
Frequently Asked Questions
Find the secondary full-load current IFL = kVA × 1000 / (√3 × VLL), then divide by per-unit impedance: Ifault = IFL × 100 / %Z. A 500 kVA, 400 V, 5% transformer gives IFL = 722 A and Ifault ≈ 14.4 kA.
%Z is the percent of rated voltage needed to drive full-load current with the secondary shorted. It is on the transformer nameplate, typically 4–6% for distribution units. Lower %Z means higher available fault current.
Breakers and fuses must have an interrupting rating (kAIC) above the available fault current, or they can fail violently during a short circuit. Busbars and cables also need a short-circuit withstand above the prospective fault current.
No. It gives the transformer source fault current assuming an infinite primary. Running motors add roughly 4–6× their full-load current for the first cycles. A full study adds motor contribution and reduces for downstream cable impedance.
Yes. It is highest at the secondary and falls with distance as cable impedance adds up. A point-to-point calculation along each run gives the lower fault current at downstream panels.