Available Short-Circuit Current from Percent Impedance

Percent impedance is a strange-looking number until you know its definition: it is the percentage of rated voltage that must be applied to a transformer with its secondary shorted to drive exactly full-load current. Turn that around and it tells you the fault current straight away. If 5% of the volts pushes full-load amps through the short, then 100% of the volts pushes twenty times that. A 500 kVA, 480 V unit with 601 A full load and 5% impedance can deliver $601 \times 20 \approx 12{,}000$ A into a bolted fault.

That number is not academic. It sets the interrupting rating of every breaker and fuse downstream, and equipment applied above its rating does not merely trip late — it can fail explosively, which is the scenario arc-flash studies exist to prevent. It is also why "buy the transformer with the lowest impedance" is bad advice: a lower %Z gives better voltage regulation and worse fault duty, and the sweet spot is a design decision, not a default.

The assumption hiding in this calculation is an infinite primary source, so it deliberately overestimates. Real utility supplies have their own impedance, and the cable between the transformer and the fault adds more, so the true available current at a panel forty metres away is lower — sometimes much lower. Overestimating is the safe direction for equipment ratings but the unsafe direction for coordination studies, where a fault current that turns out too small may fail to clear a fuse in time.