Secondary full-load current and prospective short-circuit current
by the infinite-bus method.
FLC = S ÷ (√3 · VL); PSCC = FLC × 100 ÷ %Z.
For design guidance only. Always verify with a qualified engineer.
About this transformer sizing calculator
This free transformer sizing calculator works out the secondary full-load current (FLC) and the prospective short-circuit current (PSCC) at the terminals of a distribution transformer. Enter the nameplate rating in kVA, the secondary line voltage and the short-circuit impedance (%Z), and it returns the running current, the prospective fault current in kA and the equivalent fault level in MVA. It is built for electrical designers and installers who need a quick check of switchgear and cable fault ratings during design. Everything runs in your browser. Nothing is uploaded.
How transformer FLC and PSCC are calculated
The full-load current is the standard apparent-power relation: FLC = S ÷ (√3 · VL) for a three-phase transformer (line-to-line voltage), or S ÷ VL for single-phase. The prospective short-circuit current uses the infinite-bus method: PSCC = FLC × 100 ÷ %Z, which is equivalent to a fault level of S × 100 ÷ %Z. Here %Z is the transformer's nameplate short-circuit impedance, the impedance voltage Usc referred to its own rating, as defined in IEC 60076-1 (BS EN 60076-1). The method assumes an ideal zero-impedance upstream source, so the transformer's own impedance is the sole current-limiting element. The result is therefore a conservative upper bound on the secondary fault current; the actual value is lower once utility and conductor impedance are included.
Frequently asked questions
What is %Z and where do I find it?
%Z (the impedance voltage Usc) is the percentage of rated voltage needed to circulate rated current with the secondary short-circuited, per IEC 60076-1. It is printed on the transformer nameplate. Typical distribution values run from about 4 % on smaller units up to 6 % on larger ones.
Can I use this PSCC for an arc-flash study?
No. The infinite-bus PSCC is a conservative upper bound on fault current. It is suitable for confirming the breaking capacity of switchgear and the withstand rating of cables, but it must not be used as the basis for arc-flash incident-energy calculations, where a lower real fault current can actually increase incident energy.
Why is the infinite-bus result higher than my measured fault current?
The infinite-bus method treats the upstream network as having zero impedance, so only the transformer's %Z limits the current. In reality the supply network and the conductors between the transformer and the fault add impedance, which reduces the actual prospective fault current below this figure.
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