Short Circuit Current Calculator
The worst moment on a wire — prospective fault current at a transformer's secondary, from nameplate kVA, voltage and impedance.
Short Circuit Current Calculator
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What this result does not account for
- Infinite-bus worst case — no source or cable impedance
- Symmetrical RMS; no X/R or motor-contribution term
In short: A 1,000 kVA transformer at 480 V and 5.75% impedance carries about 1,203 A of full-load current — and will deliver roughly 20,900 A (20.9 kA) into a dead short. Impedance is the only thing holding it back: at 5% the same transformer would hand a fault twenty times its full-load current. Every breaker, panel and bus below it must be rated to interrupt that number, which is why the nameplate's impedance line matters more than any other on the plate.
Formula
Isc = kVA × 1,000 ÷ (√3 × V × Z% ÷ 100)
Full-load current is kVA times a thousand over root three times the voltage. During a bolted fault the transformer's own impedance is the only limiter, so the fault current is full-load current divided by the per-unit impedance — a 5% transformer passes twenty times its rating. This is the infinite-bus worst case: the utility source is assumed bottomless, which is the conservative direction for equipment selection. IEC 60909 and the NEC's interrupting-rating duty frame the study this screen previews.
Worked Example
- Read kVA, secondary voltage and Z% off the nameplate.
- Read the prospective fault current at the terminals.
- Rate every downstream device above that number.
Defaults: 1,000 kVA at 480 V, 5.75% — about 20.9 kA at the secondary, 17.4 MVA of fault level. Re-rate the same transformer to 4% impedance and the fault climbs toward 30 kA; the copper never changed.
Strengths & Limits Of This Model
Where this engine is strong
- Nameplate-only inputs — nothing to look up twice
- The kA number is the one the gear catalog asks for
Where it stops
- Ignores utility impedance, which real studies include
Practical Use Cases
Breaker sizing
interrupting rating beats the kA
Panel schedules
the AIC line on every board
Arc-flash prep
the first number the study asks for
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Short Circuit Current Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
Why does impedance control the fault current?
During a short the load is gone; the only thing left between the coil and the fault is the transformer's own opposition. Halve the impedance and the fault doubles — a 1,000 kVA unit at 4% delivers about 30,000 A where a 5.75% unit stops near 20,900.
What does infinite bus mean here?
The formula assumes the utility side can supply unlimited current, so the transformer is the only restriction. Real grids are finite, which makes the true fault a little lower — the safe direction when picking equipment ratings.
Is the fault current the same at every panel?
No — it falls as cable adds impedance on the way downstream. The secondary-terminal number is the ceiling; long feeders can cut it by a fifth or more, but the first device after the transformer sees nearly all of it.
What is fault MVA for?
It expresses the same fault as power instead of current, which is how switchgear and utility studies are catalogued. Multiply root three by voltage by the fault current and the 480 V, 1,000 kVA example reads about 17.4 MVA.
Does the primary voltage matter?
Not to this screen — the nameplate impedance already folds the whole transformer into one percentage referred to the secondary. The utility's contribution would enter through a source impedance in a full study, not here.
Why do the standards get named at all?
Because the arithmetic is the easy part; the duty is the point. IEC 60909 wraps this calculation in correction factors, and the NEC requires every device's interrupting rating to exceed the available fault — this page prices the number those rules act on.
Single-phase transformers?
The root three drops out: fault current is kVA times a thousand over voltage over impedance. A 100 kVA, 240 V, 2% single-phase unit delivers about 20,800 A — small plate, ferocious fault.
What if the nameplate only lists kVA and Z?
That is all this needs besides the secondary voltage, which the installation fixes. Impedance tolerances run a few percent either side, so treat the answer as a ceiling estimate and let a commissioned study put the official figure on it.