Prospective Fault Current Calculator
Estimate the prospective short-circuit (fault) current available from a supply transformer. Enter the transformer's kVA rating, the line voltage and its per-unit impedance %Z, and the calculator returns the transformer-limited fault current in kilo-amps along with the rated current.
Enter Values
Before you rely on this: First-pass guide only. Verify safety-critical or regulated work against the relevant standards, your project requirements and a qualified professional.
How to use this calculator
- Enter the transformer rating S in kVA and the line voltage V in volts (line-to-line for three-phase).
- Enter the transformer impedance %Z from its nameplate — leave it blank to use the typical default of 4 %.
- Tick single-phase (enter 1) if the supply is single-phase; leave blank or 0 for three-phase. Read the prospective fault current Isc in kA.
How it works
First the rated line current is found: three-phase I_rated = S·1000/(√3·V), single-phase I_rated = S·1000/V. The prospective fault current is then Isc = I_rated / (%Z/100) — because at a bolted short circuit the only thing limiting the current is the transformer's own impedance. A lower %Z means a stiffer supply and a higher fault current. This is the fault level right at the transformer and ignores upstream source impedance and cable impedance, which both reduce it further down the installation.
Worked example
Worked example. A 500 kVA, 415 V three-phase transformer with %Z = 4: I_rated = 500,000/(√3 × 415) = 695.6 A. Isc = 695.6 / (4/100) = 17,390 A ≈ 17.39 kA. Switchgear and breakers on that board must have a short-circuit rating of at least 17.39 kA.
Common mistakes
- Treating this transformer-limited figure as the exact fault current at a distant point — cable impedance lowers the real value further down the run.
- Using the phase voltage (e.g. 230 V) for a three-phase transformer instead of the line-to-line voltage (e.g. 415 V).
- Forgetting %Z is a percentage — a 4 % impedance means dividing the rated current by 0.04, not by 4.
Frequently asked questions
Why does a lower %Z give a higher fault current?
The %Z is the fraction of rated voltage needed to drive rated current through a short-circuited transformer. A low %Z means very little impedance opposes a fault, so an enormous current flows: Isc = I_rated / (%Z/100). A 'stiffer' (low-impedance) supply therefore produces a higher prospective fault current.
Is this the actual fault current at my switchboard?
No — it is the worst-case value at the transformer terminals. It ignores the upstream network and, importantly, the resistance of the cable between the transformer and your board, both of which reduce the actual fault current. Use it as a conservative check that your equipment's kA rating is adequate; for a precise figure carry out a full fault study including all impedances.
Related tools
- Transformer Primary/Secondary Current Calculator
- Circuit Breaker Sizing Calculator
- Voltage Drop Calculator
- Conduit Fill Calculator
- Circuit Load Calculator
- Transformer Sizing Calculator
Explore more in Electrical, Electronics, Solar & Energy.
Tip: Enter any known values to calculate the remaining results.
All calculations run in your browser. Your inputs are never saved or transmitted.