Short-Circuit Current Calculator

Every breaker in a board must be able to interrupt the worst fault that can reach it. Enter the supply transformer’s rating and nameplate impedance, the HV source fault level, and optionally the cable run to the fault point, and this calculator returns the prospective three-phase fault current, the approximate fault level in MVA, the peak asymmetrical current, and the minimum interrupting capacity the switchgear needs — by the impedance method of IEC 60909.

The result feeds directly into breaker selection — see the Circuit Breaker Sizing Calculator — and the earth-conductor withstand check in the Earthing and Protective Conductor Calculator.

SheetCALC-10
TitleShort-Circuit Current Calculator
BasisImpedance method (IEC 60909)
Rev2026-07
1 · Source + Transformer
Both from the nameplate. Typical distribution units: 4–5% up to 630 kVA, 5–6.25% above.
2 · Cable to the Fault Point

Fault study CALC-10 · REV 2026-07

Source impedance (referred to LV)
Transformer impedance
Cable impedance
Total impedance to fault
Fault current at transformer LV terminals
Prospective fault current at fault point
Approximate fault level
Peak (asymmetrical) current, ip
Required breaker interrupting capacity

Approximate three-phase bolted symmetrical fault by the impedance method. A full study per IEC 60909 must also cover single-phase-to-earth faults, motor contribution, minimum fault currents for protection operation, and let-through energy coordination — work for a registered engineer.

How this calculator works

  1. Impedance chain. The fault current is set by everything between the grid and the fault: the HV source (Z = V² ÷ S, treated as pure reactance), the transformer (Z = Z% × V² ÷ S, split into R and X by the X/R ratio, typically 8 for distribution units), and the LV cable (resistance from standard conductor data, reactance about 0.08 mΩ/m, divided across parallel runs). All impedances are referred to the LV side and combined as R + jX.
  2. Fault current. Ik = c × V ÷ (√3 × |Z|), with the IEC voltage factor c = 1.05 for the maximum LV fault. The fault level is √3 × V × Ik.
  3. Peak current. The first asymmetrical peak is ip = κ√2 × Ik with κ = 1.02 + 0.98e⁻³ᴸ∕ᵀ — what the busbars and the breaker’s making capacity must survive mechanically.
  4. Interrupting capacity. The fault current is matched to the standard ratings ladder: 6 and 10 kA (MCB Icn, BS EN 60898), 16–65 kA (MCCB Icu, BS EN 60947-2), 85–100 kA (ACB class). Downstream of a cable run the fault level falls — often letting sub-boards use cheaper devices, or cascade-rated combinations verified from manufacturer tables.

Assumptions and limits

  • Three-phase bolted symmetrical fault only; single-phase-to-earth faults, motor infeed, and minimum-fault checks for protection operation need the full IEC 60909 study.
  • Conductor resistances at approximately 20 °C give the conservative (higher) fault current for rating purposes.

Need a stamped fault study, discrimination analysis or switchgear specification? See our consultancy services and design document library, or learn the method in a training program.

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