Part 2 of the short-circuit series. Adding an 80 m feeder cuts the fault level from 37.4 kA to 22.9 kA and changes the X/R ratio — which affects the peak current and equipment selection.
| Network | 250 MVA, 11 kV grid (R/X = 0.1) → 1.6 MVA, 11/0.4 kV Dyn transformer, uk = 6 %, Pk = 15 kW |
| Feeder | 2 × 185 mm² Cu cables in parallel, 80 m; r = 0.0991 Ω/km, x = 0.0739 Ω/km per cable |
| Fault | Three-phase and line-to-earth faults at the MCC |
The calculator also solves the zero-sequence network (transformer Dyn earthed at the star point, cable zero-sequence impedance about 4 × the positive-sequence).
| Check | Requirement | Actual | Status |
|---|---|---|---|
| MCC three-phase Ik″ | ≤ MCC rating 36 kA | 22.90 kA | ✓ PASS |
| MCC peak ip | ≤ making capacity | 39.6 kA | ✓ PASS |
Key insight: A short LV cable has as much impedance as the transformer. Equipment far from the transformer can often use lower-rated devices, but the minimum fault current (for protection) falls too — see Part 5.
Every input in this example is editable in the live calculator — no signup needed.
Open IEC 60909 Short-Circuit Calculator calculator →For maximum fault current IEC 60909 uses 20 °C (lower resistance, higher current). For the minimum fault current it uses the resistance at the end of the fault, typically 80 °C.
Cable resistance damps the DC offset. A higher R/X gives a lower peak factor κ.