Part 3 of the short-circuit series. Running motors feed the fault for the first cycles. 600 kW of motors add 6.8 kA to the 37.4 kA from the transformer.
| Network | 250 MVA, 11 kV grid (R/X = 0.1) → 1.6 MVA, 11/0.4 kV Dyn transformer, uk = 6 %, Pk = 15 kW |
| Motors | 600 kW total on the LV board, cos φ = 0.85, efficiency 0.94, locked-rotor current 6 × rated, R/X = 0.42 |
The grid and motor contributions are added as phasors, which gives slightly less than the arithmetic sum (37.41 + 6.83 = 44.24 kA).
Motor contribution decays quickly. IEC 60909 allows the breaking current of asynchronous motors to be reduced by factors μ and q. The calculator keeps the full motor contribution in Ib, which is on the safe side.
| Check | Requirement | Actual | Status |
|---|---|---|---|
| Without motors | — | 37.41 kA | ✓ PASS |
| With motors | ≤ board rating 50 kA | 44.06 kA | ✓ PASS |
Key insight: A quick estimate of motor contribution is the locked-rotor multiple times the total rated current of running motors — typically 4 to 7 times motor full-load current.
Every input in this example is editable in the live calculator — no signup needed.
Open IEC 60909 Short-Circuit Calculator calculator →No. IEC 60909 leaves motors out of the minimum case, because they may not be running.
Usually not significantly. A drive without regenerative capability does not feed a fault on its supply side, so drive-fed motors are normally left out.