Part 5 of the short-circuit series. Maximum currents size equipment; minimum currents prove the protection works. At a DB 150 m away the minimum earth fault is 2.85 kA — too low for a 250 A breaker set at 10 × In.
| 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 | 4-core 95 mm² Cu, 150 m; r = 0.193, x = 0.0779 Ω/km; r0 = 0.772, x0 = 0.234 Ω/km |
| Protection | 250 A MCCB at the LV board, instantaneous setting 10 × In, tolerance ±20 % |
IEC 60909 uses cmin = 0.95 at LV, leaves motors out and takes cable resistance at the end of the fault (×1.24 for copper at 80 °C).
The fault would then be cleared by the slower long-time element, which may exceed the disconnection time required by IEC 60364-4-41.
| Check | Requirement | Actual | Status |
|---|---|---|---|
| Ik1 min vs 10 × In upper limit | Ik1 min > 3,000 A | 2,845 A | ✗ FAIL |
| Ik1 min vs 8 × In upper limit | Ik1 min > 2,400 A | 2,845 A | ✓ PASS |
Key insight: Always calculate two values at every bus: the maximum current for equipment ratings and the minimum current for protection sensitivity. Long cables affect the minimum far more than the maximum.
Every input in this example is editable in the live calculator — no signup needed.
Open IEC 60909 Short-Circuit Calculator calculator →The cable's zero-sequence impedance (including the return path) is about four times its positive-sequence impedance, and the cable dominates the loop at 150 m.
A larger cable or a separate protective conductor, an RCD or earth-fault relay, or a breaker with an electronic trip unit and earth-fault (G) protection.