Short circuit basics
The key idea
A short circuit is an accidental connection that bypasses the load. With the load gone, only the network's own impedance limits the current — so the current is huge near the source, and it shrinks with every kilometer of line between the source and the fault.
The idea
In normal operation, current is small because the load resists it. A short circuit — a broken insulator, a digger through a cable, a branch across a line — replaces the load with (almost) nothing.
The circuit is still a circuit: a source voltage pushing current around a loop. But the loop impedance is now only what the network itself contributes — the source, the transformers, and the line up to the fault. That impedance is small, so the current is large: thousands of amperes where tens normally flow.
Two consequences follow, and they pull in opposite directions:
- Near the source, the danger is the size. Equipment must survive the current — thermally and mechanically — until protection clears it. This is the maximum fault current case, and it sizes switchgear.
- Far from the source, the danger is the smallness. Every kilometer adds impedance and shrinks the current. A distant fault can draw so little that protection never notices. This is the minimum fault current case, and it proves detection.
Try it
fault at 4.0 km: 2.87 kA
11 kV feeder, 150 MVA source, 0.35 Ω/km line. Drag the fault away from the source: every kilometer adds impedance, and the current falls. Keep going until it drops below the relay pickup.
The widget uses a fixed voltage and the plain impedance divide. The full IEC 60909 method refines this with the voltage factor c and defined maximum and minimum cases — the shape of the answer stays the same.
Why it matters
- Ratings are chosen against fault current, not load current. A breaker that carries 400 A in service may have to interrupt 25 kA in a fault. The rating check is a short-circuit calculation.
- Protection settings live between the two extremes. A relay must ignore the largest normal load and act on the smallest credible fault. Both numbers come from this study.
- Fault current is a property of the network, not the fault. The same fault at two locations gives two very different currents. That is why studies calculate fault duty at every bus.
The math, if you want itOptional — the page reads completely without it
For a three-phase fault, the initial symmetrical current at the fault location is:
IEC 60909 initial symmetrical current
I″k = c · Un√3 · |Ztotal|
where Un is the nominal line-to-line voltage, Ztotal is the impedance from the source to the fault, and c is the voltage factor (1.05–1.10 for maximum cases, 0.95–1.00 for minimum). The widget above uses c = 1.0 for clarity.
See it in Phasor
Phasor computes IEC 60909 fault duty at every bus — or mid-line, where long feeders make the far end marginal — and flags each switchgear item whose rating the calculated current exceeds.