X/R ratio & peak current
The key idea
A fault current cannot jump instantly to its final wave — inductance forbids it. The difference shows up as a DC offset that lifts the first peak far above the normal wave. The X/R ratio decides how slowly that offset dies, and κ turns it into a number.
The idea
An inductive circuit has a rule: current cannot change instantaneously. At the moment a fault strikes, the current is whatever it was — nearly zero — but the network is suddenly demanding a large sine wave.
The circuit reconciles the two by adding a DC offset: a one-sided component that starts exactly where it must for the total current to be continuous, and then decays away. While it lives, the AC wave rides on top of it — and the first crest of that ride is the highest instantaneous current of the entire fault: the peak current ip.
How long the offset lives is set by the impedance make-up at the fault:
- High X/R (near transformers and generators — mostly reactance): the offset decays slowly, still nearly full-size at the first crest. ip approaches twice the symmetrical peak.
- Low X/R (far along a resistive feeder): resistance kills the offset within a cycle, and the first peak is barely above normal.
The factor κ packages this: ip = κ · √2 · I″k, with κ running from about 1.0 (pure resistance) toward 2.0 (pure reactance).
Try it
κ = 1.75 · ip = 24.7 kA (I″k = 10 kA)
Move the slider: higher X/R keeps the DC offset alive longer, and the first peak grows toward twice the symmetrical value.
Why it matters
- ip is a mechanical number. The forces between conductors scale with current squared — at κ = 1.8 the first peak pushes more than three times as hard as the symmetrical wave would. Busbar bracing and breaker making-duty are rated for exactly this instant.
- The DC offset also stresses breakers. An offset that has not decayed by contact separation makes the current harder to interrupt — breaker standards account for the DC component at high X/R.
- X/R comes out of the same study. It is not an extra measurement: the fault calculation that gives I″k gives the X/R at the fault location, and κ with it.
The math, if you want itOptional — the page reads completely without it
For the worst-case switching instant, the fault current is:
the asymmetrical current
i(t) = √2 · I″k · [ sin( ωt − φ ) + sin(φ) · e−ωt·R/X ]
where φ = tan⁻¹(X/R). The second term is the DC offset — the dashed curve in the widget — with time constant X/(ωR). IEC 60909 approximates the resulting first peak with:
the peak factor and the peak
κ = 1.02 + 0.98 · e−3R/X · ip = κ · √2 · I″k
The widget above plots exactly these expressions.
See it in Phasor
Phasor reports X/R and κ at every fault location and computes ip from them — then checks it against each switchgear item's making capacity and each busbar's mechanical rating.