Fault Parameters
Results
Peak make and d.c. at break are what switchgear ratings are checked against. Verifying equipment duty across a network is a short-circuit study. Short-circuit and power system studies →
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Fault Current Components (IEC 60909)
Peak Make Current:
ip = √2 × Ik" × (1.02 + 0.98 × e-3/XR)
Maximum instantaneous current during first half-cycle
DC Time Constant:
τ = X / (2πfR) = XR / (2πf)
DC Component:
idc(t) = √2 × Ik" × e-t/τ
Decays exponentially with time constant τ
Total Fault Current:
i(t) = √2 × Ik" × sin(2πft) + idc(t)
Frequently Asked Questions
How is the peak make current calculated?
From IEC 60909-0: ip = kappa x root 2 x Ik", with kappa = 1.02 + 0.98 e^(-3R/X). At the default 13.1 kA and X/R = 10 the tool gives ip = 32.35 kA.
What is the d.c. time constant?
tau = (X/R) / (2 pi f). At X/R = 10 and 50 Hz it is 31.8 ms, so the d.c. component falls to 37 % of its initial value after 31.8 ms.
How much d.c. component is left at the break time?
idc(t) = root 2 x Ik" x e^(-t/tau). With X/R = 10 at 50 Hz and a 70 ms break time, 11.1 % of the initial d.c. component remains, which is the DC Component % result.
Why does a higher X/R ratio matter?
A higher X/R gives a longer time constant, so the d.c. component decays more slowly. That raises the peak make current and the d.c. current the breaker has to interrupt, both of which switchgear ratings have to cover.