Prospective Short-Circuit current and Fuse Selection

by VladM

12 parts

A 48V pack (20mΩ inside) feeds 5mΩ + 2µH of cable and F1, a 200A fuse. At 50ms "short" closes: a bolted short. Its symbol stays drawn open after it closes. The prospective short-circuit current: with the fuse replaced by a link is I = V/R = 48V / 26mΩ (the pack 20, the cable 5, the short's contact 1mΩ) = 1.85kA. With F1's own 1mΩ in the loop it heads for 48V / 27mΩ = 1,778A, with τ = L/R = 2µH / 27mΩ = 74µs. Choosing the fuse: a current rating above the normal load current (F1 is rated 200A), a DC voltage rating above the pack's (a DC arc has no current zero), and a breaking capacity above the prospective 1.85kA. Press Run: the short closes about 2.6 seconds later on screen at ×0.25, and the plot holds the 30µs around it. Times below are the circuit's, from the moment the solver closes the short. F1 has no I²t: it opens the instant its current crosses 200A. A real fuse carries its rating indefinitely and melts on I²t, so under this short the current would keep rising toward its prospective value until the fuse melted: which is why the breaking capacity matters. The 100 V clamp ("arc") and the snubber (Rsnub 1Ω + Csnub 10µF) stand in for the fuse's arc and the cable's damping.