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Laplace: current with S1 closing and S2 opening at 4 ms

Short answer

For 0 < t < 4 ms, i(t) = 2(1−e^(−500t)) A. For t > 4 ms, with R = 150 Ω and shifted time, i(t) = 0.6667 + 1.0626 e^(−1500(t−0.004)) A.

In the circuit shown below, switch S1 is closed at t = 0 and S2 is opened at t = 4 msec. Determine i(t) for t > 0. Assume that inductor is initially de-energized. (Use Laplace method) [Figure: 100 V source with switch S1 (t = 0) and a 50 Ω resistor in series, then a 100 Ω resistor which has switch S2 (opens at t = 4 msec) connected in parallel across it, followed by a 0.1 H inductor]

Circuit. 100 V source, 50 Ω and 0.1 H in series; a 100 Ω resistor is also in series, but S₂ is closed across it (so it is shorted) until S₂ opens at t = 4 ms. Inductor initially de-energised.

Interval 1: 0 < t < 4 ms (R = 50 Ω, L = 0.1 H)

I(s) = (100/s) / (50 + 0.1 s) = 1000 / [ s (s + 500) ] = 2/s − 2/(s + 500)

i(t) = 2 (1 − e^(−500t)) A

At t = 4 ms: i = 2 (1 − e^(−2)) = 1.7293 A.

Interval 2: t > 4 ms (R = 50 + 100 = 150 Ω)

Let t′ = t − 0.004 and i(0′) = 1.7293 A.

(150 + 0.1 s) I(s) = 100/s + 0.1 × 1.7293
I(s) = (1000/s + 1.7293) / (s + 1500)
     = 0.6667/s − 0.6667/(s + 1500) + 1.7293/(s + 1500)
     = 0.6667/s + 1.0626/(s + 1500)

i(t) = 0.6667 + 1.0626 e^(−1500(t − 0.004)) A, t > 4 ms

Result

IntervalCurrent
0 < t < 4 ms2 − 2e^(−500t) A
t > 4 ms0.6667 + 1.0626 e^(−1500(t − 0.004)) A

Continuity at 4 ms: 0.6667 + 1.0626 = 1.7293 A ✓. After S₂ opens, the current falls from 1.729 A to the new final value 100/150 = 0.667 A with τ = 0.1/150 = 0.667 ms. Verified by numerical simulation of the switched circuit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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