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Laplace: two switches, current before and after 8 ms

Short answer

For 0 < t < 8 ms, R = 5 Ω, i(t) = 2(1−e^(−25t)) A. For t > 8 ms, R = 15 Ω, shifting time gives i(t) = 0.6667 − 0.3041 e^(−75(t−0.008)) A.

In the given circuit below, switch S1 is closed at t = 0 and after 8ms, the switch S2 is opened. Find the complete expression for current in the interval 0 < t < 8ms and t > 8ms. Use Laplace Transform approach. [Figure: 10 V source with switch S1 (t = 0) and a 5 Ω resistor in series, then a 10 Ω resistor which has switch S2 connected in parallel across it, followed by a 0.2 H inductor]

Circuit. 10 V source, 5 Ω and 0.2 H in series. A 10 Ω resistor is in series, but S₂ is closed across it, so it is shorted until S₂ opens at t = 8 ms. Inductor initially de-energised.

Interval 1: 0 < t < 8 ms (S₁ closed, S₂ closed)

R = 5 Ω, L = 0.2 H.

I(s) = (10/s) / (5 + 0.2 s) = 50 / [ s (s + 25) ] = 2/s − 2/(s + 25)

i(t) = 2 (1 − e^(−25t)) A, 0 < t < 8 ms

At t = 8 ms: i = 2(1 − e^(−0.2)) = 0.3625 A.

Interval 2: t > 8 ms (S₂ open, 10 Ω in the circuit)

Use a shifted time t′ = t − 0.008. R = 5 + 10 = 15 Ω, initial current i(0′) = 0.3625 A.

(15 + 0.2 s) I(s) = 10/s + L i(0′) = 10/s + 0.2 × 0.3625
I(s) = (50/s + 0.3625) / (s + 75)
     = 0.6667/s − 0.6667/(s + 75) + 0.3625/(s + 75)
     = 0.6667/s − 0.3041/(s + 75)

i(t) = 0.6667 − 0.3041 e^(−75(t − 0.008)) A, t > 8 ms

Complete result

IntervalCurrent
0 < t < 8 ms2 − 2e^(−25t) A
t > 8 ms0.6667 − 0.3041 e^(−75(t − 0.008)) A

The current is continuous at 8 ms (0.3625 A) because of the inductor, then rises toward 10/15 = 0.667 A with τ = 0.2/15 = 13.3 ms. A numerical simulation of the switched circuit reproduces both expressions.

✓ Answer checked by Claude Opus 5.5 (high effort)

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