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Laplace: currents i1 and i2 with L and C branches

Short answer

With initial i1(0)=3.333 A and v_C(0)=66.67 V, Laplace gives i1(t)=5−1.667e^(−20t) A and i2(t)=1.667e^(−50000t) A. Branches are independent after switch closure.

In the network shown below, a steady state is reached with the switch K open with V = 100 V, R1 = 10 Ω, R2 = 20 Ω, R3 = 20 Ω, L = 1 H, and C = 1 μF. At time t = 0, the switch is closed. Evaluate the currents i1 and i2, using Laplace transform, for t > 0. [Figure: source V in series with R1, which has switch K connected in parallel across it; then two parallel branches: R2 in series with L (current i1), and R3 in series with C (current i2)]

Before t = 0 (K open, steady state). The source feeds R₁ in series with the two parallel branches. The inductor is a short, the capacitor an open circuit:

i1(0⁻) = 100 / (R1 + R2) = 100/(10 + 20) = 3.333 A,   i2(0⁻) = 0
Voltage across the branches = i1 R2 = 66.67 V  →  vC(0⁻) = 66.67 V

For t > 0 (K closed). K short-circuits R₁, so the 100 V source is applied directly across both branches. The branches are now independent.

Branch 1: R₂ + L (initial current 3.333 A)

(R2 + sL) I1 − L i1(0) = 100/s
(20 + s) I1 = 100/s + 3.333
I1 = (3.333 s + 100) / [ s (s + 20) ] = 5/s − 1.667/(s + 20)

i1(t) = 5 − 1.667 e^(−20t) A

Branch 2: R₃ + C (initial voltage 66.67 V)

I2 (R3 + 1/(sC)) = 100/s − 66.67/s = 33.33/s
I2 = 33.33 / (20 + 10⁶/s) × (1/s) = 1.667 / (s + 50 000)

i2(t) = 1.667 e^(−50 000 t) A

(time constant R₃C = 20 × 10⁻⁶ = 20 μs)

Check: i1(0⁺) = 5 − 1.667 = 3.333 A (equal to the inductor's old current) ✓; i2(0⁺) = (100 − 66.67)/20 = 1.667 A ✓; i1(∞) = 100/20 = 5 A, i2(∞) = 0 ✓. Source current for t > 0 is i1 + i2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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