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Laplace: current supplied by source, 10 V

Short answer

Find branch currents: I1(s)=10/[s(s+1)], I2(s)=5/(s+1). Summing, I(s)=10/s −5/(s+1). Inverse transform gives source current i(t)=10 −5e^(−t) A, t≥0.

In the network shown, the switch is closed at t = 0. Find the current supplied by the source using Laplace transform method. [Figure: 10 V source with a switch (t = 0) feeding two parallel branches: a 1 Ω resistor in series with a 1 H inductor, and a 2 Ω resistor in series with a ½ F capacitor]

Zero initial conditions (switch closes at t = 0). The source is 10/s. Branch impedances: Z₁ = 1 + s (R = 1 Ω, L = 1 H) and Z₂ = 2 + 2/s (R = 2 Ω, C = ½ F).

Branch currents

I1(s) = (10/s) / (1 + s)
I2(s) = (10/s) / (2 + 2/s) = 5 / (s + 1)

Source current

I(s) = I1 + I2 = 10/[s (s + 1)] + 5/(s + 1)
     = 10/s − 10/(s + 1) + 5/(s + 1)
     = 10/s − 5/(s + 1)

(equivalent to I = V/Z_eq = 5(s + 2)/[s(s + 1)])

i(t) = 10 − 5e^(−t) A, t ≥ 0

Individual branches: i₁ = 10(1 − e^(−t)) A and i₂ = 5e^(−t) A.

Check: i(0⁺) = 0 + 10/2 = 5 A (L open, C short) ✓; i(∞) = 10/1 = 10 A (C open, L short) ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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