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Laplace: inductor and capacitor currents, 12 V

Short answer

Zero initial conditions; find total impedance, then branch voltage V(s), and currents IL(s) and IC(s). Inverse transform yields underdamped responses: iL(t)=6+e^−0.9167t(−6cos0.3997t+6.255sin0.3997t) A, iC(t)=e^−0.9167t(2cos0.3997t−4.587sin0.3997t) A.

Using Laplace Transform method, find the current of inductor and capacitor for t > 0 in the circuit shown in figure below. [Figure: 12 V source with a switch (t = 0) and a 2 Ω resistor in series; then two parallel branches: a 4 Ω resistor in series with a 1/3 F capacitor, and a 1 H inductor]

Zero initial conditions (assumed, switch closes at t = 0). The source is 12/s. The 4 Ω + 1/3 F branch has impedance Z_c = 4 + 3/s, and the inductor has Z_L = s.

Total impedance and source current

Z_p = Z_c Z_L / (Z_c + Z_L) = s (4s + 3) / (s² + 4s + 3)
Z_total = 2 + Z_p = (6s² + 11s + 6) / (s² + 4s + 3)

I(s) = (12/s) / Z_total = 12 (s + 1)(s + 3) / [ s (6s² + 11s + 6) ]

Voltage across the parallel branches

V(s) = I(s) Z_p = 12 (4s + 3) / (6s² + 11s + 6)

Inductor current

IL(s) = V/s = 12 (4s + 3) / [ s (6s² + 11s + 6) ]
      = 6/s − (6s + 3) / (s² + 1.8333 s + 1)

The roots of 6s² + 11s + 6 = 0 are s = −0.9167 ± j0.3997 (complex, underdamped).

i_L(t) = 6 + e^(−0.9167t) (−6 cos 0.3997t + 6.255 sin 0.3997t) A

(final value 12/2 = 6 A, since the inductor shorts the capacitor branch at d.c.)

Capacitor current

IC(s) = V/Z_c = 12 s / (6s² + 11s + 6) = 2s / [(s + 0.9167)² + 0.3997²]

i_C(t) = e^(−0.9167t) (2 cos 0.3997t − 4.587 sin 0.3997t) A

Check: i_C(0⁺) = 12/(2 + 4) = 2 A (capacitor acts as a short, inductor open) ✓, i_L(0) = 0 ✓, i_C(∞) = 0 ✓. Sample t = 0.5 s: i_L = 3.067 A, i_C = 0.664 A, equal to the numerical Laplace inversion and to a direct numerical solution.

✓ Answer checked by Claude Opus 5.5 (high effort)

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