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Laplace: iL(t) and vc(t) after switch opens

Short answer

Determine initial conditions iL(0−)=6.857 A, vC(0−)=13.714 V from steady state with switch closed. Write s-domain equations incorporating these, solve, and inverse Laplace yields underdamped responses: iL(t)=4+e^−1.05t(2.857cos0.3122t+9.608sin0.3122t) A, vC(t)=8+e^−1.05t(5.714cos0.3122t+8.235sin0.3122t) V.

After being closed for a long time, if the switch in the circuit shown in figure below is opened at t = 0. Obtain the expressions for iL(t) and vc(t) for t > 0, using Laplace Transform Method. [Figure: 48 V source in series with a 5 Ω resistor and a second 5 Ω resistor which has a switch (t = 0) connected in parallel across it; then a 1 F capacitor with voltage vc(t) as a shunt branch; and a 2 Ω resistor in series with a 1 H inductor carrying iL(t) as a second shunt branch]

Initial conditions (t = 0⁻, switch closed, steady state)

The closed switch shorts the second 5 Ω, so one 5 Ω is in series with the source. The capacitor is open and the inductor is a short:

iL(0⁻) = 48 / (5 + 2) = 6.857 A
vc(0⁻) = 2 × iL(0⁻) = 13.714 V

For t > 0 (switch open)

The source now sees 5 + 5 = 10 Ω in series. Write the two node/branch equations (C = 1 F, L = 1 H) and transform them with the initial conditions:

C dvc/dt = (48 − vc)/10 − iL     →  s Vc − 13.714 = (48/s − Vc)/10 − IL
L diL/dt = vc − 2 iL             →  s IL − 6.857 = Vc − 2 IL

From the second equation Vc = (s + 2) IL − 6.857. Substituting in the first and collecting terms:

IL (s² + 2.1 s + 1.2) = 4.8/s + 6.857 s + 14.4
IL(s) = 6.857 (s² + 2.1 s + 0.7) / [ s (s² + 2.1 s + 1.2) ]

Partial fractions

Roots of s² + 2.1s + 1.2 = 0: s = −1.05 ± j0.3122 (underdamped).

IL(s) = 4/s + (2.857 s + 6) / (s² + 2.1 s + 1.2)
      = 4/s + 2.857 (s + 1.05) / [(s + 1.05)² + 0.3122²] + 3.0 / [(s + 1.05)² + 0.3122²]

Vc(s) = (s + 2) IL − 6.857 = 8/s + (5.714 s + 8.571) / (s² + 2.1 s + 1.2)

Inverse transform

i_L(t) = 4 + e^(−1.05t) (2.857 cos 0.3122t + 9.608 sin 0.3122t) A, t > 0

v_c(t) = 8 + e^(−1.05t) (5.714 cos 0.3122t + 8.235 sin 0.3122t) V, t > 0

Check: i_L(0) = 4 + 2.857 = 6.857 A ✓; v_c(0) = 8 + 5.714 = 13.714 V ✓; final values 48/(10 + 2) = 4 A and 4 × 2 = 8 V ✓. The expressions agree with a numerical integration of the two state equations and with a numerical inverse Laplace transform (checked at t = 0.2, 1, 3 s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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