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Classical: Vc(t) with 12 V, 0.5 H, 2 F

Short answer

Initial capacitor voltage is 10 V and inductor current is 2 A. The RLC series circuit for t>0 is critically damped. Final voltage is 12 V. v_C(t) = 12 - (2 + t) e^(-t) V.

In the following network the switch was closed for a long time before it is being opened at t = 0. Find the expression for Vc(t) for t > 0. (Use classical method). [Figure: 12 V source in series with a 1 Ω resistor and a 0.5 H inductor along the top wire; a switch (t = 0) in series with a 5 Ω resistor as a shunt branch to the bottom wire; a 2 F capacitor as a second shunt branch with voltage Vc(t)]

Initial conditions (t = 0⁻, switch closed for a long time)

Inductor acts as a short and the capacitor as an open circuit. The 12 V source drives 1 Ω and the 5 Ω shunt branch:

i_L(0⁻) = 12/(1 + 5) = 2 A

v_C(0⁻) = 5 × 2 = 10 V

Both cannot change instantly, so i_L(0⁺) = 2 A and v_C(0⁺) = 10 V.

Circuit for t > 0 (switch open)

Series loop: 12 V, R = 1 Ω, L = 0.5 H, C = 2 F.

L di/dt + R i + vC = 12,    i = C dvC/dt
LC vC'' + RC vC' + vC = 12
1·vC'' + 2 vC' + vC = 12

Solution

Characteristic equation: m² + 2m + 1 = 0, so m = −1, −1 (critically damped).

Forced response (final value) v_C(∞) = 12 V.

v_C(t) = 12 + (A + Bt) e^(−t)

  • v_C(0) = 10 → 12 + A = 10 → A = −2
  • v_C′(0) = i_L(0)/C = 2/2 = 1 V/s. Derivative: B e^(−t) − (A + Bt) e^(−t), at t = 0: B − A = 1 → B = −1

Answer: v_C(t) = 12 − (2 + t) e^(−t) V, t > 0.

Check: v_C(0) = 10 V; v_C(∞) = 12 V; i = C v_C′ = 2(1 + t)e^(−t) gives i(0) = 2 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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