---
title: "Classical: Vc(t) with 12 V, 0.5 H, 2 F"
subject: "Electrical Circuit Theory"
chapter: 3
key: "semester_3/electrical_circuit_theory:classical-vc-12v-1ohm-05h-2f"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-vc-12v-1ohm-05h-2f
site: BCT Engineering
---

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

Electrical Circuit Theory · Chapter 3: Transient analysis in RLC circuit by direct solution

- Asked in: 2068 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/classical-vc-12v-1ohm-05h-2f
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-3

## Question

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)]

## 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.

## Answer

#### 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.

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are i_L(0-) and v_C(0-)?
   A. 2 A and 10 V
   B. 12 A and 0 V
   C. 2 A and 12 V
   D. 0 A and 10 V
   Answer: A. i = 12/(1+5) = 2 A; v_C = 5 x 2 = 10 V. (Key point: Initial values)

2. What circuit remains for t > 0?
   A. Series RLC with the 5 ohm included
   B. Series RLC: 12 V, 1 ohm, 0.5 H, 2 F
   C. Source-free RLC
   D. RC only
   Answer: B. The 5 ohm branch is disconnected by the switch. (Key point: Circuit after opening)

3. What is the damping type?
   A. Overdamped
   B. Underdamped
   C. Critically damped: m = -1, -1
   D. Undamped
   Answer: C. m^2 + 2m + 1 = (m+1)^2. (Key point: Damping)

4. Why is v_C'(0) = 1 V/s?
   A. 12/2 = 6 V/s
   B. 0, since v_C is continuous
   C. 10/2 = 5 V/s
   D. i_L(0)/C = 2/2
   Answer: D. Capacitor current equals the inductor current, which is continuous. (Key point: Derivative condition)

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