---
title: "Laplace: iL(t) and vc(t) after switch opens"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-il-vc-48v-5ohm-switch"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-il-vc-48v-5ohm-switch
site: BCT Engineering
---

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

Electrical Circuit Theory · Chapter 4: Transient analysis in RLC circuit by Laplace Transform

- Asked in: 2074 Asoj, 8 marks
- Video: [Numerical on Laplace transform, finding current i(t).](https://www.youtube.com/watch?v=RwLyhIkZgbw) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-il-vc-48v-5ohm-switch
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

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]

## 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.

## Answer

#### 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:

```text
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:

```text
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:

```text
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).

```text
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).

## Check yourself

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

1. What are iL(0-) and vc(0-) with the switch closed?
   A. iL = 48/(5+2) = 6.857 A, vc = 2 x 6.857 = 13.714 V
   B. iL = 48/12 = 4 A, vc = 8 V
   C. iL = 0, vc = 48 V
   D. iL = 48/5 = 9.6 A, vc = 48 V
   Answer: A. The closed switch shorts one 5 ohm; the capacitor is open and inductor short, so only 5 and 2 ohm carry current. (Key point: Initial values)

2. What changes when the switch opens?
   A. The source is removed
   B. The second 5 ohm comes into the circuit, giving 10 ohm in series with the source
   C. The inductor is removed
   D. The capacitor is shorted
   Answer: B. The switch was shorting the second 5 ohm; opening puts it in series with the first. (Key point: Change at t = 0)

3. What are the final values and the response type?
   A. iL(inf) = 6.857 A, vc(inf) = 13.7 V; underdamped
   B. iL(inf) = 4 A, vc(inf) = 8 V; overdamped
   C. iL(inf) = 4 A, vc(inf) = 8 V; underdamped
   D. iL(inf) = 4.8 A, vc(inf) = 48 V; critically damped
   Answer: C. Final DC: 48/(10+2) = 4 A, vc = 8 V. Roots -1.05 +/- j0.3122 are complex, so underdamped. (Key point: Final and type of response)

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