---
title: "Laplace: two switches, current before and after 8 ms"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-two-switch-s1-s2-8ms"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-two-switch-s1-s2-8ms
site: BCT Engineering
---

# Laplace: two switches, current before and after 8 ms

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

- Asked in: 2068 Shrawan, 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/laplace-two-switch-s1-s2-8ms
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

In the given circuit below, switch S1 is closed at t = 0 and after 8ms, the switch S2 is opened. Find the complete expression for current in the interval 0 < t < 8ms and t > 8ms. Use Laplace Transform approach.
[Figure: 10 V source with switch S1 (t = 0) and a 5 Ω resistor in series, then a 10 Ω resistor which has switch S2 connected in parallel across it, followed by a 0.2 H inductor]

## Short answer

For 0 < t < 8 ms, R = 5 Ω, i(t) = 2(1−e^(−25t)) A. For t > 8 ms, R = 15 Ω, shifting time gives i(t) = 0.6667 − 0.3041 e^(−75(t−0.008)) A.

## Answer

**Circuit.** 10 V source, 5 Ω and 0.2 H in series. A 10 Ω resistor is in series, but S₂ is closed across it, so it is shorted until S₂ opens at t = 8 ms. Inductor initially de-energised.

#### Interval 1: 0 < t < 8 ms (S₁ closed, S₂ closed)
R = 5 Ω, L = 0.2 H.

```text
I(s) = (10/s) / (5 + 0.2 s) = 50 / [ s (s + 25) ] = 2/s − 2/(s + 25)
```
**i(t) = 2 (1 − e^(−25t)) A, 0 < t < 8 ms**

At t = 8 ms: i = 2(1 − e^(−0.2)) = 0.3625 A.

#### Interval 2: t > 8 ms (S₂ open, 10 Ω in the circuit)
Use a shifted time t′ = t − 0.008. R = 5 + 10 = 15 Ω, initial current i(0′) = 0.3625 A.

```text
(15 + 0.2 s) I(s) = 10/s + L i(0′) = 10/s + 0.2 × 0.3625
I(s) = (50/s + 0.3625) / (s + 75)
     = 0.6667/s − 0.6667/(s + 75) + 0.3625/(s + 75)
     = 0.6667/s − 0.3041/(s + 75)
```
**i(t) = 0.6667 − 0.3041 e^(−75(t − 0.008)) A, t > 8 ms**

#### Complete result
| Interval | Current |
|---|---|
| 0 < t < 8 ms | 2 − 2e^(−25t) A |
| t > 8 ms | 0.6667 − 0.3041 e^(−75(t − 0.008)) A |

The current is continuous at 8 ms (0.3625 A) because of the inductor, then rises toward 10/15 = 0.667 A with τ = 0.2/15 = 13.3 ms. A numerical simulation of the switched circuit reproduces both expressions.

## Check yourself

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

1. What is the circuit for 0 < t < 8 ms?
   A. 5 ohm and 0.2 H, because S2 shorts the 10 ohm; i = 2(1 - e^(-25t))
   B. 15 ohm and 0.2 H; i = 0.667(1 - e^(-75t))
   C. 5 ohm and 0.2 H; i = 2 e^(-25t)
   D. 5 ohm only; i = 2 A
   Answer: A. S2 closed bypasses the 10 ohm; final 10/5 = 2 A with tau = 0.2/5 = 1/25. (Key point: First interval)

2. What is the initial current for the second interval and why?
   A. 0, because the circuit restarts
   B. 0.3625 A, because inductor current is continuous across the switching
   C. 2 A, the final value of interval 1
   D. 0.667 A
   Answer: B. i(8 ms) = 2(1 - e^(-0.2)) = 0.3625 A, and an inductor current cannot jump. (Key point: Second interval initial value)

3. What are R and the final value after S2 opens?
   A. R = 5 ohm; final current 2 A
   B. R = 10 ohm; final current 1 A
   C. R = 15 ohm; final current 10/15 = 0.667 A with tau = 13.3 ms
   D. R = 15 ohm; final current 2 A
   Answer: C. The 10 ohm is now in series with 5 ohm, so 15 ohm and tau = 0.2/15. (Key point: Second interval)

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