---
title: "Laplace: current with S1 closing and S2 opening at 4 ms"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-two-switch-s1-s2-4ms"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-two-switch-s1-s2-4ms
site: BCT Engineering
---

# Laplace: current with S1 closing and S2 opening at 4 ms

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

- Asked in: 2067 Asar (old course), 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-s2-4ms
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

In the circuit shown below, switch S1 is closed at t = 0 and S2 is opened at t = 4 msec. Determine i(t) for t > 0. Assume that inductor is initially de-energized. (Use Laplace method)
[Figure: 100 V source with switch S1 (t = 0) and a 50 Ω resistor in series, then a 100 Ω resistor which has switch S2 (opens at t = 4 msec) connected in parallel across it, followed by a 0.1 H inductor]

## Short answer

For 0 < t < 4 ms, i(t) = 2(1−e^(−500t)) A. For t > 4 ms, with R = 150 Ω and shifted time, i(t) = 0.6667 + 1.0626 e^(−1500(t−0.004)) A.

## Answer

**Circuit.** 100 V source, 50 Ω and 0.1 H in series; a 100 Ω resistor is also in series, but S₂ is closed across it (so it is shorted) until S₂ opens at t = 4 ms. Inductor initially de-energised.

#### Interval 1: 0 < t < 4 ms (R = 50 Ω, L = 0.1 H)
```text
I(s) = (100/s) / (50 + 0.1 s) = 1000 / [ s (s + 500) ] = 2/s − 2/(s + 500)
```
**i(t) = 2 (1 − e^(−500t)) A**

At t = 4 ms: i = 2 (1 − e^(−2)) = 1.7293 A.

#### Interval 2: t > 4 ms (R = 50 + 100 = 150 Ω)
Let t′ = t − 0.004 and i(0′) = 1.7293 A.

```text
(150 + 0.1 s) I(s) = 100/s + 0.1 × 1.7293
I(s) = (1000/s + 1.7293) / (s + 1500)
     = 0.6667/s − 0.6667/(s + 1500) + 1.7293/(s + 1500)
     = 0.6667/s + 1.0626/(s + 1500)
```
**i(t) = 0.6667 + 1.0626 e^(−1500(t − 0.004)) A, t > 4 ms**

#### Result
| Interval | Current |
|---|---|
| 0 < t < 4 ms | 2 − 2e^(−500t) A |
| t > 4 ms | 0.6667 + 1.0626 e^(−1500(t − 0.004)) A |

Continuity at 4 ms: 0.6667 + 1.0626 = 1.7293 A ✓. After S₂ opens, the current falls from 1.729 A to the new final value 100/150 = 0.667 A with τ = 0.1/150 = 0.667 ms. Verified by numerical simulation of the switched circuit.

## Check yourself

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

1. What is i(t) for 0 < t < 4 ms?
   A. 2(1 - e^(-500t)) A
   B. 2 e^(-500t) A
   C. 0.667(1 - e^(-1500t)) A
   D. 2(1 - e^(-1500t)) A
   Answer: A. S2 shorts the 100 ohm so R = 50, tau = 0.1/50 = 1/500 and final 100/50 = 2 A. (Key point: First interval)

2. What initial current is used at 4 ms, and why?
   A. 0, as it restarts
   B. 1.7293 A, because inductor current is continuous
   C. 2 A
   D. 0.667 A
   Answer: B. i(4 ms) = 2(1 - e^(-2)) = 1.7293 A. (Key point: Continuity)

3. What happens after S2 opens?
   A. Current rises from 1.729 A to 2 A
   B. Current stays constant
   C. Current falls from 1.729 A toward 0.667 A with tau = 0.667 ms (R = 150 ohm)
   D. Current falls to 0 with tau = 2 ms
   Answer: C. R = 50 + 100 = 150, final 100/150 = 0.667 A, tau = 0.1/150. (Key point: Second interval)

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