---
title: "Laplace: iL(t) after switch opens, 4 V"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-il-4v-r-l-c"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-il-4v-r-l-c
site: BCT Engineering
---

# Laplace: iL(t) after switch opens, 4 V

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

- Asked in: 2068 Shrawan, 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-4v-r-l-c
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

The circuit shown below is in steady state with switch 'S' closed. The switch is opened at t = 0. Using Laplace Transform method, find iL(t) in the circuit. Where R1 = R2 = 2 Ω, L = 0.5 H, C = 1 F.
[Figure: 4 V source (+ at top) with switch S (t = 0) and resistor R1 in the top wire; resistor R2 as a shunt branch to the bottom wire; then inductor L in series, followed by capacitor C as a shunt with voltage vc(t)]

## Short answer

Initial capacitor voltage is 2 V, inductor current zero. With switch open, KVL in s-domain yields IL(s) = −4/(s²+4s+2). Inverse Laplace gives iL(t) = −1.414(e^(−0.586t) − e^(−3.414t)) A, t > 0.

## Answer

#### Initial conditions (t = 0⁻, S closed, steady state)
Capacitor open, inductor short. The inductor branch carries no steady current (it ends at the open capacitor), so R₁ and R₂ form a divider from the 4 V source:

```text
iL(0⁻) = 0
vc(0⁻) = 4 × R2/(R1 + R2) = 4 × 2/4 = 2 V
```

#### For t > 0 (S open)
The source and R₁ are disconnected. What remains is a source-free series loop of R₂, L and C (R₂ is the shunt resistor, which forms a loop with L and C):

R = 2 Ω, L = 0.5 H, C = 1 F, i_L(0) = 0, v_c(0) = 2 V.

Let i_L be the current from the R₂ node through L into C. KVL (the capacitor is a source of 2/s that drives current backwards):

```text
(R + sL + 1/(sC)) IL = −vc(0)/s
(2 + 0.5 s + 1/s) IL = −2/s
IL(s) = −4 / (s² + 4s + 2)
```

#### Inverse transform
Roots of s² + 4s + 2: s = −2 ± √2, i.e. −0.586 and −3.414 (overdamped).

```text
IL(s) = −4 / [(s + 0.586)(s + 3.414)]
      = −1.414/(s + 0.586) + 1.414/(s + 3.414)
```
**i_L(t) = −1.414 (e^(−0.586t) − e^(−3.414t)) A, t > 0**

The negative sign means the actual current flows backwards (from the capacitor through L and R₂), as the 2 V capacitor discharges. Its peak magnitude is about 0.81 A at t ≈ 0.62 s. i_L(0) = 0 ✓ and i_L′(0) = −4 A/s = −v_c(0)/L ✓. At t = 1 s the formula gives −0.741 A, the same as a numerical solution of the loop equations.

## 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-)?
   A. iL = 0 and vc = 4 x R2/(R1+R2) = 2 V
   B. iL = 1 A and vc = 2 V
   C. iL = 0 and vc = 4 V
   D. iL = 2 A and vc = 0
   Answer: A. With the capacitor open, no current flows through L, so R1 and R2 form a divider that sets vc. (Key point: Initial conditions)

2. What remains for t > 0?
   A. R1, L and C in series with the 4 V source
   B. A source-free series loop of R2, L and C with vc(0) = 2 V
   C. R2 alone
   D. Parallel RLC with the source
   Answer: B. The switch removes the source and R1; R2 closes a loop with L and C. (Key point: Circuit after opening)

3. What are the roots of s^2 + 4s + 2 and the response type?
   A. -2 +/- j sqrt(2): underdamped
   B. -2 repeated: critical
   C. -2 +/- sqrt(2): overdamped
   D. -4 and -2: underdamped
   Answer: C. Discriminant 16 - 8 > 0 gives real roots -0.586 and -3.414. (Key point: Response type)

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