---
title: "Laplace: inductor and capacitor currents, 12 V"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-inductor-capacitor-current-12v-2ohm"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-inductor-capacitor-current-12v-2ohm
site: BCT Engineering
---

# Laplace: inductor and capacitor currents, 12 V

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

- Asked in: 2072 Chaitra, 8 marks
- Video: [Laplace Transform Electric Circuit Example](https://www.youtube.com/watch?v=8v9oxlcOztE) — Iain Explains Signals, Systems, and Digital Comms
- 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-inductor-capacitor-current-12v-2ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

Using Laplace Transform method, find the current of inductor and capacitor for t > 0 in the circuit shown in figure below.
[Figure: 12 V source with a switch (t = 0) and a 2 Ω resistor in series; then two parallel branches: a 4 Ω resistor in series with a 1/3 F capacitor, and a 1 H inductor]

## Short answer

Zero initial conditions; find total impedance, then branch voltage V(s), and currents IL(s) and IC(s). Inverse transform yields underdamped responses: iL(t)=6+e^−0.9167t(−6cos0.3997t+6.255sin0.3997t) A, iC(t)=e^−0.9167t(2cos0.3997t−4.587sin0.3997t) A.

## Answer

Zero initial conditions (assumed, switch closes at t = 0). The source is 12/s. The 4 Ω + 1/3 F branch has impedance Z_c = 4 + 3/s, and the inductor has Z_L = s.

#### Total impedance and source current
```text
Z_p = Z_c Z_L / (Z_c + Z_L) = s (4s + 3) / (s² + 4s + 3)
Z_total = 2 + Z_p = (6s² + 11s + 6) / (s² + 4s + 3)

I(s) = (12/s) / Z_total = 12 (s + 1)(s + 3) / [ s (6s² + 11s + 6) ]
```

#### Voltage across the parallel branches
V(s) = I(s) Z_p = 12 (4s + 3) / (6s² + 11s + 6)

#### Inductor current
```text
IL(s) = V/s = 12 (4s + 3) / [ s (6s² + 11s + 6) ]
      = 6/s − (6s + 3) / (s² + 1.8333 s + 1)
```
The roots of 6s² + 11s + 6 = 0 are s = −0.9167 ± j0.3997 (complex, underdamped).

**i_L(t) = 6 + e^(−0.9167t) (−6 cos 0.3997t + 6.255 sin 0.3997t) A**

(final value 12/2 = 6 A, since the inductor shorts the capacitor branch at d.c.)

#### Capacitor current
```text
IC(s) = V/Z_c = 12 s / (6s² + 11s + 6) = 2s / [(s + 0.9167)² + 0.3997²]
```
**i_C(t) = e^(−0.9167t) (2 cos 0.3997t − 4.587 sin 0.3997t) A**

Check: i_C(0⁺) = 12/(2 + 4) = 2 A (capacitor acts as a short, inductor open) ✓, i_L(0) = 0 ✓, i_C(∞) = 0 ✓. Sample t = 0.5 s: i_L = 3.067 A, i_C = 0.664 A, equal to the numerical Laplace inversion and to a direct numerical solution.

## Check yourself

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

1. What are iC(0+) and iL(0+)?
   A. iC(0+) = 2 A, iL(0+) = 0
   B. iC(0+) = 0, iL(0+) = 6 A
   C. iC(0+) = 3 A, iL(0+) = 3 A
   D. iC(0+) = 12 A, iL(0+) = 0
   Answer: A. Capacitor shorts, inductor opens: iC = 12/(2 + 4) = 2 A. (Key point: Initial states)

2. How is IL(s) found from the parallel voltage?
   A. IL = V x sL
   B. IL = V/(sL) = V/s where V = I Zp is the common branch voltage
   C. IL = V/Zc
   D. IL = I Zp
   Answer: B. The inductor branch voltage is V, so IL = V/(sL) with L = 1. (Key point: Inductor current)

3. What do iL and iC approach at infinity?
   A. iL = 0, iC = 3 A
   B. iL = 2 A, iC = 2 A
   C. iL = 6 A (12/2, inductor shorts the capacitor branch), iC = 0
   D. iL = 3 A, iC = 3 A
   Answer: C. At DC the inductor is a short, current is 12/2 = 6 A, and the capacitor carries none. (Key point: Final values)

## Close variants set in other exams

- [Laplace: inductor current and voltage, 20 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-inductor-current-voltage-20v-2ohm) — Electrical Circuit Theory, chapter 4, asked 3×
- [Laplace: capacitor current and voltage, 10 V, 10 μF](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Alaplace-capacitor-current-voltage-10v-5ohm) — Electrical Circuit Theory, chapter 4, asked 3×

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