---
title: "Laplace: i1, i2 with 2 V charged capacitor"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-i1-i2-charged-c-2v"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-i1-i2-charged-c-2v
site: BCT Engineering
---

# Laplace: i1, i2 with 2 V charged capacitor

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

- Asked in: 2073 Shrawan
- 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-i1-i2-charged-c-2v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

Using Laplace Transform method in the circuit shown in figure below. Find the current i1 and i2 for t > 0.
[Figure: a capacitor C = 1 F charged to 2 V (+ on top plate) in the left mesh (current i1) with a switch (t = 0) in the top wire; an inductor L = 1/8 H in the middle shared branch; a resistor R = 1/9 Ω in the right mesh (current i2)]

## Short answer

Set up mesh equations in s-domain with initial capacitor voltage 2 V and zero inductor current. Solve for I1(s) and I2(s), partial fractions give overdamped poles −1 and −8. Inverse transform yields i1(t)=18.29e^−8t−0.286e^−t A, i2(t)=20.57e^−8t−2.571e^−t A.

## Answer

**Assumptions:** both mesh currents i₁ (left) and i₂ (right) are clockwise, the capacitor (C = 1 F, +ve on top plate, v_C(0) = 2 V) discharges through the circuit, and there is no initial inductor current. The source-free circuit has C in the left mesh, L = 1/8 H shared, and R = 1/9 Ω in the right mesh.

#### s-domain mesh equations
The charged capacitor becomes a source V₀/s = 2/s in series with 1/(sC) = 1/s.

```text
Mesh 1: 2/s = (1/s) I1 + (s/8)(I1 − I2)
Mesh 2: 0 = (s/8)(I2 − I1) + (1/9) I2
```
From mesh 2: I2 (s/8 + 1/9) = (s/8) I1, so I2 = 9s I1 / (9s + 8).

Substitute in mesh 1 ( (s/8)(I1 − I2) = (s/8) I1 × 8/(9s + 8) = s I1/(9s + 8) ):

```text
2/s = I1 [ 1/s + s/(9s + 8) ] = I1 (9s + 8 + s²) / [ s (9s + 8) ]
I1(s) = 2 (9s + 8) / (s² + 9s + 8) = 2 (9s + 8) / [(s + 1)(s + 8)]
I2(s) = 18 s / [(s + 1)(s + 8)]
```

#### Partial fractions
```text
I1 = −(2/7)/(s + 1) + (128/7)/(s + 8)
I2 = −(18/7)/(s + 1) + (144/7)/(s + 8)
```

#### Time domain
**i₁(t) = 18.29 e^(−8t) − 0.286 e^(−t) A, t > 0**

**i₂(t) = 20.57 e^(−8t) − 2.571 e^(−t) A, t > 0**

Check: i₁(0) = 18.29 − 0.286 = 18 A and i₂(0) = 20.57 − 2.571 = 18 A. At t = 0⁺ the inductor carries no current, so the 2 V capacitor drives 2/(1/9) = 18 A through R ✓. The system is overdamped (roots −1 and −8). Values at t = 0.5 s: i₁ = 0.1616 A, i₂ = −1.183 A, matching the numerical inverse transform.

## Check yourself

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

1. How is the charged capacitor modelled?
   A. As a series source V0/s = 2/s with impedance 1/(sC)
   B. As a source of 2 V constant
   C. As an open circuit
   D. As a source of 2s
   Answer: A. Initial voltage V0 becomes V0/s in series with 1/(sC). (Key point: Charged capacitor in s-domain)

2. Why is i1(0+) = i2(0+) = 18 A?
   A. The capacitor acts as a short, so 18 A comes from the source
   B. The inductor carries no current at 0+, so the 2 V capacitor drives 2/(1/9) = 18 A through R
   C. The inductor acts as a short
   D. The resistor is 1/9 ohm and the source is 18 V
   Answer: B. L is open at 0+ so the loop is C and R only: 2 V / (1/9 ohm) = 18 A. (Key point: Initial currents)

3. Why is the response overdamped?
   A. The roots are complex
   B. The roots are equal
   C. The denominator s^2 + 9s + 8 has real distinct roots -1 and -8
   D. There are no roots
   Answer: C. (s+1)(s+8) gives real negative roots, so the currents are sums of decaying exponentials. (Key point: Response type)

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