---
title: "Laplace: loop currents with 10 V charged capacitor"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-loop-currents-10v-charged-capacitor"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-loop-currents-10v-charged-capacitor
site: BCT Engineering
---

# Laplace: loop currents with 10 V charged capacitor

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

- Asked in: 2071 Chaitra, 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-loop-currents-10v-charged-capacitor
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

Using Laplace transform method, find the loop currents i1 and i2 for t > 0 in the circuit shown in the following figure.
[Figure: a capacitor C = 1 F charged to 10 V (+ on top plate) in the left mesh (current i1) with a switch (t = 0) in the top wire; an inductor L = 1/6 H in the middle shared branch; a resistor R = 1/7 Ω in the right mesh (current i2). The scan prints R = 1/7 F, taken here as 1/7 Ω.]

## Short answer

Mesh analysis in s-domain with initial capacitor voltage 10 V and zero inductor current. Poles at s=−1,−6 (overdamped). Inverse transform yields i1(t)=72e^−6t−2e^−t A, i2(t)=84e^−6t−14e^−t A.

## Answer

**Assumptions:** the resistor value is taken as R = 1/7 Ω (the scan shows 1/7 "F"); both mesh currents are clockwise; the capacitor (C = 1 F, v_C(0) = 10 V, + on top) discharges; the inductor L = 1/6 H is the shared branch with zero initial current.

#### s-domain mesh equations
```text
Mesh 1: 10/s = (1/s) I1 + (s/6)(I1 − I2)
Mesh 2: 0 = (s/6)(I2 − I1) + (1/7) I2
```
Mesh 2: I2 (s/6 + 1/7) = (s/6) I1 → I2 = 7s I1 / (7s + 6).

Then (s/6)(I1 − I2) = s I1/(7s + 6), and mesh 1 gives

```text
10/s = I1 (7s + 6 + s²) / [ s (7s + 6) ]
I1(s) = 10 (7s + 6) / (s² + 7s + 6) = 10 (7s + 6) / [(s + 1)(s + 6)]
I2(s) = 70 s / [(s + 1)(s + 6)]
```

#### Partial fractions
```text
I1 = −2/(s + 1) + 72/(s + 6)
I2 = −14/(s + 1) + 84/(s + 6)
```

#### Time domain
**i₁(t) = 72 e^(−6t) − 2 e^(−t) A, t > 0**

**i₂(t) = 84 e^(−6t) − 14 e^(−t) A, t > 0**

Check: i₁(0) = i₂(0) = 70 A = 10/(1/7), which is correct because the inductor blocks current changes at t = 0⁺ and the capacitor voltage is across R. Roots −1 and −6 (overdamped). At t = 0.5 s: i₁ = 2.372 A, i₂ = −4.309 A, equal to the numerical inverse Laplace transform.

## Check yourself

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

1. From mesh 2, how are I2 and I1 related?
   A. I2 = 7s I1/(7s + 6)
   B. I2 = I1/2
   C. I2 = 6 I1/(7s + 6)
   D. I2 = I1 (7s + 6)/7s
   Answer: A. I2 (s/6 + 1/7) = (s/6) I1, so I2 = 7s I1/(7s + 6). (Key point: Mesh 2 relation)

2. Why are i1(0+) and i2(0+) both 70 A?
   A. The capacitor is a short
   B. L blocks any current change, so the 10 V capacitor sits across R = 1/7 ohm: 10/(1/7) = 70 A
   C. The inductor is a short
   D. The source is 70 V
   Answer: B. No current in L at 0+, so the capacitor drives only R. (Key point: Initial currents)

3. What do the roots of s^2 + 7s + 6 tell you?
   A. Complex roots: underdamped
   B. Equal roots: critical
   C. Roots -1 and -6: overdamped
   D. Positive roots: unstable
   Answer: C. (s+1)(s+6) = s^2 + 7s + 6 has real negative distinct roots. (Key point: Roots)

## Close variants set in other exams

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