---
title: "Laplace: i1, i2 in unenergised two-mesh network"
subject: "Electrical Circuit Theory"
chapter: 4
key: "semester_3/electrical_circuit_theory:laplace-two-mesh-unenergised-100v-10ohm"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/laplace-two-mesh-unenergised-100v-10ohm
site: BCT Engineering
---

# Laplace: i1, i2 in unenergised two-mesh network

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

- Asked in: 2072 Chaitra, 8 marks; 2069 Chaitra, 8 marks; 2068 Baisakh, 8 marks (asked 3 times)
- Video: [Determine the current i(t) assuming zero initial conditions in the network elements.](https://www.youtube.com/watch?v=RWwxbWDy-4w) — MANTHAN FOUNDATION
- 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-mesh-unenergised-100v-10ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-4

## Question

Using Laplace transform method, find the currents i1(t) and i2(t) for t > 0 in the circuit shown in the figure below. Assume the network is unenergised before the switch is closed.
[Figure: 100 V source with a switch (closes at t = 0), a 10 Ω resistor and a 1 H inductor in the top wire of the left mesh (i1); a 10 Ω resistor in the middle shared vertical branch; a 1 H inductor and a 10 Ω resistor in the right mesh (i2)]

## Short answer

Solving s-domain mesh equations yields i1(t)=6.667−5e^(−10t)−1.667e^(−30t) A and i2(t)=3.333−5e^(−10t)+1.667e^(−30t) A, found by partial fractions and inverse Laplace transform.

## Answer

Initial currents and energy are zero, so only the s-domain impedances are needed: R = 10 Ω, sL = s Ω, source 100/s. Both mesh currents are taken clockwise.

#### s-domain mesh equations
```text
Mesh 1: (10 + s + 10) I1 − 10 I2 = 100/s
Mesh 2: −10 I1 + (10 + s + 10) I2 = 0
```
Mesh 2 gives I2 = 10 I1/(s + 20). Substituting in mesh 1:

```text
[(s + 20) − 100/(s + 20)] I1 = 100/s
```
Using the determinant Δ = (s + 20)² − 100 = (s + 10)(s + 30):

```text
I1(s) = 100 (s + 20) / [ s (s + 10)(s + 30) ]
I2(s) = 1000 / [ s (s + 10)(s + 30) ]
```

#### Partial fractions
```text
I1 = 6.667/s − 5/(s + 10) − 1.667/(s + 30)
     (K1 = 100×20/(10×30);  K2 = 100×10/((−10)(20));  K3 = 100×(−10)/((−30)(−20)))
I2 = 3.333/s − 5/(s + 10) + 1.667/(s + 30)
```

#### Inverse Laplace transform
**i1(t) = 6.667 − 5e^(−10t) − 1.667e^(−30t) A, t > 0**

**i2(t) = 3.333 − 5e^(−10t) + 1.667e^(−30t) A, t > 0**

Check: i1(0) = 0 and i2(0) = 0 (inductors open the circuit initially); final values i1 = 100/(10 + 5) = 6.667 A and i2 = i1/2 = 3.333 A ✓. Values at t = 0.1 s: i1 = 4.744 A, i2 = 1.577 A, which match a numerical inverse Laplace transform.

## Check yourself

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

1. Why are no initial-condition sources needed here?
   A. The network is unenergised, so all initial inductor currents are zero
   B. Inductors are replaced by open circuits
   C. The source is 0 at t = 0
   D. Laplace transforms ignore initial conditions
   Answer: A. With zero initial currents each inductor is just sL. (Key point: s-domain model)

2. Which is mesh 1 in the s-domain?
   A. (10 + s) I1 - 10 I2 = 100
   B. (10 + s + 10) I1 - 10 I2 = 100/s
   C. (10 + s + 10) I1 + 10 I2 = 100/s
   D. (10 + s + 10) I1 - 10 I2 = 100
   Answer: B. Mesh 1 has 10 + s in its own path plus the shared 10 ohm, and the step source is 100/s. (Key point: Mesh equation)

3. What are the steady-state currents?
   A. i1 = 5 A and i2 = 5 A
   B. i1 = 10 A and i2 = 0
   C. i1 = 6.667 A and i2 = 3.333 A
   D. i1 = 3.333 A and i2 = 6.667 A
   Answer: C. Inductors short at DC: i1 = 100/(10 + 10 parallel 10) = 100/15 = 6.667 A, and it splits equally. (Key point: Final values)

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