---
title: "Mesh analysis: find Vs for 6.76 A in 3 Ω"
subject: "Electrical Circuit Theory"
chapter: 1
key: "semester_3/electrical_circuit_theory:mesh-analysis-find-vs-6-76a"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/mesh-analysis-find-vs-6-76a
site: BCT Engineering
---

# Mesh analysis: find Vs for 6.76 A in 3 Ω

Electrical Circuit Theory · Chapter 1: Network Analysis of AC circuit & Dependent Sources

- Asked in: 2067 Asar (old course), 8 marks
- Video: [Mesh Analysis Example Problem #1](https://www.youtube.com/watch?v=jewkmjVjbmw) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/mesh-analysis-find-vs-6-76a
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-1

## Question

Using mesh analysis, determine the value of Vs so that the current through the 3 Ω resistor is 6.76 Amp as shown in the following figure.
[Figure: voltage source Vs (+ at top) in series with a 5 Ω resistor on the left side, feeding a top-left node A. From A: a 3 Ω resistor (carrying the 6.76 A, direction left to right) and then a 2 Ω resistor along the top wire to the top-right node C; a 3 A current source (arrow pointing down-right) from A diagonally to the middle node M; from C a 10 Ω resistor diagonally to M; a 10 A current source (arrow pointing down) from C to the bottom wire; a 1 Ω resistor from M to the bottom wire. Exact connection of the 3 Ω and 2 Ω junction is not clear in the scan.]

## Short answer

Mesh currents: I2 = 6.76 A, I3 = 10 A, 3 A source gives I1 = I2 + 3 = 9.76 A. Supermesh KVL yields Vs = 5·9.76 + 5·6.76 + 10·(6.76 − 10) + 1·(9.76 − 10) = 49.96 V ≈ 50 V.

## Answer

**Assumptions (figure unclear in the scan):** Vs with the 5 Ω resistor feeds node A. The 3 Ω and 2 Ω resistors are in series from A to C and both carry the 6.76 A. The 3 A source goes from A to M, a 10 Ω resistor from C to M, a 10 A source from C to ground, and a 1 Ω resistor from M to ground.

#### Mesh currents (clockwise)

- I₁: left mesh (Vs, 5 Ω, 3 A source, 1 Ω)
- I₂: top mesh (3 Ω, 2 Ω, 10 Ω, 3 A source), so I₂ = 6.76 A
- I₃: right mesh (10 Ω, 10 A source, 1 Ω), so I₃ = 10 A (the 10 A source is only in this mesh)

#### Current source constraint

The 3 A source is shared by I₁ and I₂ (arrow A to M):

```text
I₁ − I₂ = 3  →  I₁ = 6.76 + 3 = 9.76 A
```

#### Supermesh of I₁ and I₂ (excluding the 3 A source)

KVL around Vs, 5 Ω, 3 Ω, 2 Ω, 10 Ω, 1 Ω:

```text
Vs = 5·I₁ + (3 + 2)·I₂ + 10(I₂ − I₃) + 1(I₁ − I₃)
   = 5(9.76) + 5(6.76) + 10(6.76 − 10) + (9.76 − 10)
   = 48.8 + 33.8 − 32.4 − 0.24
   = 49.96 V
```

**Answer: Vs ≈ 50 V (49.96 V).**

## Check yourself

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

1. Given the 3 A source, what is I1?
   A. I1 = I2 + 3 = 9.76 A
   B. I2 − 3 = 3.76 A
   C. 3 A
   D. 10 A
   Answer: A. The 3 A source is shared by I1 and I2 giving I1 − I2 = 3. (Key point: Current source constraint)

2. What does the supermesh KVL leave out?
   A. The 5 Ω resistor
   B. The branch containing the 3 A source
   C. The 10 A source
   D. Vs
   Answer: B. The unknown voltage across a current source is avoided by omitting it. (Key point: Supermesh KVL)

3. What is Vs?
   A. About 82 V
   B. About 33 V
   C. About 50 V (49.96 V)
   D. About 98 V
   Answer: C. 48.8 + 33.8 − 32.4 − 0.24 = 49.96 V. (Key point: Result)

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