---
title: "Mesh analysis with 3 A source"
subject: "Basic Electrical and Electronics Engineering"
chapter: 1
key: "semester_1/basic_electrical_and_electronics_engineering:mesh-analysis-3a-source"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/mesh-analysis-3a-source
site: BCT Engineering
---

# Mesh analysis with 3 A source

Basic Electrical and Electronics Engineering · Chapter 1: Fundamentals of Electrical and Electronics Circuits

- Asked in: 2081 Baishakh (new course), 4 marks
- Video: [Mesh Analysis with Current Source](https://www.youtube.com/watch?v=s5svOTeKnOE) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_and_electronics_engineering/mesh-analysis-3a-source
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_and_electronics_engineering/chapter-1

## Question

Determine the mesh current using Mesh Analysis for the network shown in figure below. [Figure: left mesh has 5 Ω in the top branch, a 12 V source on the left side, and 2 Ω in the middle vertical branch; top branch carries a 3 A current source (arrow pointing left) across the top; right mesh has 6 Ω in its top branch, a 6 V source on the right side, and 1 Ω in the bottom branch shared with the bottom rail.]

## Short answer

Right mesh current is forced to -3 A by the 3 A source. KVL on left mesh: 12 = 7I₁ – 2(-3) gives I₁ = 0.857 A. Thus mesh currents are I₁ = 0.857 A (clockwise), I₂ = -3 A (3 A anticlockwise).

## Answer

**Assumptions (from the figure):** the 3 A current source is in the top branch of the right mesh in series with the 6 Ω resistor, and it forces 3 A to flow leftwards there. Mesh currents $I_1$ (left) and $I_2$ (right) are both taken clockwise. The 12 V source aids $I_1$ and the 6 V source opposes $I_2$.

**Step 1: Right mesh (current source in the branch).** Clockwise $I_2$ flows to the right along the top branch, but the source forces 3 A to the left, so

$$
I_2=-3\ \text{A}
$$

A current source in a mesh needs no KVL equation for that mesh; its current fixes the mesh current directly.

**Step 2: KVL in the left mesh.**

$$
\begin{aligned} 12 &= 5I_1 + 2(I_1-I_2)\\ 12 &= 7I_1 - 2I_2\\ 7I_1 &= 12 + 2(-3) = 6\\ I_1 &= \frac{6}{7} = 0.857\ \text{A} \end{aligned}
$$

**Step 3: Current in the 2 Ω shared branch.**

$$
I_{2\Omega}=I_1-I_2 = 0.857-(-3)=3.857\ \text{A}
$$

**Answer: $I_1 = 0.857$ A (clockwise), $I_2 = -3$ A (i.e. 3 A anticlockwise, fixed by the source); current in the 2 Ω resistor = 3.857 A.**

## Check yourself

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

1. How is the right mesh handled when a 3 A current source lies in its branch?
   A. Write KVL assuming zero voltage across the source
   B. Short-circuit the source and solve normally
   C. Fix the mesh current by the source (I2 = -3 A) and skip the KVL for that mesh
   D. Convert the source to a voltage source of 3 V
   Answer: C. A current source in an outer branch of a mesh sets that mesh current directly, so no KVL equation is needed for it. (Key point: Current source in a mesh)

2. Why is I2 = -3 A when I2 is taken clockwise?
   A. The source forces 3 A leftwards in the top branch, opposite to clockwise flow there
   B. The 6 V source is reversed
   C. Mesh currents are always negative for the right mesh
   D. The 1 Ω resistor reverses the current
   Answer: A. Clockwise I2 would go right along the top, but the source forces 3 A to the left, so I2 = -3 A. (Key point: Sign of I2)

3. Which KVL equation applies to the left mesh?
   A. 12 = 5 I1 + 2 (I1 + I2)
   B. 12 = 5 I1 + 2 (I1 - I2)
   C. 12 = 5 I1 - 2 (I1 - I2)
   D. 12 = 7 I1 + 2 I2 + 6
   Answer: B. The 2 Ω is shared, so its current is I1 - I2, giving 12 = 5I1 + 2(I1 - I2). (Key point: KVL left mesh)

4. What is the current in the 2 Ω shared resistor?
   A. 0.857 A
   B. 2.143 A
   C. 3 A
   D. 3.857 A
   Answer: D. I1 = 6/7 = 0.857 A, so I(2 Ω) = I1 - I2 = 0.857 + 3 = 3.857 A. (Key point: Shared branch current)

## Related in Basic Electrical and Electronics Engineering

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- [Nodal analysis current in 6 Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Anodal-analysis-6-ohm-current) — Basic Electrical and Electronics Engineering, chapter 1

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