---
title: "Loop current method with 3 A source"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:loop-current-2-ohm-3a-9v-12v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/loop-current-2-ohm-3a-9v-12v
site: BCT Engineering
---

# Loop current method with 3 A source

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2074 Chaitra, 6 marks
- Video: [Mesh Current Method (Loop Current Method) | Electrical Engineering](https://www.youtube.com/watch?v=0nv0MjWzGrc) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/loop-current-2-ohm-3a-9v-12v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Use loop current method to calculate the current through the 2Ω resistance for the network shown below. [Figure: 3 A current source across the top; 9 V source with 1 Ω in the left branch; 2 Ω in the middle vertical branch; 3 Ω with a 12 V source in the right branch.]

## Short answer

The 3 A source is across fixed battery voltages and does not affect the 2 Ω. Using loop currents with meshes, solve 3I1 − 2I2 = 9 and −2I1 + 5I2 = −12. Current through 2 Ω is I1 − I2 = 39/11 ≈ 3.55 A downward.

## Answer

#### Reading the figure
The 3 A current source is connected across the top between the left node L (+ of the 9 V battery) and the right node R (+ of the 12 V battery), direction left to right. Middle node M joins the 1 Ω (to L), 3 Ω (to R) and 2 Ω (to ground).

The 3 A source is in parallel with the (1 Ω + 3 Ω) path between L and R, but L and R are fixed at 9 V and 12 V by the batteries. So the source does not change the current in the 2 Ω.

#### Loop currents (clockwise)
- I₁: left loop (9 V, 1 Ω, 2 Ω)
- I₂: right loop (2 Ω, 3 Ω, 12 V)

Left loop: 9 = 1 × I₁ + 2 (I₁ − I₂)
⇒ 3 I₁ − 2 I₂ = 9

Right loop (going down through the 12 V battery from + to −): 0 = 3 I₂ + 12 + 2 (I₂ − I₁)
⇒ −2 I₁ + 5 I₂ = −12

Solving: I₁ = 1.909 A, I₂ = −1.636 A

#### Current through 2 Ω
I = I₁ − I₂ = 1.909 + 1.636 = 3.545 A (downwards)

**Answer: I₂Ω = 3.55 A (39/11 A)**

## Check yourself

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

1. Why does the 3 A source not change the current in the 2 Ω resistor?
   A. A current source is always ignored in loop analysis
   B. Nodes L and R are held at 9 V and 12 V by the batteries, so the source only adds current in the outer path
   C. The 3 A source is in series with the 2 Ω so it cancels
   D. The 3 A source has zero internal voltage so it is open circuit
   Answer: B. L and R are fixed by the batteries, so the 3 A source cannot alter the voltage across or current in the 2 Ω. (Key point: Effect of the current source)

2. Which is the left loop equation for I1 and I2 (clockwise)?
   A. 3 I1 - 2 I2 = 9
   B. 3 I1 + 2 I2 = 9
   C. I1 + 2 I2 = 9
   D. 3 I1 - 2 I2 = -9
   Answer: A. KVL: 9 = 1(I1) + 2(I1 - I2), which gives 3 I1 - 2 I2 = 9. (Key point: Left loop equation)

3. The 2 Ω is common to both loops. Its current is
   A. I1 + I2
   B. I1 only
   C. I2 only
   D. I1 - I2
   Answer: D. Both loop currents are clockwise, so in the shared branch they flow in opposite directions and the net current is the difference. (Key point: Current in shared resistor)

4. With I1 = 1.909 A and I2 = -1.636 A, the current in the 2 Ω is
   A. 0.27 A upwards
   B. 1.909 A
   C. 3.55 A (39/11 A) downwards
   D. 1.636 A
   Answer: C. I = I1 - I2 = 1.909 + 1.636 = 3.545 A, downwards. (Key point: Final result)

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