---
title: "Superposition, current in branch AB"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:superposition-branch-ab-6v-12v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-branch-ab-6v-12v
site: BCT Engineering
---

# Superposition, current in branch AB

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2071 Chaitra, 8 marks
- Video: [Superposition Theorem Solved Example Problem | Electrical Engineering](https://www.youtube.com/watch?v=4dsNAiZYwDw) — 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/superposition-branch-ab-6v-12v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Use the superposition theorem to determine the current in the branch 'AB' of the network shown below: [Figure: branch AB contains R = 6 Ω; 6 Ω and 24 Ω resistors with a 6 V source on the left; 24 Ω resistors in the middle and right; 6 Ω with a 12 V source on the right; 2 Ω resistor in the bottom branch.]

## Short answer

Use superposition: with 6 V source alone, current is 0.25 A from A to B; with 12 V source alone, it is 0.5 A from B to A. Net current in branch AB is 0.25 A from B to A.

## Answer

#### Reading the figure
Terminals A and B are joined by the 6 Ω resistor (R_L). The bottom-left corner is wired to G₁ (the bottom of the left 24 Ω resistor) and the bottom-right corner is wired to G₂; G₁ and G₂ are joined by the 2 Ω resistor. Take G₂ as the reference (0 V).

- A to G₁: 24 Ω, 24 Ω (diagonal) and the branch 6 Ω in series with the 6 V battery (+ on the 6 Ω side).
- B to G₂: 24 Ω, 24 Ω (diagonal) and the branch 6 Ω in series with the 12 V battery (+ on the 6 Ω side).
- 24 Ω ∥ 24 Ω = 12 Ω on each side.

#### Case 1: 6 V alone (12 V source shorted)
Right side from B to ground: 12 Ω ∥ 6 Ω = 4 Ω.

Node A: (V_A − V_G1 − 6)/6 + (V_A − V_G1)/12 + (V_A − V_B)/6 = 0
Node B: V_B/4 + (V_B − V_A)/6 = 0
Node G₁: V_G1/2 = (V_A − V_G1 − 6)/6 + (V_A − V_G1)/12

Solving: V_A = 2.5 V, V_B = 1.0 V, V_G1 = −0.5 V

I′(AB) = (2.5 − 1.0)/6 = **0.25 A** from A to B.

#### Case 2: 12 V alone (6 V source shorted)
Left side from A to G₁: 12 Ω ∥ 6 Ω = 4 Ω.

Node A: (V_A − V_G1)/4 + (V_A − V_B)/6 = 0
Node B: (V_B − 12)/6 + V_B/12 + (V_B − V_A)/6 = 0
Node G₁: V_G1/2 = (V_A − V_G1)/4

Solving: V_A = 3.0 V, V_B = 6.0 V, V_G1 = 1.0 V

I″(AB) = (3 − 6)/6 = **−0.5 A**, that is 0.5 A from B to A.

#### Total
I(AB) = 0.25 − 0.5 = −0.25 A

The current in branch AB is **0.25 A flowing from B to A**.

Check with both sources together: V_A = 5.5 V and V_B = 7.0 V, so I = (5.5 − 7.0)/6 = −0.25 A.

**Answer: 0.25 A from B to A**

Assumption: the polarities of both batteries are read from the figure with the + terminal on the 6 Ω side.

## Check yourself

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

1. With 6 V acting alone, what is done with the 12 V source?
   A. Shorted, keeping its 6 ohm
   B. Opened
   C. Removed with its 6 ohm
   D. Doubled
   Answer: A. Voltage sources replaced by shorts. (Key point: Source handling)

2. What is I' in AB from the 6 V source alone?
   A. 0.5 A from B to A
   B. 0.25 A from A to B
   C. 0.25 A from B to A
   D. 1 A
   Answer: B. (2.5 - 1.0)/6. (Key point: Partial 1)

3. What is I'' from the 12 V alone?
   A. 0.5 A A to B
   B. 0.25 A
   C. 0.5 A from B to A
   D. 1 A
   Answer: C. (3 - 6)/6 = -0.5. (Key point: Partial 2)

4. What is the net current in AB?
   A. 0.75 A from A to B
   B. 0.25 A from A to B
   C. 0.75 A from B to A
   D. 0.25 A from B to A
   Answer: D. 0.25 - 0.5 = -0.25 A, confirmed by the full circuit. (Key point: Total)

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