---
title: "Superposition: branch currents, 6 V and 12 V"
subject: "Basic Electronics Engineering"
chapter: 1
key: "semester_2/basic_electronics_engineering:superposition-6v-12v"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/superposition-6v-12v
site: BCT Engineering
---

# Superposition: branch currents, 6 V and 12 V

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2073 Bhadra, 6 marks
- Video: [How to Use Superposition to Solve Circuits | Engineering Circuit Analysis | (Solved Examples)](https://www.youtube.com/watch?v=uoZ0E6zxoi0) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/superposition-6v-12v
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

Calculate the current flowing in each branch using superposition theorem. [Figure: 6 V source in series with 3 Ω on the left; 6 Ω resistor in the middle branch; 3 Ω in series with 12 V source on the right.]

## Short answer

Using superposition, the branch currents are: left branch (3 Ω with 6 V) 0.4 A into the source (charging), middle 6 Ω branch 1.2 A downward, and right branch (3 Ω with 12 V) 1.6 A from the 12 V source.

## Answer

**Assumption:** the left branch is the 6 V source with 3 Ω, the middle branch is 6 Ω, and the right branch is 3 Ω with the 12 V source. Both sources have their positive terminals toward the top node. Currents are taken as positive in these directions: left branch up into the top node (I1), right branch up into the top node (I3), and middle branch down through the 6 Ω (I2).

#### Case 1: 6 V alone (12 V shorted)
6 Ω ∥ 3 Ω = 2 Ω; R_total = 3 + 2 = 5 Ω
I1' = 6/5 = 1.2 A (up into the node)
Through 6 Ω: I2' = 1.2 × 3/(6 + 3) = 0.4 A (down)
Right branch: 1.2 × 6/9 = 0.8 A flows down through its 3 Ω, so I3' = −0.8 A.

#### Case 2: 12 V alone (6 V shorted)
6 Ω ∥ 3 Ω = 2 Ω; R_total = 3 + 2 = 5 Ω
I3'' = 12/5 = 2.4 A (up into the node)
Through 6 Ω: I2'' = 2.4 × 3/9 = 0.8 A (down)
Left branch: I1'' = −1.6 A (1.6 A flows down into the 6 V branch).

#### Superposition
| Branch | 6 V alone | 12 V alone | Total |
|---|---|---|---|
| Left (3 Ω, 6 V) | +1.2 | −1.6 | −0.4 A |
| Middle (6 Ω) | 0.4 | 0.8 | 1.2 A |
| Right (3 Ω, 12 V) | −0.8 | +2.4 | +1.6 A |

Check: 1.6 − 0.4 = 1.2 A at the node. The voltage across the 6 Ω is 7.2 V; (12 − 7.2)/3 = 1.6 A and (6 − 7.2)/3 = −0.4 A, which matches.

**Answer: left branch 0.4 A (flowing into the 6 V source, i.e. charging it), middle 6 Ω branch 1.2 A downward, right branch 1.6 A delivered by the 12 V source.**

## Check yourself

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

1. With 6 V alone, what resistance does the 6 V source see?
   A. 3 + (6 ∥ 3) = 5 Ω
   B. 6 + 3 + 3 = 12 Ω
   C. 3 ∥ 6 ∥ 3 = 1.2 Ω
   D. 6 + (3 ∥ 3) = 7.5 Ω
   Answer: A. 6 Ω ∥ 3 Ω = 2 Ω plus 3 Ω = 5 Ω, so I1' = 1.2 A. (Key point: Case 1 total resistance)

2. In the 12 V alone case, why is the left branch current −1.6 A?
   A. It flows down into the 6 V branch, opposite to the positive reference (up into the node)
   B. The 6 V source is faulty
   C. The 12 V source is reversed
   D. Current cannot be negative
   Answer: A. The reference is up into the node; 1.6 A flows down through the 6 V branch, so it is negative. (Key point: Sign of the branch current)

3. What is the total current in the 6 Ω middle branch?
   A. 0.4 A
   B. 0.8 A
   C. 1.2 A
   D. 2.4 A
   Answer: C. 0.4 A + 0.8 A = 1.2 A, both downward. (Key point: Total middle branch)

4. The left branch total is −0.4 A. What does this mean?
   A. The 6 V source delivers 0.4 A
   B. The 6 V source is being charged by 0.4 A
   C. No current flows
   D. The 6 V source is shorted
   Answer: B. The net current flows into the 6 V source, so it is charged by the 12 V source. (Key point: Meaning of left branch result)

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