---
title: "Superposition: current in 4 Ω"
subject: "Basic Electronics Engineering"
chapter: 1
key: "semester_2/basic_electronics_engineering:superposition-14v-12v-4ohm"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/superposition-14v-12v-4ohm
site: BCT Engineering
---

# Superposition: current in 4 Ω

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2070 Bhadra, 2+4 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-14v-12v-4ohm
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

State superposition theorem. In the following figure find the current flowing in the 4 Ω resistor using superposition theorem. [Figure: 14 V source with 5 Ω resistor in the left branch; 12 V source in the middle branch with 3 Ω resistor; 4 Ω resistor in the right branch.]

## Short answer

Superposition theorem: in a linear network, the current in any branch equals the sum of currents produced by each independent source acting alone, with other voltage sources shorted. Using this method, the current in the 4 Ω resistor is found to be 2.17 A.

## Answer

**Superposition theorem:** in a linear network containing two or more independent sources, the current through (or voltage across) any element is the algebraic sum of the currents (or voltages) produced by each source acting alone, while the other voltage sources are replaced by short circuits and current sources by open circuits.

**Assumption:** the 14 V source with 5 Ω, the 12 V source with 3 Ω and the 4 Ω resistor are three parallel branches; both sources have their positive terminals toward the top node.

#### Case 1: 14 V alone (12 V source shorted)
3 Ω ∥ 4 Ω = 12/7 = 1.714 Ω
R_total = 5 + 1.714 = 6.714 Ω
I = 14/6.714 = 2.085 A
Current in 4 Ω: I' = 2.085 × 3/(3 + 4) = 0.894 A

#### Case 2: 12 V alone (14 V source shorted)
5 Ω ∥ 4 Ω = 20/9 = 2.222 Ω
R_total = 3 + 2.222 = 5.222 Ω
I = 12/5.222 = 2.298 A
Current in 4 Ω: I'' = 2.298 × 5/(5 + 4) = 1.277 A

#### Total
I(4 Ω) = 0.894 + 1.277

**Answer: I = 2.17 A through the 4 Ω resistor**

## Check yourself

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

1. Superposition applies to which kind of network?
   A. Any non-linear network
   B. Linear networks with two or more independent sources
   C. Only a single-source network
   D. Only ac circuits
   Answer: B. The theorem holds for linear networks with multiple independent sources. (Key point: Superposition applicability)

2. With the 14 V alone, what is the total resistance?
   A. 5 + (3 ∥ 4) = 6.714 Ω
   B. 5 + 3 + 4 = 12 Ω
   C. 3 + (5 ∥ 4) = 5.222 Ω
   D. 5 ∥ 3 ∥ 4 = 1.5 Ω
   Answer: A. The 12 V source is shorted, so 3 Ω ∥ 4 Ω = 1.714 Ω in series with 5 Ω. (Key point: Case 1 equivalent)

3. With 14 V alone, current in the 4 Ω is 2.085 A × ?
   A. 4/7
   B. 3/7 (the other branch resistance over the sum)
   C. 5/9
   D. 1/2
   Answer: B. The current divider gives the 4 Ω a share of 3/(3+4): 0.894 A. (Key point: Current divider)

4. The two contributions are 0.894 A and 1.277 A in the same direction. The total is
   A. 0.383 A
   B. 2.17 A
   C. 1.15 A
   D. 2.30 A
   Answer: B. I = 0.894 + 1.277 = 2.17 A. (Key point: Total)

## Related in Basic Electronics Engineering

- [Superposition: branch currents, 6 V and 12 V](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Asuperposition-6v-12v) — Basic Electronics Engineering, chapter 1
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- [Thevenin: 4 kΩ resistor, 12 V, 10 mA](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-12v-10ma) — Basic Electronics Engineering, chapter 1, asked 2×
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## Related in other subjects

- [Superposition, current in branch AB](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-branch-ab-6v-12v) — Basic Electrical Engineering, chapter 3
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- [Superposition theorem, 5 Ω current](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-5-ohm-15v-20v) — Basic Electrical Engineering, chapter 3
- [Superposition, current in 0.5Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-0-5-ohm-15v-20v) — Basic Electrical Engineering, chapter 3