---
title: "Superposition theorem, 5 Ω current"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:superposition-5-ohm-15v-20v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-5-ohm-15v-20v
site: BCT Engineering
---

# Superposition theorem, 5 Ω current

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2073 Shrawan, 8 marks
- Video: [Superposition Theorem Solved Example Problem | Electrical Engineering](https://www.youtube.com/watch?v=4dsNAiZYwDw) — Electrical and Electronics Engineering
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-5-ohm-15v-20v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Find the current in 5-ohm resistor in the network shown below by using superposition theorem. [Figure: 15 V source with 1 Ω on the left; 0.5 Ω on top; resistors 1 Ω, 2 Ω, 5 Ω and 1 Ω in a bridge arrangement; 20 V source on the right.]

## Short answer

The resistor in question is actually 0.5 Ω. Using superposition: with 15 V alone, current through 0.5 Ω is 4.5 A P→Q; with 20 V alone, 8 A Q→P. Net current is 3.5 A from right to left through the 0.5 Ω resistor.

## Answer

#### Reading the figure
The figure contains no 5 Ω resistor. The branch between the two top corners P and Q is **0.5 Ω**, and the question is the same circuit as the 2069 Asar question on the 0.5 Ω resistor. So the required current is the current in the **0.5 Ω** resistor.

Circuit: node P (top left) has 1 Ω in series with the 15 V battery (+ on top) on the left, and a 1 Ω going to the centre point O. Node Q (top right) has a 2 Ω going to O and a 1 Ω in series with the 20 V battery (+ on the 1 Ω side). The 0.5 Ω joins P and Q. O and the bottom rail are the reference.

#### Case 1: 15 V alone (20 V shorted)
Right side from Q to ground: 2 Ω ∥ 1 Ω = 0.667 Ω.
Path P to ground through the 0.5 Ω: 0.5 + 0.667 = 1.167 Ω.

V_P (1/1 + 1/1 + 1/1.167) = 15/1 → V_P = 15/2.857 = 5.25 V
V_Q = 5.25 × 0.667/1.167 = 3.0 V

I′ = (5.25 − 3.0)/0.5 = **4.5 A** from P to Q.

#### Case 2: 20 V alone (15 V shorted)
Left side from P to ground: 1 Ω ∥ 1 Ω = 0.5 Ω.

V_Q (1/1 + 1/2 + 1/(0.5 + 0.5)) = 20/1 → V_Q = 20/2.5 = 8 V
V_P = 8 × 0.5/1.0 = 4 V

I″ = (8 − 4)/0.5 = **8 A** from Q to P.

#### Total
The two currents oppose, so

I = 8 − 4.5 = **3.5 A from Q to P (right to left)**

Check by solving the full circuit: V_P = 9.25 V and V_Q = 11 V, so I = 1.75/0.5 = 3.5 A.

**Answer: I(0.5 Ω) = 3.5 A, from the right node to the left node.**

## Check yourself

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

1. For the 15 V case, what happens to the 20 V source?
   A. It is removed with its 1 ohm
   B. It is replaced by a short, keeping its 1 ohm
   C. Opened
   D. Doubled
   Answer: B. Voltage sources become shorts; their series resistance stays. (Key point: Turning off sources)

2. Why are the two partial currents subtracted?
   A. They are in the same direction
   B. One is a power
   C. I' = 4.5 A flows P to Q and I'' = 8 A flows Q to P, so they oppose
   D. They are perpendicular
   Answer: C. Superposition adds algebraically with direction. (Key point: Directions)

3. What is the final current?
   A. 12.5 A from P to Q
   B. 3.5 A from P to Q
   C. 8 A from Q to P
   D. 3.5 A from Q to P
   Answer: D. 8 - 4.5 = 3.5 A in the direction of the larger one. (Key point: Total)

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