---
title: "Superposition, current in 0.5Ω"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:superposition-0-5-ohm-15v-20v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/superposition-0-5-ohm-15v-20v
site: BCT Engineering
---

# Superposition, current in 0.5Ω

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2069 Asar, 6 marks
- Video: [Superposition Theorem](https://www.youtube.com/watch?v=ZJ8zD8m-B1Q) — Neso Academy
- 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-0-5-ohm-15v-20v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

State superposition theorem. Find the current in 0.5Ω resistor in the following network shown, by using superposition theorem. [Figure: RL = 0.5 Ω in the top branch; 1 Ω with a 15 V source on the left; 1 Ω and 2 Ω in the middle; 1 Ω with a 20 V source on the right.]

## Short answer

With 15 V source alone, current is 4.5 A from P to Q; with 20 V source alone, it is 8 A from Q to P. Superposition gives net current of 3.5 A from Q to P in the 0.5 Ω resistor.

## Answer

#### Statement
**Superposition theorem:** in a linear bilateral network with more than one independent source, the response (current or voltage) in any element is the algebraic sum of the responses due to each source acting alone, with the other voltage sources replaced by short circuits and the other current sources by open circuits.

#### Reading the figure
Terminals P (top left) and Q (top right) are joined by R_L = 0.5 Ω. At P: 1 Ω in series with the 15 V battery (+ on top) and a 1 Ω to the centre point O. At Q: 2 Ω to O and 1 Ω in series with the 20 V battery (+ on the 1 Ω side). 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 from P: 0.5 + 0.667 = 1.167 Ω.

V_P (1/1 + 1/1 + 1/1.167) = 15 → V_P = 5.25 V
V_Q = 5.25 × 0.667/1.167 = 3.0 V
I′ = (5.25 − 3)/0.5 = **4.5 A** (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 + 1/1.0) = 20 → V_Q = 8 V
V_P = 8 × 0.5/1.0 = 4 V
I″ = (8 − 4)/0.5 = **8 A** (Q to P)

#### Total
The two currents are opposite:

I = 8 − 4.5 = **3.5 A from Q to P**

**Answer: Current in the 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. Which statement of the superposition theorem is correct?
   A. Response is the algebraic sum of responses due to each source acting alone, with other voltage sources shorted and current sources opened
   B. Response is the sum of the squares of each source's response
   C. Response is found with all sources acting together and ignoring resistors
   D. Response is the product of the individual responses
   Answer: A. Applies to linear bilateral networks; each independent source acts alone. (Key point: Statement of superposition)

2. With only the 15 V source active, what is the current in the 0.5 ohm and its direction?
   A. 3.5 A from P to Q
   B. 4.5 A from P to Q
   C. 8 A from Q to P
   D. 4.5 A from Q to P
   Answer: B. V_P = 5.25 V, V_Q = 3 V, so I' = (5.25 - 3)/0.5 = 4.5 A from P to Q. (Key point: Case 1 result)

3. Why are the two contributions subtracted?
   A. The 15 V source is smaller than the 20 V source
   B. Superposition always subtracts the second case
   C. They flow through the 0.5 ohm in opposite directions (4.5 A P to Q, 8 A Q to P)
   D. Because the 20 V source is shorted
   Answer: C. Contributions add algebraically; opposite directions give 8 - 4.5 = 3.5 A from Q to P. (Key point: Combining the effects)

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