---
title: "Superposition: 90, 50, 130 ohm circuit"
subject: "Basic Electronics Engineering"
chapter: 1
key: "semester_2/basic_electronics_engineering:superposition-90-50-130"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/superposition-90-50-130
site: BCT Engineering
---

# Superposition: 90, 50, 130 ohm circuit

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2067 Chaitra, 5 marks
- Video: [Superposition Theorem Solved Example Problem | Circuit Analysis](https://www.youtube.com/watch?v=auZ3aXOR9-I) — Electrical and Electronics Engineering
- 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-90-50-130
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

Describe the principle of superposition theorem by solving the following problem. [Figure: E1 = 100 V with R1 = 90 Ω in series on the left; R2 = 50 Ω with E2 = 60 V in the middle branch; load RL = 130 Ω.] Find the current I through RL.

## Short answer

Superposition theorem states that in a linear network with multiple sources, the response in any element is the algebraic sum of the responses caused by each source acting alone (other voltage sources shorted). For this circuit, analyzing each source separately and summing gives I = 0.458 A through RL.

## Answer

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

**Assumption:** both batteries have their positive ends toward the top node, so both drive current through RL. E1 = 100 V, R1 = 90 Ω, E2 = 60 V, R2 = 50 Ω, RL = 130 Ω.

#### Case 1: E1 acting alone (E2 shorted)
R2 ∥ RL = (50 × 130)/180 = 36.11 Ω
R_total = 90 + 36.11 = 126.11 Ω
I1 (from E1) = 100/126.11 = 0.793 A
Current in RL (current divider): I' = 0.793 × 50/(50 + 130) = 0.2203 A

#### Case 2: E2 acting alone (E1 shorted)
R1 ∥ RL = (90 × 130)/220 = 53.18 Ω
R_total = 50 + 53.18 = 103.18 Ω
I2 (from E2) = 60/103.18 = 0.5815 A
Current in RL: I'' = 0.5815 × 90/(90 + 130) = 0.2379 A

Both currents flow in the same direction through RL.

#### Total
I = I' + I'' = 0.2203 + 0.2379

**Answer: I = 0.458 A through RL (130 Ω)**

## Check yourself

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

1. What does the superposition theorem state?
   A. Response in any element is the algebraic sum of the responses due to each source acting alone
   B. Response is the product of the responses of each source
   C. Only the largest source matters
   D. Power is additive for sources
   Answer: A. The total response is the algebraic sum of the responses from each independent source acting alone. (Key point: Statement)

2. When E1 acts alone, how is E2 treated?
   A. Opened
   B. Replaced by a short circuit
   C. Doubled
   D. Left in the circuit
   Answer: B. Other independent voltage sources are replaced by short circuits. (Key point: Treatment of other sources)

3. With E1 alone, how is the RL current found?
   A. R2 ∥ RL in series with R1 gives the total current, then a current divider gives the RL share: 0.2203 A
   B. Divide E1 by RL
   C. Add R1, R2 and RL in series
   D. Use E1/R1 directly
   Answer: A. Total current 0.793 A divides as 50/(50+130) into RL: 0.2203 A. (Key point: Case 1 method)

4. Both currents flow the same way through RL. What is the total?
   A. 0.017 A
   B. 0.458 A
   C. 0.793 A
   D. 0.238 A
   Answer: B. I = 0.2203 + 0.2379 = 0.458 A. (Key point: Total)

## Close variants set in other exams

- [Thevenin: 90, 50, 130 ohm circuit](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-90-50-130) — Basic Electronics Engineering, chapter 1, asked 3×

## Related in Basic Electronics Engineering

- [Thevenin: 100, 60, 150 ohm circuit](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-100-60-150) — Basic Electronics Engineering, chapter 1, asked 2×
- [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

## Related in other subjects

- [Superposition theorem, current in RL](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-load-2-ohm-10v-5a) — Basic Electrical Engineering, chapter 3
- [Superposition theorem and maximum power to RL](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Asuperposition-max-power-transfer) — Fundamental of Electrical and Electronics Engineering, chapter 1
- [Superposition, current IL through 2Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asuperposition-il-2-ohm-5a-10v-20v) — Basic Electrical Engineering, chapter 3
- [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