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

# Thevenin: 90, 50, 130 ohm circuit

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2072 Magh, 4 marks; 2067 Mangsir, 7 marks (asked 2 times)
- Video: [Thevenin’s Theorem | Electric Circuits | Example 4.8 | Electrical Engineering](https://www.youtube.com/watch?v=4hr5YqcSeqY) — 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/thevenin-90-50-130
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

Describe the principle of Thevenin's theorem by solving the following problem. [Figure: E1 = 100 V in series with R1 = 90 Ω on the left branch; R2 = 50 Ω in series with E2 = 60 V in the middle branch (E2 polarity as in the figure); R3 = 130 Ω in the right branch.] Find the current I through R3 (130 Ω).

## Short answer

Thevenin's theorem replaces a linear network with Vth in series with Rth. Removing R3, Vth = (E1−E2)/(R1+R2) gives 74.29 V, Rth = R1∥R2 = 32.14 Ω. Load current I = 74.29/(32.14+130) = 0.458 A.

## Answer

**Thevenin's theorem:** any linear two-terminal network containing sources and resistors can be replaced by a single voltage source Vth in series with a single resistance Rth. Vth is the open-circuit voltage across the two terminals and Rth is the resistance seen from the terminals with all independent sources made zero (voltage sources shorted, current sources opened).

**Principle shown by the problem:** the load R3 is removed, the rest of the circuit is replaced by Vth and Rth, and then the load current is found from one simple series loop.

**Assumption:** the low-resolution figure does not show the polarity of E2, so both batteries are taken with their positive terminals toward the top node (both help to drive current through R3). E1 = 100 V, R1 = 90 Ω, E2 = 60 V, R2 = 50 Ω, R3 = 130 Ω. The three branches are in parallel across the same two nodes.

#### Step 1: remove R3 and find Vth
With R3 removed, only a circulating current flows through E1, R1, R2 and E2:

I' = (E1 − E2)/(R1 + R2) = (100 − 60)/(90 + 50) = 0.2857 A

Vth = E1 − I'R1 = 100 − 0.2857 × 90 = 74.29 V

(Check: 60 + 0.2857 × 50 = 74.29 V.)

#### Step 2: find Rth
Short both batteries. R1 and R2 are then in parallel as seen from the open terminals:

Rth = (90 × 50)/(90 + 50) = 4500/140 = 32.14 Ω

#### Step 3: Thevenin equivalent
```text
     Rth = 32.14 ohm
 +----/\/\/\----+
 |              |
 Vth = 74.29 V  R3 = 130 ohm
 |              |
 +--------------+
```

I = Vth/(Rth + R3) = 74.29/(32.14 + 130) = 74.29/162.14

**Answer: I = 0.458 A (about 458 mA) through R3 (130 Ω)**

## Check yourself

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

1. Thevenin's theorem replaces a linear two-terminal network by
   A. A current source in parallel with a resistance
   B. A voltage source Vth in series with a resistance Rth
   C. A resistance equal to the sum of all resistors
   D. A voltage source in parallel with Rth
   Answer: B. The Thevenin equivalent is Vth in series with Rth. (Key point: Thevenin's theorem)

2. How is Vth found in this problem?
   A. Short R3 and find the current
   B. Remove R3 and find the open-circuit voltage across its terminals
   C. Add all battery voltages
   D. Divide E1 by R1
   Answer: B. Vth is the open-circuit voltage with the load removed; here Vth = E1 − I'R1 = 74.29 V. (Key point: Finding Vth)

3. Rth is found with both batteries shorted. What is Rth here?
   A. R1 + R2 = 140 Ω
   B. R1 ∥ R2 = 32.14 Ω
   C. R1 ∥ R2 ∥ R3 = 25.4 Ω
   D. R3 = 130 Ω
   Answer: B. With sources shorted, R1 and R2 are in parallel: 90×50/140 = 32.14 Ω. (Key point: Finding Rth)

4. What is the current through R3 using the Thevenin equivalent?
   A. 0.458 A
   B. 0.571 A
   C. 0.200 A
   D. 2.31 A
   Answer: A. I = Vth/(Rth + R3) = 74.29/162.14 = 0.458 A. (Key point: Load current)

## Close variants set in other exams

- [Superposition: 90, 50, 130 ohm circuit](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Asuperposition-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×

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