---
title: "Thevenin: 100, 60, 150 ohm circuit"
subject: "Basic Electronics Engineering"
chapter: 1
key: "semester_2/basic_electronics_engineering:thevenin-100-60-150"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/thevenin-100-60-150
site: BCT Engineering
---

# Thevenin: 100, 60, 150 ohm circuit

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2069 Poush, 4 marks; 2070 Magh, 6 marks (asked 2 times)
- Video: [Thevenin's Theorem | Thevenin equivalent circuit | Thevenin's Theorem example | Electrical Circuits](https://www.youtube.com/watch?v=cilFp7Zmzuo) — Electrical and Electronics Engineering
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/thevenin-100-60-150
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

Find the current I in the 150 Ω resistor R3 using Thevenin's theorem. [Figure: E1 = 100 V with R1 = 100 Ω in series in the left branch; E2 = 60 V in the middle branch (with R2 = 60 Ω in series between the middle branch and the load branch); R3 = 150 Ω load carrying current I.]

## Short answer

Thevenin's theorem: replace network with Vth and Rth. Open-circuit voltage Vth = (100−60)/(100+60) gives 75 V; Rth = 100∥60 = 37.5 Ω. Current through 150 Ω is I = 75/(37.5+150) = 0.4 A.

## Answer

**Thevenin's theorem:** a linear two-terminal network can be replaced by a voltage source Vth in series with a resistance Rth, where Vth is the open-circuit voltage and Rth is the resistance seen from the terminals with all independent sources set to zero.

**Assumption:** from the description, E1 = 100 V with R1 = 100 Ω, and E2 = 60 V with R2 = 60 Ω are two parallel source branches across the nodes to which the 150 Ω load R3 is connected. Both batteries are taken with their positive ends toward the top node.

#### Step 1: remove R3, find Vth
Loop current with R3 open:

I' = (E1 − E2)/(R1 + R2) = (100 − 60)/(100 + 60) = 0.25 A

Vth = E1 − I'R1 = 100 − 0.25 × 100 = 75 V

(Check: E2 + I'R2 = 60 + 0.25 × 60 = 75 V.)

#### Step 2: find Rth
Short both sources: Rth = R1 ∥ R2 = (100 × 60)/(100 + 60) = 37.5 Ω

#### Step 3: load current
```text
 Rth = 37.5 ohm
 +--/\/\/\--+
 |            |
 Vth = 75 V   R3 = 150 ohm  ->  I
 |            |
 +------------+
```

I = Vth/(Rth + R3) = 75/(37.5 + 150) = 75/187.5

**Answer: I = 0.4 A through the 150 Ω resistor**

## Check yourself

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

1. With R3 removed, what current circulates through E1, R1, R2, E2?
   A. 0.625 A
   B. 0.25 A
   C. 1.0 A
   D. 0.4 A
   Answer: B. I' = (100 − 60)/(100 + 60) = 0.25 A. (Key point: Open-circuit loop current)

2. What is Vth for this circuit?
   A. 60 V
   B. 100 V
   C. 75 V
   D. 85 V
   Answer: C. Vth = E1 − I'R1 = 100 − 0.25×100 = 75 V (check: 60 + 0.25×60 = 75 V). (Key point: Vth calculation)

3. How is Rth found and what is its value?
   A. R1 + R2 = 160 Ω
   B. R1 ∥ R2 = 37.5 Ω with sources shorted
   C. R1 ∥ R2 ∥ R3 = 29.8 Ω
   D. R1 ∥ R2 = 37.5 Ω with sources left in place
   Answer: B. Voltage sources are shorted, leaving R1 ∥ R2 = 6000/160 = 37.5 Ω. (Key point: Rth calculation)

4. What is the current in R3 = 150 Ω?
   A. 0.5 A
   B. 0.4 A
   C. 0.25 A
   D. 0.75 A
   Answer: B. I = 75/(37.5 + 150) = 0.4 A. (Key point: Load current)

## Related in Basic Electronics Engineering

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