---
title: "Thevenin: current through 10 Ω"
subject: "Basic Electronics Engineering"
chapter: 1
key: "semester_2/basic_electronics_engineering:thevenin-10v-5a"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/thevenin-10v-5a
site: BCT Engineering
---

# Thevenin: current through 10 Ω

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2071 Bhadra, 4 marks
- Video: [Thevenin's Theorem - Circuit Analysis](https://www.youtube.com/watch?v=zTDgziJC-q8) — The Organic Chemistry Tutor
- 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-10v-5a
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

Determine the current through the 10 Ω resistance using Thevenin's theorem. [Figure: 10 V source in series with 15 Ω, then 10 Ω resistor in a vertical branch, then 5 Ω in series, and a 5 A current source in the last branch.]

## Short answer

Thevenin's theorem: Vth = 10V + (5A×15Ω) = 85V, Rth = 15Ω (current source opened). Current through 10Ω is I = 85/(15+10) = 3.4 A.

## Answer

**Assumed circuit:** the 10 V source with 15 Ω in series connects to node A. The 10 Ω load is the vertical branch from A to ground. The last branch is the 5 A current source with a 5 Ω resistor in series, pushing 5 A into node A (direction taken as upward toward A).

#### Step 1: remove the 10 Ω and find Vth
With the load open, the 5 A from the current source must return through the 15 Ω and the 10 V source. The 5 Ω is in series with an ideal current source, so it has no effect on the voltage at A.

Vth = 10 + 5 × 15 = 85 V

#### Step 2: find Rth
Short the 10 V source and open the current source (its branch, including the 5 Ω, becomes open):

Rth = 15 Ω

#### Step 3: load current
I = Vth/(Rth + RL) = 85/(15 + 10) = 85/25

**Answer: I = 3.4 A through the 10 Ω resistor.** (If the current source points the other way, Vth = 10 − 75 = −65 V and I = −2.6 A.)

## Check yourself

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

1. Why does the 5 Ω in series with the 5 A source not affect Vth?
   A. A series resistor with an ideal current source does not change the branch current, so it does not change the voltage at A
   B. The 5 Ω is shorted
   C. The 5 Ω is in parallel with the load
   D. It is cancelled by the 10 V source
   Answer: A. An ideal current source fixes the branch current regardless of series elements. (Key point: Current source in series with resistor)

2. What is Vth?
   A. 10 V
   B. 75 V
   C. 85 V
   D. 25 V
   Answer: C. The 5 A flows through 15 Ω and the 10 V source: Vth = 10 + 5×15 = 85 V. (Key point: Vth calculation)

3. What is Rth, with the current source opened?
   A. 15 Ω
   B. 20 Ω
   C. 5 Ω
   D. 3.75 Ω
   Answer: A. Opening the current source removes its branch (including the 5 Ω), leaving Rth = 15 Ω. (Key point: Rth)

4. What is the current in the 10 Ω?
   A. 8.5 A
   B. 3.4 A
   C. 5.67 A
   D. 2.6 A
   Answer: B. I = 85/(15 + 10) = 3.4 A. (Key point: Load current)

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