---
title: "Thevenin: Vth, Rth and 5 kΩ current"
subject: "Basic Electronics Engineering"
chapter: 1
key: "semester_2/basic_electronics_engineering:thevenin-48v-12k"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/thevenin-48v-12k
site: BCT Engineering
---

# Thevenin: Vth, Rth and 5 kΩ current

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2075 Baisakh, 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-48v-12k
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

Determine Vth, Rth and the current through the 5 kΩ resistor using Thevenin's theorem. [Figure: 48 V source; 12 kΩ and 8 kΩ resistors in series along the top; a 4 kΩ resistor and the 5 kΩ resistor in vertical branches.]

## Short answer

Thevenin's theorem: Vth = 48×4/(12+4) = 12 V, Rth = 8k + (12k∥4k) = 11 kΩ. Current through 5kΩ is I = 12/(11k+5k) = 0.75 mA.

## Answer

**Assumed circuit:** the 48 V source feeds a 12 kΩ resistor in series along the top, reaching node A. A 4 kΩ resistor connects A to ground (vertical branch). An 8 kΩ resistor in series along the top connects A to node B, and the 5 kΩ load is connected from B to ground.

#### Step 1: remove the 5 kΩ and find Vth
With the 5 kΩ open, no current flows in the 8 kΩ, so Vth is the voltage across the 4 kΩ:

Vth = 48 × 4/(12 + 4) = 12 V

#### Step 2: find Rth
Short the 48 V source. Looking into terminals B-ground:

Rth = 8 kΩ + (12 kΩ ∥ 4 kΩ) = 8 + (12 × 4)/16 = 8 + 3 = 11 kΩ

#### Step 3: current through 5 kΩ
I = Vth/(Rth + RL) = 12 V/(11 kΩ + 5 kΩ) = 12/16 mA

**Answer: Vth = 12 V, Rth = 11 kΩ, I(5 kΩ) = 0.75 mA**

## Check yourself

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

1. Why does Vth equal the voltage across the 4 kΩ with the 5 kΩ removed?
   A. Because the 8 kΩ carries no current when the load is open, so no drop appears across it
   B. Because the 12 kΩ is shorted
   C. Because the 4 kΩ and 8 kΩ are in parallel
   D. Because the source is shorted
   Answer: A. No current flows in the 8 kΩ at open circuit, so VB = voltage across 4 kΩ = 48×4/16 = 12 V. (Key point: Vth by voltage divider)

2. What is Vth?
   A. 48 V
   B. 12 V
   C. 9.6 V
   D. 36 V
   Answer: B. Vth = 48 × 4/(12 + 4) = 12 V. (Key point: Value of Vth)

3. What is Rth seen from B to ground with the source shorted?
   A. 8 + (12 ∥ 4) = 11 kΩ
   B. 8 + 12 + 4 = 24 kΩ
   C. 12 ∥ 4 = 3 kΩ
   D. 8 ∥ 3 = 2.18 kΩ
   Answer: A. The 8 kΩ is in series with 12k∥4k = 3 kΩ, giving 11 kΩ. (Key point: Rth)

4. What is the current in the 5 kΩ?
   A. 2.4 mA
   B. 0.75 mA
   C. 1.09 mA
   D. 1.2 mA
   Answer: B. I = 12 V/(11k + 5k) = 0.75 mA. (Key point: Load current)

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