---
title: "Thevenin: 4 kΩ resistor, 12 V, 10 mA"
subject: "Basic Electronics Engineering"
chapter: 1
key: "semester_2/basic_electronics_engineering:thevenin-12v-10ma"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/thevenin-12v-10ma
site: BCT Engineering
---

# Thevenin: 4 kΩ resistor, 12 V, 10 mA

Basic Electronics Engineering · Chapter 1: Basic Circuits Concepts

- Asked in: 2073 Magh, 1+5 marks; 2068 Magh, 2+6 marks (asked 2 times)
- 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-12v-10ma
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-1

## Question

State Thevenin's theorem and use it to find the current through the 4 kΩ resistor. [Figure: 12 V source feeding a 3 kΩ series resistor, then 6 kΩ and 2 kΩ resistors in parallel branches, then a 4 kΩ series resistor to a 10 mA current source in the last branch.]

## Short answer

Thevenin's theorem: Vth is open-circuit voltage, Rth is resistance with sources zeroed. Here Vth = 8V − 20V = −12V, Rth = (3k∥6k)+2k = 4kΩ. Current through 4kΩ is 1.5 mA from B to A.

## Answer

**Statement:** Thevenin's theorem says that any linear two-terminal network containing independent sources and resistors can be replaced by one voltage source Vth in series with one resistance Rth. Vth is the open-circuit voltage across the terminals and Rth is the resistance seen at the terminals with voltage sources shorted and current sources opened. The load current is then I = Vth/(Rth + RL).

**Assumed circuit (the figure was unclear):** the 12 V source feeds node A through 3 kΩ. A 6 kΩ resistor is connected from A to ground. The 4 kΩ load connects A to node B. At node B, the 2 kΩ resistor goes to ground, and the 10 mA source (pointing into B) is also connected to ground. Load: 4 kΩ between A and B.

#### Step 1: remove the 4 kΩ load, find Vth = VA − VB
- Left side (A): VA = 12 × 6/(3 + 6) = 8 V
- Right side (B): the 10 mA flows through 2 kΩ: VB = 10 mA × 2 kΩ = 20 V

Vth = VA − VB = 8 − 20 = −12 V

#### Step 2: find Rth
Short the 12 V source and open the 10 mA source.
- Left: 3 kΩ ∥ 6 kΩ = 2 kΩ
- Right: 2 kΩ

Rth = 2 kΩ + 2 kΩ = 4 kΩ

#### Step 3: load current
I = Vth/(Rth + RL) = −12 V/(4 kΩ + 4 kΩ) = −1.5 mA

**Answer: I = 1.5 mA through the 4 kΩ resistor, flowing from B to A (opposite to the A to B reference), under the stated assumption.**

## Check yourself

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

1. When finding Rth, how are the sources treated?
   A. Voltage sources opened, current sources shorted
   B. Both shorted
   C. Voltage sources shorted, current sources opened
   D. Both left in place
   Answer: C. Independent voltage sources become short circuits and current sources become open circuits. (Key point: Setting sources to zero)

2. With the 4 kΩ removed, VA = 8 V and VB = 20 V. How is Vth obtained?
   A. Vth = VA + VB = 28 V
   B. Vth = VA − VB = −12 V
   C. Vth = VB only = 20 V
   D. Vth = 12 V of the source
   Answer: B. Vth is the open-circuit voltage between A and B: 8 − 20 = −12 V. (Key point: Vth by node voltages)

3. What is Rth for this circuit?
   A. 2 kΩ
   B. 4 kΩ (3k∥6k = 2k, plus 2k)
   C. 11 kΩ
   D. 6 kΩ
   Answer: B. Left: 3k∥6k = 2 kΩ; right: 2 kΩ; they are in series, so Rth = 4 kΩ. (Key point: Rth)

4. Why is the load current negative (−1.5 mA)?
   A. Because Vth is negative with respect to the A to B reference, so current flows from B to A
   B. Because the load is faulty
   C. Because Rth is negative
   D. Because the 10 mA source was shorted
   Answer: A. I = −12/(4 + 4) = −1.5 mA; a negative sign means it flows opposite to the reference direction. (Key point: Current direction)

## Related in Basic Electronics Engineering

- [Thevenin: current through 10 Ω](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-10v-5a) — Basic Electronics Engineering, chapter 1
- [Thevenin: Vth, Rth and 5 kΩ current](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-48v-12k) — Basic Electronics Engineering, chapter 1
- [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
- [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×
- [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×
- [Thevenin's theorem and steps](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-steps) — Basic Electronics Engineering, chapter 1

## Related in other subjects

- [Thevenin theorem, current in 10Ω](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Athevenin-current-10-ohm-20v-12v) — Basic Electrical Engineering, chapter 3
- [Thevenin current through RL and maximum power](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Athevenin-max-power-load) — Fundamental of Electrical and Electronics Engineering, chapter 1
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