---
title: "Norton theorem with example"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:state-explain-nortons-theorem"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/state-explain-nortons-theorem
site: BCT Engineering
---

# Norton theorem with example

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2074 Chaitra, 4 marks; 2072 Chaitra, 4 marks (asked 2 times)
- Video: [Norton's Theorem and Thevenin's Theorem - Electrical Circuit Analysis](https://www.youtube.com/watch?v=-kkvqr1wSwA) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/state-explain-nortons-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

State and explain Norton's theorem with suitable example.

## Short answer

Norton’s theorem: any linear two-terminal network can be replaced by a current source I_N in parallel with resistance R_N. I_N is the short-circuit current across terminals, R_N is the equivalent resistance with all independent sources off. Example: a 12 V source with 4 Ω series feeding a 6 Ω load gives I_N=3 A, R_N=4 Ω, load current 1.2 A.

## Answer

#### Statement
**Norton's theorem:** any linear two-terminal network containing sources and resistors can be replaced, as seen from its two terminals, by a **current source I_N in parallel with a resistance R_N**.
- I_N = the short-circuit current between the terminals.
- R_N = the resistance seen from the terminals with all independent sources set to zero (voltage sources shorted, current sources opened). R_N equals the Thevenin resistance.

```text
 Linear network          Norton equivalent
 o--------+-----o a       o----+--------+---o a
          |                   (I_N)    [R_N]   load
 (with    |                    |        |
 sources) |                 o--+--------+---o b
 o--------+-----o b
```

#### Steps
1. Remove the load and short the two terminals.
2. Find the short-circuit current I_N.
3. Set all sources to zero and find R_N across the terminals.
4. Connect the load across the Norton source: I_L = I_N × R_N/(R_N + R_L).

#### Example
A 12 V source with 4 Ω in series feeds a 6 Ω load (terminals a, b across the 6 Ω).
- Short a-b: I_N = 12/4 = 3 A.
- Source off (shorted): R_N = 4 Ω.
- Load current: I_L = 3 × 4/(4 + 6) = 1.2 A.
- Load voltage: V_L = 1.2 × 6 = 7.2 V.

Check by the original circuit: I = 12/(4 + 6) = 1.2 A. The results are the same.

## Check yourself

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

1. What does Norton's theorem replace a linear two-terminal network with?
   A. A voltage source in series with R
   B. A current source in series with R_N
   C. A current source I_N in parallel with R_N
   D. A single resistor only
   Answer: C. The equivalent is a current source in parallel with a resistor. (Key point: Norton equivalent)

2. How is I_N found?
   A. Open the terminals and measure the voltage
   B. Set all sources to zero
   C. Divide the source voltage by the load
   D. Short the terminals and measure the short-circuit current
   Answer: D. I_N is the short-circuit current between the terminals. (Key point: Finding I_N)

3. When finding R_N, how are sources treated?
   A. Voltage sources shorted, current sources opened
   B. Voltage sources opened, current sources shorted
   C. All sources left in place
   D. Both shorted
   Answer: A. Deactivating independent sources: voltage -> short, current -> open. (Key point: Finding R_N)

4. In the example (12 V, 4 ohm, 6 ohm load), what is the load current?
   A. 3 A
   B. 1.2 A = 3 x 4/(4+6)
   C. 2 A
   D. 0.5 A
   Answer: B. Current divides between R_N and the load: I_L = I_N R_N/(R_N + R_L). (Key point: Example)

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