---
title: "Power in 3Ω using Norton theorem (10 V, 10 A)"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:norton-power-3-ohm-10v-10a-1-ohm"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/norton-power-3-ohm-10v-10a-1-ohm
site: BCT Engineering
---

# Power in 3Ω using Norton theorem (10 V, 10 A)

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2071 Shrawan, 8 marks
- Video: [Norton's Theorem Example | Electric Circuits | Network Analysis | Network Theory](https://www.youtube.com/watch?v=LUOcKhaDiCQ) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/norton-power-3-ohm-10v-10a-1-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Determine power dissipated in 3Ω resitor in the circuit shown in figure below using Norton's theorem. [Figure: 10 V source on the left; 2 Ω in the top series branch; 2 Ω shunt; 10 A current source; 1 Ω in series on top towards terminal a; load 3 Ω between terminals a and b.]

## Short answer

Norton current is 7.5 A and Norton resistance is 2 Ω. Load current is 7.5 × 2/(2+3) = 3 A. The power dissipated in the 3 Ω resistor is 3² × 3 = 27 W.

## Answer

#### Reading the figure
Bottom rail = ground. 10 V source (+ on top) with 2 Ω in series feeds node N. At N: a 2 Ω shunt to ground and a 10 A source (arrow up, into N). From N, a 1 Ω in series goes to terminal a. The load 3 Ω is between a and b (b on the bottom rail).

#### Step 1: Norton current (short a-b)
With a shorted to ground, the 1 Ω is connected directly between N and ground.

Node N: (V_N − 10)/2 + V_N/2 + V_N/1 = 10
0.5 V_N − 5 + 0.5 V_N + V_N = 10 → 2 V_N = 15 → V_N = 7.5 V

I_N = V_N/1 = **7.5 A**

#### Step 2: Norton resistance (sources off)
10 V shorted, 10 A opened: R_N = 1 + (2 ∥ 2) = 1 + 1 = **2 Ω**

(Check: V_th = I_N R_N = 15 V, which is the open-circuit voltage of node N.)

#### Step 3: Load current and power
I_L = I_N × R_N/(R_N + R_L) = 7.5 × 2/(2 + 3) = 3 A

P = I_L² R = 3² × 3 = **27 W**

**Answer: Power in the 3 Ω = 27 W**

## Check yourself

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

1. How is I_N found?
   A. Short a-b and solve node N with the 1 ohm to ground
   B. Open a-b
   C. Remove the 10 A
   D. Divide 10 V by 2 ohm
   Answer: A. I_N is the short-circuit current. (Key point: I_N method)

2. What is I_N?
   A. 15 A
   B. 7.5 A
   C. 10 A
   D. 3 A
   Answer: B. V_N = 7.5 V across 1 ohm. (Key point: I_N value)

3. What is R_N?
   A. 5 ohm
   B. 1 ohm
   C. 1 + (2 || 2) = 2 ohm
   D. 4 ohm
   Answer: C. Sources off. (Key point: R_N)

4. What is power in the 3 ohm?
   A. 7.5^2 x 3 = 168.75 W
   B. 15 W
   C. 3 W
   D. I_L = 3 A so P = 27 W
   Answer: D. I_L = 7.5 x 2/5 = 3 A, P = 9 x 3. (Key point: Power)

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