---
title: "Power in 5Ω using Norton theorem (2 A, 10 V)"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:norton-power-5-ohm-2a-10v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/norton-power-5-ohm-2a-10v
site: BCT Engineering
---

# Power in 5Ω using Norton theorem (2 A, 10 V)

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2067 Asoj, 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-5-ohm-2a-10v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Find the power dissipated by 5 ohm resistor of the following circuit using Norton's theorem. [Figure: 2 A current source; 1 Ω in series on top; 10 V source in the top branch; 2 Ω shunt; 2 Ω shunt; 5 Ω load.]

## Short answer

Norton current with 5Ω shorted is 7 A, Norton resistance is 1 Ω. Load current = 7 × 1/(1+5) = 1.167 A. Power dissipated in 5Ω = 1.167² × 5 = 6.81 W.

## Answer

#### Reading the figure
Bottom rail = ground. The 2 A source (arrow up) feeds terminal a, then a 1 Ω in series to node N₁. N₁ has a 2 Ω shunt. Between N₁ and N₂ is the 10 V battery (+ on the N₂ side). N₂ has a 2 Ω shunt and the 5 Ω load.

```text
 (2A)--[1 Ω]--N1--(10V)--N2
               |          |   |
              [2]        [2] [5]
               |          |   |
 --------------+----------+---+--
```

#### Step 1: Norton current (5 Ω shorted)
With N₂ shorted to ground, the battery makes V_N1 = −10 V (V_N2 = V_N1 + 10 = 0). The 2 Ω shunt at N₂ carries nothing.

Current from ground into N₁ through its 2 Ω shunt: 10/2 = 5 A
Current from the 2 A source: 2 A

I_N = 2 + 5 = **7 A** (through the battery into the short)

#### Step 2: Norton resistance (sources off)
2 A opened, 10 V shorted: N₁ and N₂ are joined, so the two 2 Ω shunts are in parallel (the 1 Ω is in series with the open source and carries nothing):

R_N = 2 ∥ 2 = **1 Ω**

#### Step 3: Current and power in the 5 Ω
I₅ = I_N × R_N/(R_N + 5) = 7 × 1/6 = 1.167 A

P = I₅² × 5 = 1.167² × 5 = **6.81 W**

**Answer: Power in the 5 Ω = 6.81 W**

## Check yourself

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

1. Why is I_N = 7 A when the 5 ohm is shorted?
   A. The 2 A source plus 10/2 = 5 A through the 2 ohm shunt at N1 (battery makes V_N1 = -10 V)
   B. The two 2 ohm shunts are in series
   C. The 1 ohm carries 10 A
   D. Only the 2 A source contributes
   Answer: A. Shorting the load makes V_N2 = 0, so V_N1 = -10 V; the shunt supplies 5 A, which adds to the 2 A. (Key point: Shorted-load current)

2. What is R_N?
   A. 2 + 2 + 1 = 5 ohm
   B. 2 || 2 = 1 ohm, because the 10 V is shorted and 2 A opened
   C. 5 || 2 = 1.43 ohm
   D. 1 ohm in series with 2 ohm = 3 ohm
   Answer: B. With sources off, N1 and N2 are joined; the 1 ohm carries nothing since the current source is open. (Key point: R_N)

3. What gives the power in the 5 ohm?
   A. I = 7/5 = 1.4 A, P = 9.8 W
   B. I = 7 A, P = 245 W
   C. P = 7 x 5 = 35 W
   D. I = 7 x 1/6 = 1.167 A, P = I^2 x 5 = 6.81 W
   Answer: D. Current divider with R_N = 1 ohm and R_L = 5 ohm. (Key point: Power)

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