---
title: "Thevenin theorem, power in 10Ω"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:thevenin-power-10-ohm-24v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/thevenin-power-10-ohm-24v
site: BCT Engineering
---

# Thevenin theorem, power in 10Ω

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2067 Asoj, 8 marks
- Video: [Thevenin's Theorem | Engineering Circuit Analysis | Solved Examples](https://www.youtube.com/watch?v=jpO37lVBe-o) — Electrical Electronics Applications
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/thevenin-power-10-ohm-24v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Find the power dissipated in 10 Ohm resistor using Thevenin theorem of the following network. Note that all resistances are in Ohm. [Figure: 5, 6, 8 and 2 Ω resistors around the outside with a 10 Ω resistor in the middle vertical branch; 24 V source at the bottom.]

## Short answer

Open-circuit voltage across 10Ω terminals is 8.29 V, Thevenin resistance is 4.327 Ω. Current through 10Ω = 8.29/(4.327+10) = 0.5787 A. Power dissipated = I²×10 = 3.35 W.

## Answer

#### Reading the figure
The 24 V source is connected between the left node P and the right node Q. Two paths join P to Q: P–5 Ω–T–6 Ω–Q (upper) and P–8 Ω–M–2 Ω–Q (lower). The 10 Ω load joins T and M.

```text
   P --[5]-- T --[6]-- Q
   |         |         |
   |       [10]        |
   |         |         |
   P --[8]-- M --[2]-- Q      (24 V source between P and Q)
```

The polarity of the source does not change the power in the 10 Ω. Take V_P = 0 and V_Q = 24 V.

#### Step 1: Thevenin voltage (10 Ω removed)
V_T = 24 × 5/(5 + 6) = 10.909 V
V_M = 24 × 8/(8 + 2) = 19.2 V

V_th = V_M − V_T = **8.29 V**

#### Step 2: Thevenin resistance (source shorted, so P = Q)
From T: 5 Ω ∥ 6 Ω = 2.727 Ω
From M: 8 Ω ∥ 2 Ω = 1.6 Ω

R_th = 2.727 + 1.6 = **4.327 Ω**

#### Step 3: Current and power in the 10 Ω
I = V_th/(R_th + 10) = 8.291/14.327 = 0.5787 A

P = I² × 10 = 0.5787² × 10 = **3.35 W**

**Answer: Power dissipated in the 10 Ω = 3.35 W**

## Check yourself

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

1. How is V_th found for this bridge-type circuit?
   A. Remove the 10 ohm and take the difference between the two divider voltages V_M and V_T
   B. Use 24 V directly
   C. Short the 24 V source
   D. Add the two divider voltages
   Answer: A. V_T = 24 x 5/11 = 10.909 V and V_M = 24 x 8/10 = 19.2 V; V_th = V_M - V_T = 8.29 V. (Key point: V_th method)

2. With the source shorted (P = Q), what is R_th seen from T and M?
   A. 5 + 6 + 8 + 2 = 21 ohm
   B. (5 + 8) || (6 + 2) = 4.95 ohm
   C. (5 || 6) + (8 || 2) = 2.727 + 1.6 = 4.327 ohm
   D. 5 || 6 || 8 || 2
   Answer: C. Shorting the source joins P and Q, so the two parallel pairs are in series between T and M. (Key point: R_th method)

3. How is the power in the 10 ohm obtained?
   A. P = V_th^2/(4 R_th) = 3.97 W
   B. P = V_th^2/10 = 6.9 W
   C. I = V_th/(R_th + 10) = 0.5787 A, then P = I^2 x 10 = 3.35 W
   D. P = 24^2/10
   Answer: C. Use the Thevenin equivalent with the load: current then I^2R. (Key point: Load power)

## Related in Basic Electrical Engineering

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