---
title: "Maximum power transfer, find R"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:max-power-r-5-10-10v-20v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-r-5-10-10v-20v
site: BCT Engineering
---

# Maximum power transfer, find R

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2074 Chaitra, 6 marks
- Video: [Maximum Power Transfer Theorem Solved Example | Electrical Engineering](https://www.youtube.com/watch?v=Sck5IpxLHKI) — 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/max-power-r-5-10-10v-20v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

Calculate the value of 'R' such that maximum power will be absorbed by it in the given circuit. [Figure: 5 Ω and 10 Ω resistors in the top branch; 10 V source on the left; R in the middle vertical branch; 20 V source on the right.]

## Short answer

Thevenin voltage V_th = 13.33 V, Thevenin resistance R_th = 5 Ω∥10 Ω = 3.33 Ω. R should equal R_th = 3.33 Ω for maximum power; maximum power P_max = V_th²/(4R_th) = 13.33 W.

## Answer

#### Reading the figure
Left branch: 10 V battery (+ on top). Right branch: 20 V battery (+ on top). Top: 5 Ω from the left top node to the middle node M, and 10 Ω from M to the right top node. R is the vertical branch at M. Bottom rail common.

So R is connected at M to two source branches: (10 V, 5 Ω) and (20 V, 10 Ω).

#### Step 1: Thevenin voltage (R removed)
V_th = V_M = (10/5 + 20/10)/(1/5 + 1/10) = 4/0.3 = **13.33 V**

#### Step 2: Thevenin resistance
V sources shorted: R_th = 5 ∥ 10 = 50/15 = **3.33 Ω**

#### Step 3: Maximum power
R = R_th = **3.33 Ω**

P_max = V_th²/(4 R_th) = 13.33²/(4 × 3.333) = 177.8/13.33 = **13.33 W**

Check: current in R = 13.33/(2 × 3.333) = 2 A, so P = 2² × 3.333 = 13.33 W.

**Answer: R = 3.33 Ω (10/3 Ω); P_max = 13.33 W**

## Check yourself

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

1. How is V_th at M found?
   A. Average of 10 and 20 = 15 V
   B. Millman/nodal: (10/5 + 20/10)/(1/5 + 1/10) = 13.33 V
   C. 10 + 20 = 30 V
   D. 10 x 20/30 = 6.67 V
   Answer: B. Two source branches meet at M; weighted by conductance gives 13.33 V. (Key point: V_th method)

2. What is R_th?
   A. 15 ohm
   B. 7.5 ohm
   C. 5 || 10 = 3.33 ohm
   D. 5 ohm
   Answer: C. Voltage sources shorted leaves 5 and 10 ohm in parallel. (Key point: R_th)

3. What is R and P_max?
   A. R = 15 ohm, 13.33 W
   B. R = 3.33 ohm, 26.7 W
   C. R = 13.33 ohm, 3.33 W
   D. R = 3.33 ohm, P_max = 13.33 W
   Answer: D. P_max = V_th^2/(4 R_th) = 177.8/13.33. (Key point: Maximum power)

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