---
title: "Find R for maximum power, 10 A and 48 V"
subject: "Basic Electrical Engineering"
chapter: 3
key: "semester_1/basic_electrical_engineering:max-power-r-10a-48v"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/max-power-r-10a-48v
site: BCT Engineering
---

# Find R for maximum power, 10 A and 48 V

Basic Electrical Engineering · Chapter 3: Network Theorems

- Asked in: 2073 Chaitra, 6 marks
- Video: [GATE 2012 ECE Find the value of R for which maximum power is transferred from circuit A to circuit B](https://www.youtube.com/watch?v=RYUCtIAHMkE) — GATE Paper Satish Bojjawar
- 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-10a-48v
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3

## Question

What is the value of R such that maximum power is transferred to it? Find the value of this maximum power for the circuit shown below. [Figure: 10 A current source on the left; 4 Ω and 10 Ω shunt resistors; R in the top branch; 20 Ω resistor and 48 V source on the right.]

## Short answer

Maximum power transfer occurs when R equals the Thevenin resistance, 10.67 Ω. The maximum power delivered is 13.5 W.

## Answer

#### Reading the figure
Left node A: 10 A source (up) into A and a 4 Ω shunt to ground. R joins A to node B. Node B: 10 Ω shunt to ground, and 20 Ω in series with the 48 V battery (+ on top) to ground.

R is the load; find its Thevenin equivalent.

#### Step 1: Open-circuit voltage (R removed)
- V_A = 10 × 4 = 40 V (all the 10 A flows through the 4 Ω).
- V_B = 48 × 10/(10 + 20) = 16 V (voltage divider).

V_th = V_A − V_B = 40 − 16 = **24 V**

#### Step 2: Thevenin resistance (sources off)
10 A opened, 48 V shorted:
R_th = 4 + (10 ∥ 20) = 4 + 6.667 = **10.67 Ω**

#### Step 3: Maximum power
R = R_th = **10.67 Ω** (32/3 Ω)

P_max = V_th²/(4 R_th) = 24²/(4 × 10.667) = 576/42.67 = **13.5 W**

Check: current in R = 24/(2 × 10.667) = 1.125 A; P = 1.125² × 10.667 = 13.5 W.

**Answer: R = 10.67 Ω; P_max = 13.5 W**

## Check yourself

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

1. How is V_A found with R removed?
   A. All 10 A flows through the 4 ohm: 40 V
   B. 10 V
   C. 48 V
   D. 24 V
   Answer: A. R removed, so the current source flows in the 4 ohm. (Key point: V_th)

2. What is V_th?
   A. 40 V
   B. V_A - V_B = 40 - 16 = 24 V
   C. 16 V
   D. 56 V
   Answer: B. V_B = 48 x 10/30 = 16 V, by divider. (Key point: V_th value)

3. What is R_th?
   A. 4 + 10 + 20 = 34 ohm
   B. 10 || 20 = 6.67 ohm
   C. 4 + (10 || 20) = 10.67 ohm
   D. 4 ohm
   Answer: C. Sources off: 4 ohm in series with parallel 10 and 20. (Key point: R_th)

4. What is P_max?
   A. 54 W
   B. 27 W
   C. 24^2/10.67 = 54 W
   D. 24^2/(4 x 10.67) = 13.5 W
   Answer: D. V_th^2/(4 R_th). (Key point: P_max)

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