---
title: "Power in 75Ω across XY"
subject: "Basic Electrical Engineering"
chapter: 2
key: "semester_1/basic_electrical_engineering:power-75-ohm-xy-200v-network"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/power-75-ohm-xy-200v-network
site: BCT Engineering
---

# Power in 75Ω across XY

Basic Electrical Engineering · Chapter 2: DC circuits

- Asked in: 2069 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/power-75-ohm-xy-200v-network
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-2

## Question

Calculate the power which would be dissipated in a 75 Ω resistor connected across XY in the network shown below. [Figure: 200 V source; 80 Ω shunt; 20 Ω and 20 Ω series resistors along the top to terminal X; 10 Ω shunt in the middle; 20 Ω and 20 Ω along the bottom to terminal Y; 75 Ω connected across X and Y.]

## Short answer

The load sees 115 Ω, paralleled with 10 Ω gives 9.2 Ω; voltage across 10 Ω is 37.4 V. Current in 75 Ω is 0.325 A, dissipated power = 0.325² × 75 ≈ 7.93 W.

## Answer

#### Reading the figure
The 200 V source feeds a ladder: 80 Ω across the source, then 20 Ω (top) and 20 Ω (bottom) in series, a 10 Ω shunt, another 20 Ω (top) and 20 Ω (bottom), and finally the 75 Ω load across X and Y.

#### Step 1: Resistance looking from the 10 Ω towards the load
R = 20 + 20 + 75 = 115 Ω

#### Step 2: Combine with the 10 Ω
10 ∥ 115 = 10 × 115/125 = 9.2 Ω

#### Step 3: Voltage across the 10 Ω
The 80 Ω shunt is directly across the source, so the rest of the ladder receives 200 V through the first 20 + 20 Ω:

V₁₀ = 200 × 9.2/(9.2 + 40) = 37.40 V

#### Step 4: Load current and power
I₇₅ = 37.40/115 = 0.3252 A

P = I² R = 0.3252² × 75 = 7.93 W

**Answer: Power in the 75 Ω ≈ 7.93 W**

## Check yourself

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

1. The 80 Ω across the source
   A. halves the voltage
   B. is in series with 20 Ω
   C. adds to the load
   D. does not affect the ladder because it is directly across 200 V
   Answer: D. The source holds 200 V across it. (Key point: 80 Ω shunt)

2. From the 10 Ω towards the load, the resistance is
   A. 20 || 20 || 75
   B. 75 Ω
   C. 20 + 20 + 75 = 115 Ω
   D. 95 Ω
   Answer: C. Top and bottom 20 Ω resistors are in series with the load. (Key point: Resistance looking in)

3. V across the 10 Ω is
   A. 200 V
   B. 200 x 9.2/(9.2 + 40) = 37.4 V
   C. 200 x 10/50 V
   D. 18.7 V
   Answer: B. The first 20 + 20 Ω is in series with the 10||115 = 9.2 Ω. (Key point: Voltage divider)

4. P in 75 Ω = I^2 R with I = 0.3252 A is
   A. 7.93 W
   B. 12.2 W
   C. 24.4 W
   D. 0.1 W
   Answer: A. 0.3252^2 x 75 = 7.93 W. (Key point: Power)

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