---
title: "Resistor and capacitor with half voltage"
subject: "Basic Electrical Engineering"
chapter: 6
key: "semester_1/basic_electrical_engineering:adjustable-resistor-25uf-0-8a"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/adjustable-resistor-25uf-0-8a
site: BCT Engineering
---

# Resistor and capacitor with half voltage

Basic Electrical Engineering · Chapter 6: Single-phase AC Circuits

- Asked in: 2069 Asar, 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/adjustable-resistor-25uf-0-8a
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-6

## Question

An adjustable resistor R in series with a capacitance of 25μF draws a current of 0.8A, when connected across 50Hz supply. Calculate: i) The value of resistor so that the voltage across the capacitor is 'half' of the supply voltage. ii) Power consumed and iii) The power factor

## Answer

Given C = 25 μF, I = 0.8 A, f = 50 Hz (series R–C circuit).

X_C = 1/(2πfC) = 1/(314.16 × 25 × 10⁻⁶) = 127.3 Ω

#### i) Value of R
The voltage across the capacitor is V_C = I X_C = 0.8 × 127.3 = 101.9 V. For V_C = V/2, the supply voltage is V = 203.7 V, so the impedance is

Z = V/I = 203.7/0.8 = 254.6 Ω = 2X_C

Z² = R² + X_C² ⟹ R = √(Z² − X_C²) = √(4X_C² − X_C²) = √3 X_C

**R = 1.732 × 127.3 = 220.5 Ω**

#### ii) Power consumed
P = I²R = 0.8² × 220.5 = **141.1 W**

#### iii) Power factor
cos φ = R/Z = 220.5/254.6 = **0.866 leading** (φ = 30°)

(Check: sin φ = X_C/Z = 1/2, so φ = 30°.)

## Check yourself

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

1. X_C for 25 μF at 50 Hz is:
   A. 31.8 Ω
   B. 7.85 Ω
   C. 127.3 Ω
   D. 254.6 Ω
   Answer: C. X_C = 1/(2πfC) = 1/(314.16 × 25×10^-6). (Key point: Capacitor reactance)

2. If V_C is half the supply, impedance Z equals:
   A. 4 X_C
   B. X_C/2
   C. X_C
   D. 2 X_C
   Answer: D. V = 2V_C means Z = 2X_C = 254.6 Ω. (Key point: Condition V_C = V/2)

3. Using Z^2 = R^2 + X_C^2, R is:
   A. 3 X_C
   B. √3 X_C = 220.5 Ω
   C. 2 X_C
   D. X_C = 127.3 Ω
   Answer: B. R^2 = 4X_C^2 − X_C^2 = 3X_C^2. (Key point: Resistor value)

4. pf is:
   A. 0.707
   B. 0.866 leading (φ = 60°)
   C. 0.5 leading
   D. 0.866 lagging
   Answer: B. cos φ = R/Z = 220.5/254.6 and the circuit is capacitive. (Key point: Power factor)

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