---
title: "Capacitor for power factor correction of lamp"
subject: "Basic Electrical Engineering"
chapter: 7
key: "semester_1/basic_electrical_engineering:fluorescent-lamp-pf-capacitor"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/fluorescent-lamp-pf-capacitor
site: BCT Engineering
---

# Capacitor for power factor correction of lamp

Basic Electrical Engineering · Chapter 7: Power in AC Circuits

- Asked in: 2074 Asoj, 6 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/fluorescent-lamp-pf-capacitor
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-7

## Question

A fluorescent lamp takes a current of 0.75A when connected across a 240V, 50Hz a.c supply. The power consumed by the lamp is 80 watt. Calculate the value of the capacitance to be connected in parallel with the lamp to improve the power factor to (i) unity (ii) 0.95 lagging.

## Short answer

Lamp: S=180VA, P=80W, cos φ₁=0.444 lag, Q₁=161.2VAR. For unity pf, Qc=161.2VAR, C=8.91µF. For 0.95 lag, Qc=134.95VAR, C=7.46µF. Capacitor in parallel.

## Answer

Given V = 240 V, f = 50 Hz, I = 0.75 A, P = 80 W.

#### Lamp data
- Apparent power S = VI = 240 × 0.75 = 180 VA
- Power factor cos φ₁ = P/S = 80/180 = 0.444 lagging (φ₁ = 63.6°)
- Reactive power Q₁ = √(S² − P²) = √(180² − 80²) = 161.2 VAR

A capacitor in parallel must supply Q_c = Q₁ − Q₂ so that the net reactive power Q₂ gives the required power factor. Then C = Q_c / (ωV²), with ω = 2π × 50 = 314.16 rad/s and ωV² = 314.16 × 240² = 1.8096 × 10⁷.

#### (i) Unity power factor
Q₂ = 0, so Q_c = 161.2 VAR.

C = 161.2 / (1.8096 × 10⁷) = 8.91 × 10⁻⁶ F = **8.91 μF**

#### (ii) 0.95 lagging
tan φ₂ = tan(cos⁻¹ 0.95) = 0.3287

Q₂ = P tan φ₂ = 80 × 0.3287 = 26.3 VAR

Q_c = 161.2 − 26.3 = 134.95 VAR

C = 134.95 / (1.8096 × 10⁷) = 7.46 × 10⁻⁶ F = **7.46 μF**

**Answer: 8.91 μF for unity, 7.46 μF for 0.95 lagging.** The supply current falls from 0.75 A to 0.333 A (unity) or 0.351 A (0.95).

## Check yourself

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

1. P = 80 W, S = 240 × 0.75 = 180 VA. pf is:
   A. 0.444 lagging
   B. 0.444 leading
   C. 0.8 lagging
   D. 2.25
   Answer: A. cos φ = P/S. (Key point: Lamp pf)

2. Lamp reactive power is:
   A. 80 VAR
   B. S − P = 100
   C. √(S^2 − P^2) = 161.2 VAR
   D. P tan φ
   Answer: C. From the power triangle. (Key point: Reactive power)

3. Required capacitor is C = Q_c/(ωV^2), where Q_c is:
   A. The reactive power to be cancelled, Q1 − Q2
   B. The active power
   C. The apparent power
   D. Q2 only
   Answer: A. The capacitor supplies reactive power without changing P. (Key point: Capacitor formula)

4. For unity and 0.95 lagging, C is:
   A. 7.46 and 8.91
   B. 16 μF both
   C. 80 μF
   D. 8.91 μF and 7.46 μF
   Answer: D. Q_c = 161.2 and 134.95 VAR; more correction needed for unity. (Key point: Values)

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