---
title: "Capacitor for single-phase motor pf correction"
subject: "Basic Electrical Engineering"
chapter: 7
key: "semester_1/basic_electrical_engineering:single-phase-motor-capacitor-40a-0-7"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/single-phase-motor-capacitor-40a-0-7
site: BCT Engineering
---

# Capacitor for single-phase motor pf correction

Basic Electrical Engineering · Chapter 7: Power in AC Circuits

- Asked in: 2068 Chaitra, 6 marks
- Video: [Power factor correction calculations](https://www.youtube.com/watch?v=WdJVXbyTwP4) — Prof MAD
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/single-phase-motor-capacitor-40a-0-7
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-7

## Question

A single phase motor takes a current of 40A at pf 0.7 lagging from a 440V, 50HZ supply. What value must a shunting capacitor have to raise the power factor to 0.9 lagging.

## Short answer

Compute active power P = 440×40×0.7 = 12320 W. Then Qc = P (tan(arccos 0.7) – tan(arccos 0.9)) = 6602 VAR. C = Qc/(2πf V²) = 6602/(314.16×440²) = 108.5 μF.

## Answer

Given V = 440 V, f = 50 Hz, I = 40 A, cos φ₁ = 0.7 lagging, target cos φ₂ = 0.9 lagging.

#### Motor power
- Active power P = V I cos φ₁ = 440 × 40 × 0.7 = 12 320 W
- φ₁ = cos⁻¹ 0.7 = 45.57°, tan φ₁ = 1.0202
- φ₂ = cos⁻¹ 0.9 = 25.84°, tan φ₂ = 0.4843

#### Reactive power
- Q₁ = P tan φ₁ = 12 320 × 1.0202 = 12 569 VAR
- Q₂ = P tan φ₂ = 12 320 × 0.4843 = 5 967 VAR

The capacitor must supply Q_c = Q₁ − Q₂ = 6 602 VAR.

#### Capacitance
Q_c = V²/X_C = ωCV²

C = Q_c / (ωV²) = 6602 / (314.16 × 440²) = 6602 / (6.0820 × 10⁷)

**C = 108.5 μF**

The supply current after correction is P/(V × 0.9) = 12 320/(440 × 0.9) = 31.1 A.

## Check yourself

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

1. How is the active power of the motor found?
   A. P = V I = 17 600 W
   B. P = V I cos φ1 = 440 x 40 x 0.7 = 12 320 W
   C. P = V I sin φ1
   D. P = I²/V
   Answer: B. Active power is V I cos φ; sin φ would give reactive power. (Key point: Active power)

2. How is the capacitor reactive power found?
   A. Q_c = P(tan φ1 - tan φ2)
   B. Q_c = P(cos φ1 - cos φ2)
   C. Q_c = P(tan φ1 + tan φ2)
   D. Q_c = V I
   Answer: A. The capacitor must cancel the reduction in reactive power: Q1 - Q2 = P(tan φ1 - tan φ2) = 6602 VAR. (Key point: Capacitor VAR)

3. Which formula gives the capacitance from Q_c?
   A. C = Q_c x ω x V²
   B. C = Q_c / (ω V²)
   C. C = Q_c / V
   D. C = V² / (ω Q_c)
   Answer: B. Q_c = ωCV², so C = Q_c/(ωV²) = 108.5 μF. (Key point: Capacitance formula)

4. After correction to pf 0.9, what is the supply current?
   A. 40 A
   B. About 31.1 A, since P stays 12 320 W
   C. About 51 A
   D. About 22 A
   Answer: B. Capacitors do not change P, so I = 12 320/(440 x 0.9) = 31.1 A, lower than before. (Key point: Supply current after correction)

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