---
title: "Capacitor for 6 kW load pf improvement"
subject: "Basic Electrical Engineering"
chapter: 7
key: "semester_1/basic_electrical_engineering:inductive-load-6kw-pf-capacitor"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/inductive-load-6kw-pf-capacitor
site: BCT Engineering
---

# Capacitor for 6 kW load pf improvement

Basic Electrical Engineering · Chapter 7: Power in AC Circuits

- Asked in: 2067 Asoj, 4 marks
- Video: [Capacitor Bank Sizing (KVAR) for Power Factor Improvement](https://www.youtube.com/watch?v=qrJDMBR7i1s) — ElectricalEngineeringPlanet
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/inductive-load-6kw-pf-capacitor
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-7

## Question

An inductive load of 6 kW at power factor of 0.6 is connected across 220V, 50Hz supply. Calculate the value of capacitance to be connected in parallel with the load to bring the resultant power factor of 0.95 lagging.

## Short answer

Active power is 6 kW. Required reactive compensation Qc = 6000×(tan(arccos 0.6) – tan(arccos 0.95)) = 6028 VAR. Capacitance C = Qc/(2πf V²) = 6028/(314.16×220²) = 396 μF.

## Answer

Given P = 6 kW, V = 220 V, f = 50 Hz, cos φ₁ = 0.6 lagging, cos φ₂ = 0.95 lagging.

- tan φ₁ = tan(cos⁻¹ 0.6) = 1.3333
- tan φ₂ = tan(cos⁻¹ 0.95) = 0.3287

#### Reactive power
- Before: Q₁ = P tan φ₁ = 6000 × 1.3333 = 8000 VAR
- After: Q₂ = P tan φ₂ = 6000 × 0.3287 = 1972 VAR
- Capacitor: Q_c = 8000 − 1972 = 6028 VAR

#### Capacitance
C = Q_c / (ωV²) = 6028 / (314.16 × 220²) = 6028 / (1.5205 × 10⁷)

**C = 396 μF**

Supply current falls from 6000/(220 × 0.6) = 45.5 A to 6000/(220 × 0.95) = 28.7 A.

## Check yourself

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

1. For P = 6 kW at pf 0.6, what is Q1?
   A. 4800 VAR
   B. 8000 VAR
   C. 3600 VAR
   D. 10 000 VAR
   Answer: B. tan φ1 = 1.3333, so Q1 = 6000 x 1.3333 = 8000 VAR. (Key point: Reactive power before)

2. What reactive power must the capacitor supply?
   A. Q1 + Q2 = 9972 VAR
   B. Q1 - Q2 = 6028 VAR
   C. Q2 = 1972 VAR
   D. P = 6000 VAR
   Answer: B. Capacitor VAR is the reduction in reactive power: 8000 - 1972 = 6028 VAR. (Key point: Capacitor VAR)

3. What happens to the supply current after the capacitor is added?
   A. It rises to 45.5 A
   B. It stays at 45.5 A
   C. It falls from 45.5 A to about 28.7 A
   D. It falls to zero
   Answer: C. The real power is unchanged, so I = 6000/(220 x 0.95) = 28.7 A. (Key point: Effect on current)

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