---
title: "Capacitor bank for 3-phase motor pf"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:capacitor-bank-60kw-induction-motor"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/capacitor-bank-60kw-induction-motor
site: BCT Engineering
---

# Capacitor bank for 3-phase motor pf

Basic Electrical Engineering · Chapter 8: Three-Phase Circuit Analysis

- Asked in: 2074 Chaitra, 8 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/capacitor-bank-60kw-induction-motor
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

A 400V, 50 Hz, 3 phase induction motor takes 60 KW power from supply mains at 0.8 power factor lagging. Calculate the capacitance per phase and KVAR rating per phase of capacitor in order to improve the power factor to 0.9 lagging using (i) star connected capacitor bank and (ii) Delta connected capacitor bank.

## Short answer

Reactive power required to improve pf from 0.8 to 0.9 lag is 15.94 kVAR total, 5.31 kVAR per phase. Star capacitor per phase: 317 μF; delta capacitor per phase: 105.7 μF.

## Answer

Given P = 60 kW, V_L = 400 V, f = 50 Hz, cos φ₁ = 0.8 lag, cos φ₂ = 0.9 lag.

#### Reactive power
- tan φ₁ = tan(36.87°) = 0.75, so Q₁ = 60 × 0.75 = 45.0 kVAR
- tan φ₂ = tan(25.84°) = 0.4843, so Q₂ = 60 × 0.4843 = 29.06 kVAR
- Total capacitor kVAR: Q_c = 45.0 − 29.06 = **15.94 kVAR**
- Per phase: 15.94/3 = **5.31 kVAR** (the rating per phase is the same for star or delta bank)

#### (i) Star-connected bank
Voltage across each capacitor = 400/√3 = 230.9 V.

C_Y = (Q_c/3) / (ω V_ph²) = 5314 / (314.16 × 230.9²) = 5314 / (1.6753 × 10⁷)

**C_Y = 317 μF per phase**

#### (ii) Delta-connected bank
Voltage across each capacitor = 400 V.

C_Δ = 5314 / (314.16 × 400²) = 5314 / (5.0265 × 10⁷)

**C_Δ = 105.7 μF per phase**

(C_Y = 3 C_Δ, so the delta bank needs only one third of the capacitance.)

Line current of the motor falls from 60 000/(√3 × 400 × 0.8) = 108.3 A to 96.2 A.

## Check yourself

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

1. What total capacitor kVAR is needed?
   A. 45.0 kVAR
   B. 29.06 kVAR
   C. 15.94 kVAR
   D. 74.06 kVAR
   Answer: C. Q_c = Q1 - Q2 = 45.0 - 29.06 = 15.94 kVAR. (Key point: Total capacitor kVAR)

2. How does the per-phase kVAR rating compare for star and delta banks?
   A. Star is 3 times delta
   B. Delta is 3 times star
   C. They are the same, 5.31 kVAR per phase
   D. Star is zero
   Answer: C. The total kVAR is split equally into three, regardless of the connection. (Key point: Rating per phase)

3. What voltage appears across each capacitor in a star bank on a 400 V line?
   A. 400 V
   B. 230.9 V
   C. 692 V
   D. 133 V
   Answer: B. In star the capacitor sees the phase voltage 400/√3 = 230.9 V. (Key point: Star voltage)

4. How do C_Y and C_Δ compare?
   A. C_Y = 3 C_Δ
   B. C_Δ = 3 C_Y
   C. They are equal
   D. C_Y = √3 C_Δ
   Answer: A. C = Q/(ωV²); the delta voltage is √3 times bigger, so delta needs one third of the capacitance. (Key point: Capacitance comparison)

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