---
title: "Delta capacitor per phase for 50 kW motor"
subject: "Basic Electrical Engineering"
chapter: 8
key: "semester_1/basic_electrical_engineering:capacitor-delta-415v-50kw-0-72"
source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/capacitor-delta-415v-50kw-0-72
site: BCT Engineering
---

# Delta capacitor per phase for 50 kW motor

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

- Asked in: 2071 Shrawan, 8 marks
- Video: [3 Phase Power Capacitor connection For Power factor Correction @ElectricalTechnician](https://www.youtube.com/watch?v=XewovcWd7wE) — Electrical Technician
- 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-delta-415v-50kw-0-72
- Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-8

## Question

A 415 V, 3 phase, 50 HZ induction motor takes 50 KW power from supply mains at 0.72 power factor lagging. A bank of capacitors is connected in delta across the line to improve the overall power factor. Calculate the capacitance per phase in order to raise the power factor to 0.9 lagging.

## Short answer

Reactive power reduction from 48.19 to 24.21 kVAR, requiring 23.98 kVAR total. Delta-connected capacitors see 415 V; capacitance per phase is 147.7 μF.

## Answer

Given P = 50 kW, V_L = 415 V, f = 50 Hz, cos φ₁ = 0.72 lagging, cos φ₂ = 0.9 lagging. The capacitors are delta connected.

#### Reactive power
- tan φ₁ = tan(cos⁻¹ 0.72) = 0.9638, so Q₁ = 50 × 0.9638 = 48.19 kVAR
- tan φ₂ = tan(cos⁻¹ 0.9) = 0.4843, so Q₂ = 50 × 0.4843 = 24.21 kVAR
- Capacitor reactive power: Q_c = 48.19 − 24.21 = 23.98 kVAR

Per phase: Q_c/3 = 7.99 kVAR.

#### Capacitance per phase (delta)
Voltage across each capacitor = line voltage = 415 V.

Q_phase = ω C V² ⟹ C = 7992 / (314.16 × 415²) = 7992 / (5.411 × 10⁷)

**C = 147.7 μF per phase**

(Check of line current: 50 000/(√3 × 415 × 0.72) = 96.6 A before and 77.3 A after correction.)

## Check yourself

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

1. What reactive power must the capacitor bank supply?
   A. Q1 + Q2 = 72.4 kVAR
   B. Q1 - Q2 = 23.98 kVAR
   C. 48.19 kVAR
   D. 24.21 kVAR
   Answer: B. Q_c = P(tan φ1 - tan φ2) = 48.19 - 24.21 = 23.98 kVAR. (Key point: Capacitor VAR)

2. What is the kVAR per phase of the delta bank?
   A. 23.98 kVAR
   B. 7.99 kVAR
   C. 13.8 kVAR
   D. 71.9 kVAR
   Answer: B. Total Q_c is split across three phases: 23.98/3 = 7.99 kVAR. (Key point: Per phase)

3. In the delta bank, what voltage is across each capacitor?
   A. 239.6 V
   B. 415 V (the line voltage)
   C. 719 V
   D. 138 V
   Answer: B. In delta, each element sees the line voltage. (Key point: Voltage across capacitor)

4. Which formula gives the capacitance per phase?
   A. C = Q_phase x ω x V²
   B. C = Q_phase/(ω V²), about 147.7 μF
   C. C = V²/Q_phase
   D. C = Q_phase/V
   Answer: B. Q_phase = ωCV², so C = 7992/(314.16 x 415²) = 147.7 μF. (Key point: Result)

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