---
title: "PF correction for 1250 kVA generator (+150 kW)"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:generator-capacitor-150kw"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/generator-capacitor-150kw
site: BCT Engineering
---

# PF correction for 1250 kVA generator (+150 kW)

Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry

- Asked in: 2073 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/generator-capacitor-150kw
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10

## Question

A 1250 kVA turbine generator set is operating at rated load with a power factor of 0.8. An additional load of 150 kW at 0.85 power factor is to be added. What kVAR value of capacitors is required so that the generator is not loaded?

## Short answer

Capacitor required is 353.06 kVAr. Generator then supplies 1150 kW at 0.92 pf, exactly 1250 kVA, not overloaded.

## Answer

The generator must not exceed 1250 kVA after the extra load is added, so capacitors must supply enough kVAR to keep the total kVA at 1250.

**Existing load (rated, pf 0.8):**
```text
P1 = 1250 × 0.8          = 1000 kW
Q1 = 1250 × sin(cos⁻¹0.8) = 1250 × 0.6 = 750 kVAR
```

**Additional load (150 kW, pf 0.85):**
```text
tan(cos⁻¹0.85) = 0.6197
Q2 = 150 × 0.6197        = 92.96 kVAR
```

**Total load:**
```text
P  = 1000 + 150   = 1150 kW
Q  = 750 + 92.96  = 842.96 kVAR
```

**Reactive power allowed on the generator at 1250 kVA:**
```text
Q_allowed = √(1250² − 1150²)
          = √(1 562 500 − 1 322 500)
          = √240 000 = 489.90 kVAR
```

**Capacitor kVAR required:**
```text
Qc = 842.96 − 489.90 = 353.06 kVAR
```

New power factor of generator = 1150 / 1250 = 0.92 lagging.

**Answer: About 353 kVAR of capacitors is required; the generator then supplies 1150 kW at 0.92 pf, i.e. exactly 1250 kVA (not overloaded).**

## Check yourself

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

1. What must the capacitors achieve for the generator?
   A. Raise the total kW to 1250
   B. Keep the total kVA at or below 1250
   C. Increase generator voltage
   D. Make the pf leading
   Answer: B. The generator kVA rating is the limit, so reactive power is cut to keep S at 1250 kVA. (Key point: Constraint)

2. What is the reactive power of the added 150 kW at 0.85 pf?
   A. About 92.96 kVAr
   B. About 150 kVAr
   C. About 127.5 kVAr
   D. About 750 kVAr
   Answer: A. Q2 = 150 x tan(cos^-1 0.85) = 150 x 0.6197 = 92.96 kVAr. (Key point: Additional load Q)

3. With total P = 1150 kW, what reactive power is allowed at 1250 kVA?
   A. 842.96 kVAr
   B. 600 kVAr
   C. 353 kVAr
   D. 489.90 kVAr
   Answer: D. sqrt(1250^2 - 1150^2) = sqrt(240000) = 489.90 kVAr. (Key point: Allowed Q)

4. What capacitor kVAr is required?
   A. About 353 kVAr
   B. About 489.9 kVAr
   C. About 842.96 kVAr
   D. About 92.96 kVAr
   Answer: A. Total Q = 750 + 92.96 = 842.96; Qc = 842.96 - 489.90 = 353.06 kVAr, giving pf 0.92. (Key point: Capacitor size)

## Related in Industrial Power Distribution and Illumination

- [PF correction for 1250 kVA generator (+170 kW)](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Agenerator-capacitor-170kw) — Industrial Power Distribution and Illumination, chapter 10

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