---
title: "PF correction for 1250 kVA generator (+170 kW)"
subject: "Industrial Power Distribution and Illumination"
chapter: 10
key: "semester_6/industrial_power_distribution_and_illumination:generator-capacitor-170kw"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/generator-capacitor-170kw
site: BCT Engineering
---

# PF correction for 1250 kVA generator (+170 kW)

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

- Asked in: 2079 Chaitra, 2+6 marks
- Video: [Power Factor Explained - The basics what is power factor pf](https://www.youtube.com/watch?v=Tv_7XWf96gg) — The Engineering Mindset
- 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-170kw
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10

## Question

Explain the need for power correction. A 1250 kVA generator set is operating at rated load with a pf of 0.8. An additional load of 170 kW at 0.85 pf is to be added. What value of capacitors is required so that the generator is not overloaded?

## Short answer

Existing load 1000 kW, 750 kVAr; extra 170 kW adds 105.36 kVAr, total 1170 kW, 855.36 kVAr overloads. Allowed Q at 1250 kVA is 440 kVAr, so 415.36 kVAr capacitors are required, resulting in 0.936 pf and exactly 1250 kVA.

## Answer

#### Need for power factor correction
Most industrial loads (induction motors, transformers, welding sets, fluorescent ballasts) draw lagging reactive power. Low pf means:
- Higher current for the same kW → higher I²R losses in cables, transformers and generators.
- Larger kVA rating of generators, transformers and switchgear is needed; less kW can be supplied from a given kVA.
- Larger voltage drop and poor regulation.
- Utility penalties / higher kVA demand charges.

Correcting pf with capacitors supplies the reactive power locally, reducing current, losses and kVA demand, and releasing capacity for extra load.

#### Numerical
**Given:** generator 1250 kVA at rated load, pf 0.8; extra load 170 kW at pf 0.85.

**Existing load**
```text
P₁ = 1250 × 0.8 = 1000 kW
Q₁ = 1250 × 0.6 = 750 kVAr
```
**Additional load**
```text
P₂ = 170 kW
Q₂ = 170 × tan(cos⁻¹0.85) = 170 × 0.6197
   = 105.36 kVAr
```
**Total load without capacitor**
```text
P = 1000 + 170    = 1170 kW
Q = 750 + 105.36  = 855.36 kVAr
S = √(1170² + 855.36²) = 1449.3 kVA  (> 1250, overload)
```
**Maximum kVAr allowed so that S = 1250 kVA**
```text
Q_allowed = √(1250² − 1170²) = √(193 600) = 440.0 kVAr
New pf    = 1170 / 1250 = 0.936 lag
```
**Capacitor required**
```text
Qc = 855.36 − 440.0 = 415.36 kVAr
```

**Answer: about 415.4 kVAr of capacitors is needed; the generator then supplies 1170 kW at 0.936 pf, i.e. exactly 1250 kVA.**

## Check yourself

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

1. Why is low power factor undesirable?
   A. It lowers current for the same kW
   B. It reduces losses
   C. It increases the available kW from the generator
   D. It increases current, I^2R losses, voltage drop and kVA demand charges
   Answer: D. A low pf means higher current for the same kW, so more loss and larger equipment. (Key point: Need for pf correction)

2. What are P and Q of the 1250 kVA generator at rated load and 0.8 pf?
   A. P = 1000 kW, Q = 750 kVAr
   B. P = 750 kW, Q = 1000 kVAr
   C. P = 1250 kW, Q = 0
   D. P = 1000 kW, Q = 1250 kVAr
   Answer: A. P = 1250 x 0.8 = 1000 kW; Q = 1250 x 0.6 = 750 kVAr. (Key point: Existing load)

3. After adding 170 kW, total P = 1170 kW. What reactive power is allowed so S = 1250 kVA?
   A. 855.36 kVAr
   B. 415.36 kVAr
   C. 440 kVAr
   D. 1250 kVAr
   Answer: C. Q_allowed = sqrt(1250^2 - 1170^2) = 440 kVAr. (Key point: Allowed reactive power)

4. What capacitor rating is needed?
   A. 440 kVAr
   B. 415.4 kVAr
   C. 855.4 kVAr
   D. 105.4 kVAr
   Answer: B. Total Q = 750 + 105.36 = 855.36; Qc = 855.36 - 440 = 415.36 kVAr, giving pf 0.936. (Key point: Capacitor size)

## Related in Industrial Power Distribution and Illumination

- [PF correction for 1250 kVA generator (+150 kW)](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Agenerator-capacitor-150kw) — Industrial Power Distribution and Illumination, chapter 10

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