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Power Factor Calculator

Calculate power factor, real power (kW), apparent power (kVA), and reactive power (kVAR) in AC systems.

kW
kVA
PF

Optional: size the capacitor bank needed to correct to this PF.

Power Factor & Triangle Results

Power Factor (cos θ)
0.8

Phase Angle: 36.87°

Reactive Power (Q)
60 kVAR
Real Power (P)
80 kW
Apparent Power (S)
100 kVA

Capacitor Bank for Correction

Capacitor Bank Required
33.71 kVAR

Raises PF from 0.8 to the target value

kVA Demand After Correction
84.21 kVA

Down from 100 kVA (15.8% less demand)

Reactive Power After
26.29 kVAR

Step-by-Step Engineering Calculation

  1. 1Present condition: 80 kW80\text{ kW} at PF 0.80.8 → S1=100.00 kVAS_1 = 100.00\text{ kVA}, Q1=60.00 kVARQ_1 = 60.00\text{ kVAR}
  2. 2Target condition: PF 0.950.95 → S2=84.21 kVAS_2 = 84.21\text{ kVA}, Q2=26.29 kVARQ_2 = 26.29\text{ kVAR}
  3. 3Capacitor bank required: Qc=P×(tan⁡θ1−tan⁡θ2)=80×(0.7500−0.3287)=33.71 kVARQ_c = P \times (\tan\theta_1 - \tan\theta_2) = 80 \times (0.7500 - 0.3287) = \mathbf{33.71\text{ kVAR}}
  4. 4Correcting to 0.95 PF frees 15.79 kVA of system capacity (15.8% reduction in kVA demand).

Step-by-Step Engineering Calculation

  1. 1Power Factor: PF = 80 kW / 100 kVA = 0.8000
  2. 2Reactive Power: Q = √(100² - 80²) = 60.000 kVAR

Formula & Theory

PF = P (kW) / S (kVA) = cos(θ) | Q (kVAR) = √(S² - P²)

Computes the efficiency of electrical power delivery in AC networks.

Variable Definitions
PF:
Power Factor (Ratio (0 - 1.0)) – Ratio of real power to apparent power
P:
Real Power (kW) – Working power consumed
S:
Apparent Power (kVA) – Total power supplied
Q:
Reactive Power (kVAR) – Magnetizing/reactive power

Worked Engineering Example

Factory Induction Motor

A commercial facility with 120 kW real load drawing 150 kVA apparent power.

Inputs:
  • Real Power: 120 kW
  • Apparent Power: 150 kVA
Result: Power Factor = 0.8000 | Reactive Power = 90.00 kVAR (Phase Angle: 36.87°)

Key Assumptions

  • •Sinusoidal AC waveforms.

Limitations & Safety Scope

  • •Does not calculate displacement power factor with high harmonic content.

Frequently Asked Questions

Why is a low power factor penalized by utilities?

A low power factor requires the utility to generate and transmit higher total current (kVA) to deliver the same useful work (kW), causing higher transmission losses and requiring oversized transformers and cables.

Related Calculators & Engineering Tools

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Breaker Size Calculator
Watts to VA Converter
VA to Watts Converter
Power Factor Theory Reference

Educational & Reference Notice: This calculator provides technical estimates based on standard physical formulas. Actual electrical installation, protective device sizing, and conductor selection must be verified against the applicable electrical code (NEC, IEC, CEC) by a licensed electrician or qualified professional engineer.