About the Power Factor Calculator
This power factor calculator takes the measured voltage, current and real power of a single-phase or three-phase load and works out the power factor, phase angle, apparent power (kVA) and reactive power (kvar). It then sizes the power factor correction capacitor needed to raise the power factor to a target such as 0.95, in kvar and in microfarads, and shows how much the line current drops afterwards.
It is useful for electricians and facility engineers dealing with utility power factor penalties, for sizing capacitor banks for motors and compressors, and for students learning the power triangle. Lower current after correction also means lower cable losses and voltage drop, and can free up transformer capacity.
The calculation assumes a sinusoidal supply and a lagging (inductive) load, which is typical of motors and transformers. For three-phase systems, enter line-to-line voltage; capacitance is given per phase for a delta-connected bank. Loads with heavy harmonics (VFDs, rectifiers) need a harmonic study before adding capacitors.
With the default inputs, the power factor is 0.802. Change any value above to recalculate instantly.
How to use the power factor calculator
- 1Choose single-phase or three-phase.
- 2Enter the voltage (line-to-line for three-phase), measured current and real power in kW.
- 3Set the power factor you want to reach and your supply frequency.
- 4Read the present power factor, kVA and kvar.
- 5Use the kvar and µF results to choose a correction capacitor bank.
Formula and method
Apparent power S is what the supply delivers in volt-amperes: V × I for single-phase, √3 × V(line) × I for three-phase. Power factor is the share of that doing useful work, P ÷ S, which for sinusoidal waveforms equals the cosine of the angle between voltage and current. Reactive power Q completes the power triangle, S² = P² + Q².
To raise the power factor from cos φ₁ to cos φ₂ without changing the real power, capacitors must supply Qc = P × (tan φ₁ − tan φ₂) kvar. A capacitor’s reactive power is 2πf·C·V², so the capacitance is Qc ÷ (2πf·V²); for a three-phase delta bank each of the three capacitors supplies a third of Qc at line voltage. The corrected current is P ÷ (PF₂ × V) or P ÷ (√3 × V × PF₂).
- P
- Real power (kW)
- S
- Apparent power (kVA)
- Q
- Reactive power (kvar)
- φ
- Phase angle between voltage and current
- Qc
- Capacitor reactive power required (kvar)
- f
- Supply frequency (Hz)
- C
- Capacitance (µF)
Worked examples
480 V three-phase motor load: 60 A, 40 kW
S = √3 × 480 × 60 ÷ 1000 ≈ 49.88 kVA, so PF = 40 ÷ 49.88 ≈ 0.802 and Q ≈ 29.8 kvar. Reaching 0.95 needs 40 × (tan φ₁ − tan φ₂) ≈ 16.7 kvar, about 63.9 µF per phase in delta at 60 Hz, and current falls to about 50.6 A.
Single-phase 230 V, 50 Hz: 20 A, 3.2 kW
S = 230 × 20 = 4.6 kVA, so PF = 3.2 ÷ 4.6 ≈ 0.696. Correcting to 0.95 needs about 2.25 kvar, which at 230 V and 50 Hz is a capacitor of roughly 136 µF.
Load already above target
120 × 10 = 1.2 kVA and 1.17 ÷ 1.2 = 0.975, which is already above 0.95, so no capacitor is required.
Frequently asked questions
How do you calculate power factor?+
Divide real power in kW by apparent power in kVA. Apparent power is volts × amps for single-phase, or √3 × line volts × amps for three-phase. For example 40 kW on a 50 kVA supply is a 0.80 power factor.
How do I calculate kvar for power factor correction?+
Use Qc = kW × (tan(acos PF₁) − tan(acos PF₂)). Raising 100 kW from 0.80 to 0.95 needs 100 × (0.750 − 0.329) ≈ 42 kvar of capacitance.
What is a good power factor?+
Above about 0.95 is generally considered good, and 1.0 is ideal. Many utilities charge commercial customers penalties or demand surcharges when power factor falls below roughly 0.90 to 0.95.
Does power factor correction save money on a home electric bill?+
Usually not. Residential customers are billed for real energy in kWh, not kVA, so correction rarely lowers a home bill. Savings come for commercial and industrial customers billed on kVA or with power factor penalties.
What causes low power factor?+
Inductive loads such as lightly loaded induction motors, transformers, welders and older fluorescent ballasts draw magnetizing current that is out of phase with the voltage. Motors running well below full load have particularly poor power factor.