acalculator

What is the power factor?

Enter the real power with the volts and amps (or the kVA). The power factor calculator shows the power factor, phase angle and reactive power, and the capacitor kVAR that would raise it to your target.

Your numbers

Units
Supply
I know
Power factor
0.8

The power factor is 0.8 (phase angle 36.87°), with 12 kVA and 7.2 kVAR.

Phase angle (°)
36.87
Apparent power (kVA)
12
Reactive power (kVAR)
7.2
Capacitor kVAR to reach the target
4.04463
kVA after correction
10.1053

Power factor: 0.8. The power factor is 0.8 (phase angle 36.87°), with 12 kVA and 7.2 kVAR.

How to calculate

Works out the power factor from real power and volts and amps (single or three-phase) or kVA, with the phase angle, reactive power, and the capacitor kVAR needed to correct it to a target.

Example with the default inputs (Supply Single-phase, I know kW, volts and amps, Real power 9.6 kW, Voltage 240 V, Current 50 A, Target power factor 0.95): The power factor is 0.8 (phase angle 36.87°), with 12 kVA and 7.2 kVAR.

Method: S = V × I ÷ 1,000 kVA (single-phase), √3 × V × I ÷ 1,000 (three-phase), or as typed. pf = P ÷ S; φ = arccos pf; Q = P × √(1 − pf²) ÷ pf. Correction to a target pf₂: Qc = Q − P × √(1 − pf₂²) ÷ pf₂; kVA after = P ÷ pf₂.

  • RMS values on a sinusoidal supply; three-phase means a balanced load with line-to-line volts and line amps.
  • The correction assumes a lagging (inductive) load such as a motor, corrected with capacitors. A target at or below the present power factor shows no correction.
  • Steps are exact on the typed values except √3 and the angle.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. I know kW, volts and amps, Supply Single-phase, Real power 9.6 kW, Voltage 240 V, Current 50 A, Target power factor 0.95 gives Power factor 0.8, Phase angle (°) 36.869898, Apparent power (kVA) 12, Reactive power (kVAR) 7.2, Capacitor kVAR to reach the target 4.044633, kVA after correction 10.105263.Source: OpenStax, University Physics Volume 2, §15.4 Power in an AC Circuit (P_ave = I_rms V_rms cos φ; cos φ is the power factor), https://openstax.org/books/university-physics-volume-2/pages/15-4-power-in-an-ac-circuit (retrieved 2026-10-05); Wikipedia, "Power factor" (pf = P ÷ S; S² = P² + Q²; correction by capacitors that supply reactive power), https://en.wikipedia.org/wiki/Power_factor (retrieved 2026-10-05)
  2. I know kW and kVA, Supply Single-phase, Real power 67.5 kW, Apparent power (kVA) 75 gives Power factor 0.9, Reactive power (kVAR) 32.691742.Source: Wikipedia, "Power factor" (pf = P ÷ S; S² = P² + Q²; correction by capacitors that supply reactive power), https://en.wikipedia.org/wiki/Power_factor (retrieved 2026-10-05)
  3. I know kW, volts and amps, Supply Three-phase, Real power 60 kW, Voltage 480 V, Current 90 A gives Power factor 0.801875, Apparent power (kVA) 74.824595.Source: Wikipedia, "Three-phase electric power" (balanced load: S = √3 × line voltage × line current), https://en.wikipedia.org/wiki/Three-phase_electric_power (retrieved 2026-10-05)

How it works

With P the real power in kW and S the apparent power in kVA:

  • S = V × I ÷ 1,000 for single-phase, √3 × V × I ÷ 1,000 for balanced three-phase (V line-to-line), or S as typed
  • power factor pf = P ÷ S
  • phase angle φ = arccos(pf), in degrees
  • reactive power Q (kVAR) = P × √(1 − pf²) ÷ pf, which equals √(S² − P²)

Correction to a target power factor pf₂

  • Q after = P × √(1 − pf₂²) ÷ pf₂
  • capacitor kVAR = Q − Q after = P × (tan φ1 − tan φ2)
  • kVA after = P ÷ pf₂

The correction is shown only when the target is above the present power factor.

Rules

  • The real power cannot be above the apparent power; if it is, the page gives no answer.
  • Real power from 1 mW to 10¹⁰ W; voltage from 1 mV to 1,000 kV; current from 1 µA to 10⁶ A; kVA up to 10⁷; target from 0.01 to 1.
  • The steps are exact on the typed values except √3, the square roots that are not exact, and the angle. The power factor is shown to 4 decimals, the angle to 2, and the powers to 6 significant figures.
  • The load is taken as lagging (inductive), corrected with capacitors.

Worked examples by hand

9.6 kW at 240 V and 50 A, single-phase, target 0.95. S = 240 × 50 ÷ 1,000 = 12 kVA. pf = 9.6 ÷ 12 = 0.8. φ = arccos 0.8 = 36.87°. Q = 9.6 × 0.6 ÷ 0.8 = 7.2 kVAR. At 0.95, Q after = 9.6 × √(1 − 0.9025) ÷ 0.95 = 9.6 × 0.312250 ÷ 0.95 = 3.15537 kVAR, so the capacitors supply 7.2 − 3.15537 = 4.0446 kVAR. kVA after = 9.6 ÷ 0.95 = 10.1053 kVA.

67.5 kW out of 75 kVA. pf = 67.5 ÷ 75 = 0.9. Q = √(75² − 67.5²) = √1,068.75 = 32.6917 kVAR.

60 kW at 480 V and 90 A, three-phase. S = 1.7320508 × 480 × 90 ÷ 1,000 = 74.8246 kVA. pf = 60 ÷ 74.8246 = 0.8019.

Other questions people ask

How do I calculate power factor?

Divide the real power in kW by the apparent power in kVA. Apparent power is volts × amps ÷ 1,000 for single-phase, and √3 × volts × amps ÷ 1,000 for three-phase. A load using 9.6 kW at 240 V and 50 A has 12 kVA, so its power factor is 9.6 ÷ 12 = 0.8.

What is a good power factor?

Closer to 1 is better: the supply then carries no more current than the work needs. Heaters and filament bulbs are near 1. Motors are lower, and lower still at part load. Some utilities charge extra when it falls below a level set in their tariff.

How do I size a capacitor for power factor correction?

The capacitor must supply the drop in reactive power: kVAR = kW × (tan φ1 − tan φ2), where cos φ1 is the present power factor and cos φ2 the target. For 9.6 kW from 0.8 to 0.95: 9.6 × (0.75 − 0.3287) = 4.04 kVAR.

What is the phase angle?

In an AC circuit with coils or capacitors, the current peaks a little before or after the voltage. The phase angle φ measures that shift, and the power factor is its cosine. A power factor of 0.8 is a phase angle of 36.87°.

What is the difference between leading and lagging power factor?

Lagging means the current peaks after the voltage, as in motors and other coils. Leading means it peaks before, as with too many capacitors. The number alone does not tell you which; the correction on this page assumes a lagging load.

Why does a low power factor matter?

For the same real power, a low power factor means more current. More current heats cables and transformers, causes larger voltage drops, and uses up capacity, which is why utilities may charge for it.