Power Factor Calculator

Written byOleh Soloninupdated 25 September 2026

Real power, kW

Apparent power, kVA

Target power factor

Advanced options

Voltage, V

Frequency, Hz

Result

Power factor0.8
VerdictBelow the target of 0.95: correct 3.371 kVAr
Phase angle, °36.87
Reactive power, kVAr6
Correction needed, kVAr3.371
Capacitance, µF74.3
How to connect the capacitors
74.3 µF — one capacitor across 380 V, or one per phase in star; in delta, three of 24.8 µF
Working
Q = √(10² − 8²) = 6 kVAr; Q target = 8 × tan(arccos 0.95) = 2.629 kVAr

How much current works and how much just circulates

Computes the power factor, the reactive power behind it and the capacitance needed to reach a target figure.

The power factor is the share of real power in apparent power: PF = P ÷ S. With sinusoidal current and voltage it equals cos φ; with a distorted current, as from switch-mode power supplies, P ÷ S falls below cos φ, and the capacitor sizing below does not apply to such a load. With 8 kW real and 10 kVA apparent it is 0.8, and the reactive power Q = √(S² − P²) is 6 kVAr.

Motors' reactive power is corrected with capacitors. To bring the power factor up to a target, they have to supply the gap between today's reactive power and what would remain at the target: Q − P × tan(arccos PF target).

How to use

Enter the real power in kilowatts and the apparent power in kilovolt-amperes, from the rating plate or the meter readings.
Set the target power factor; the advanced fields hold the supply voltage and frequency.
Read the power factor, the phase angle, the reactive power, how much to correct and the capacitance, with a note on how to connect it.

Correcting a 10 kW load to 0.95

Each row is a separate run: 10 kW, target 0.95, 380 V, 50 Hz. The capacitance is for one capacitor across 380 V, or for each of three in star.

All rows come from the calculator on this page.
Apparent, kVA	Power factor	Reactive, kVAr	Correct, kVAr	Capacitance
10.5	0.952	3.202	0 — not needed	—
11.76	0.85	6.189	2.902	64 µF
12.5	0.8	7.5	4.213	92.9 µF
14.29	0.7	10.208	6.921	152.6 µF
16.67	0.6	13.337	10.051	221.6 µF

Star or delta

A capacitor's reactive power is Q = 2πf × C × V². In star each of the three capacitors sits across the phase voltage V ÷ √3, and together they give what one capacitor across V would. In delta each sits across the full V, so for the same correction each needs a third of the capacitance: 92.9 µF in star against 31 µF in delta.

What this calculator shows

The power factor and the phase angle in degrees
Reactive power and how much of it must be corrected to hit the target
The capacitance in microfarads to do that correction

What to keep in mind

The capacitance assumes a lagging, inductive load with sinusoidal voltage and current: Q = 2πf × C × V². From real and apparent power alone the calculator cannot tell an inductive load from a capacitive one, and capacitors do not remove harmonic distortion.
The calculator does not design a correction bank or its steps — only the reactive power and capacitance needed.

FAQs

How do I calculate the power factor?

Divide real power by apparent power: 8 kW ÷ 10 kVA = 0.8.

How many kVAr raise the power factor from 0.8 to 0.95?

For 8 kW, 3.371 kVAr: the 6 kVAr of reactive power less 8 × tan(arccos 0.95) = 2.629 kVAr.

What if I set the target to 1?

The calculator will correct all the reactive power: for 8 kW and 10 kVA that is 6 kVAr and 132.3 µF at 380 V.

How do I work out the capacitance?

C = Q ÷ (2πf × V²). 3.371 kVAr at 380 V and 50 Hz is 74.3 µF.

How are the capacitors connected on three phases?

In star, three capacitors of the full calculated value; in delta, three of a third of it. The calculator gives both.

What does the angle φ mean?

The phase shift between current and voltage: 36.87° at a power factor of 0.8 and 53.13° at 0.6.

Worked example

An 8 kW workshop

Input: 8 kW, 10 kVA, target 0.95

Output: Power factor 0.8, correct 3.37 kVAr

Note: At 380 V that is one 74.3 µF capacitor, or three of 24.8 µF in delta.

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Result

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