How to Calculate the Size of a Power Factor Correction Capacitor (kvar) with an Example

Correcting the PF means installing capacitors that supply the reactive power (kvar) to inductive loads, instead of drawing it from the grid. The size you need follows from a single formula and can be checked with the power triangle.

By Ajarn ModUpdated 16 Oct 2026

30-second summary

$Q_c=P\,(\tan\varphi_1-\tan\varphi_2)$ , where $\varphi_1$ is the angle before correction and $\varphi_2$ the target angle
Real power P does not change; what falls is Q and S, so the current falls
Choose the nearest standard size, and check harmonics before installing

Where the formula comes from

From the power triangle, $Q=P\tan\varphi$ . Before correction the load has $Q_1=P\tan\varphi_1$ ; after correction we want $Q_2=P\tan\varphi_2$ . The capacitor must supply the difference.

Q_c=Q_1-Q_2=P\,(\tan\varphi_1-\tan\varphi_2),\qquad \varphi=\cos^{-1}(\text{PF})

Then find the capacitance per phase from $Q_c=3\,\omega C V^2$ for a delta connection, where $V$ is the line voltage and $\omega=2\pi f$ .

Example: a 100 kW, 400 V three-phase load, PF 0.75 → 0.95

Find the angles and tangents before and after
$\tan\varphi_1=\tan\!\big(\cos^{-1}0.75\big)=0.8819,\qquad \tan\varphi_2=\tan\!\big(\cos^{-1}0.95\big)=0.3287$
The reactive power to compensate
$Q_c=100\times(0.8819-0.3287)\approx55.3\ \text{kvar}$
Line current before and after correction ( $I=\dfrac{P}{\sqrt{3}\,V\,\text{PF}}$ )
$I_1=\frac{100000}{\sqrt{3}\,(400)(0.75)}\approx192.5\ \text{A},\qquad I_2=\frac{100000}{\sqrt{3}\,(400)(0.95)}\approx151.9\ \text{A}$
Capacitance per phase for a delta connection at 50 Hz
$C=\frac{Q_c}{3\,\omega V^2}=\frac{55323}{3\,(2\pi\cdot50)(400)^2}\approx367\ \mu\text{F}$

About 55.3 kvar of capacitors is needed. Choose the next standard size up, for example 60 kvar in several switched steps. The line current falls from 192.5 A to 151.9 A (about 21 % lower). If connected in wye, the capacitance per phase would be three times larger, about 1,101 µF.

Multiplier table: Qc = P × k for a target PF of 0.95

Multiply the real power in kW by $k=\tan\varphi_1-\tan\varphi_2$ to get kvar directly.

Original PF	$k$	Example P = 100 kW
0.70	0.692	69.2 kvar
0.75	0.553	55.3 kvar
0.80	0.421	42.1 kvar
0.85	0.291	29.1 kvar
0.90	0.156	15.6 kvar

Cautions before installing

Use the measured P, not the motor nameplate power, because at light load the PF is lower and P is below the rating.
Do not over-correct. If the capacitor is larger than the load's Q at light load, you get a leading PF and a higher voltage. Use an automatic bank that switches in steps.
Harmonics: resonance between the capacitors and the system inductance can amplify harmonic currents. Measure THD, and choose a detuned type with reactors when there are many electronic loads.
Voltage rating: choose capacitors with a voltage rating comfortably above the system voltage, as the manufacturer recommends.

Keep learning in the library

The full lessons are in Thai, with English technical terms and the same formulas.

Frequently asked questions

What power factor should I correct to?

A common target is 0.95 or above, because it allows for light-load periods when a motor's PF drops, and it stops short of 1.0, which needs a lot more capacitance for only a small further reduction in current. An automatic installation that switches steps with the load prevents over-compensation into a leading PF.

Should capacitors be connected in delta or wye?

For the same kvar, a delta connection needs one third of the capacitance per phase of a wye connection (because the voltage across each unit is $\sqrt{3}$ times higher). Low-voltage capacitor banks are usually connected in delta.

What should I watch for after installing capacitors?

If the system has a lot of harmonics, the capacitors may resonate with the transformer inductance and amplify harmonic currents. Measure THD before installing, and use a capacitor bank in series with reactors (detuned) when necessary.

Which part of the electricity bill do capacitors reduce?

They reduce the PF penalty on the bill under the utility's criteria, and they reduce current in cables and transformers, which lowers losses. They do not reduce the kWh the load actually uses.