What Is Power Factor? The cos φ Formula and the Relationship between kW, kVA and kvar

Power factor tells you how much of the electrical power supplied to a load is turned into real work. A low value means the cables and transformers carry current that does no work. This article explains it with the power triangle and numeric examples.

By Ajarn ModUpdated 16 Oct 2026

30-second summary

$\text{PF}=\dfrac{P}{S}=\cos\varphi$ when voltage and current are sinusoidal
P = real power (kW) · Q = reactive power (kvar) · S = apparent power (kVA), and $S^2=P^2+Q^2$
Same real power but a low PF ⇒ the current rises by a factor of $1/\text{PF}$

What is power factor

In an AC circuit, voltage and current do not necessarily rise and fall together. If they are out of phase by an angle $\varphi$ , the power converted to real work is only a fraction of $V\times I$ .

P = V I \cos\varphi, \qquad S = V I, \qquad \text{PF} = \frac{P}{S} = \cos\varphi

Here $V$ and $I$ are RMS values. A purely resistive load (a heater, an incandescent lamp) has $\varphi=0$ and PF = 1. Induction motors, transformers and iron-core ballasts contain coils, which make the current lag the voltage, so their PF is below 1.

The P Q S power triangle

The three kinds of power are related as a right triangle.

Power	Symbol · unit	Meaning	Formula
Real power	$P$ · W, kW	Power converted to work, heat, light or torque	$VI\cos\varphi$
Reactive power	$Q$ · var, kvar	Power that shuttles back and forth between the source and the magnetic or electric field in the load	$VI\sin\varphi$
Apparent power	$S$ · VA, kVA	The burden that cables, transformers and generators must support	$VI=\sqrt{P^2+Q^2}$

For a balanced three-phase system, using line voltage and line current, $P=\sqrt{3}\,V_L I_L\cos\varphi$ and $S=\sqrt{3}\,V_L I_L$ .

Example: a 230 V, 10 A load with a lagging PF of 0.8

Apparent power and real power
$S=VI=230\times10=2300\ \text{VA},\qquad P=S\cos\varphi=2300\times0.8=1840\ \text{W}$
Reactive power from the power triangle
$Q=\sqrt{S^2-P^2}=\sqrt{2300^2-1840^2}=1380\ \text{var}$
If the PF is corrected to 1.0 with the same real power, the current needed is
$I'=\frac{P}{V}=\frac{1840}{230}=8\ \text{A}$
Compare the line loss $\propto I^2$
$\left(\frac{10}{8}\right)^2=1.5625$

At PF 0.8 the load draws 10 A, but at PF 1.0 only 8 A for the same 1840 W of work. The current differs by 25 % and the line loss is about 56 % higher (2300 : 1840 : 1380 is a 5 : 4 : 3 triangle).

Lagging and leading PF

Lagging: the current lags the voltage. Inductive loads (motors, transformers), which make up most of a factory's load.
Leading: the current leads the voltage. Capacitive loads, or PF-correction capacitors installed beyond what is needed. A long unloaded cable has a similar effect.

A PF value alone does not say whether it is lagging or leading, so state it explicitly, for example "0.8 lagging".

With harmonics: the true PF is lower than cos φ

Electronic loads such as switching power supplies and rectifiers draw non-sinusoidal current, which lowers the true PF even when the fundamental current is in phase with the voltage.

\text{PF}_{true}=\underbrace{\cos\varphi_1}_{\text{displacement}}\times\frac{1}{\sqrt{1+\text{THD}_I^2}}

Example: $\cos\varphi_1=1$ and $\text{THD}_I=30\,\%$ give $\text{PF}_{true}=1/\sqrt{1.09}\approx0.958$ . Installing PF-correction capacitors alone is therefore not enough, and may cause resonance with the harmonics, so THD must be considered as well.

Keep learning in the library

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

Frequently asked questions

What is a good power factor?

The closer to 1.0 the better. In Thai practice, customers in tariff classes that carry a penalty usually correct the PF to above 0.85 under the utility's criterion, and design for about 0.95 to allow for light-load periods. See the power factor and electricity bill page for the criteria.

What is the difference between lagging and leading PF?

Lagging means the current lags the voltage, caused by inductive loads such as motors and transformers. Leading means the current leads the voltage, caused by capacitive loads or by too many capacitors installed.

What harm does a low PF cause?

More current is needed for the same real power, so cables, transformers and breakers carry a heavier load, line losses ( $I^2R$ ) and voltage drop increase, and some customer classes must pay a penalty.

How does true PF differ from cos φ?

cos φ is only the displacement PF of the fundamental frequency. If the current is distorted by harmonics, the true PF is $\text{PF}=\cos\varphi_1/\sqrt{1+\text{THD}_I^2}$ , which is lower than cos φ.