Definition of reactive power

For most electrical loads like motors, the current I is lagging behind the voltage V by an angle φ.

If currents and voltages are perfectly sinusoidal signals, a vector diagram can be used for representation.

In this vector diagram, the current vector can be split into two components: one in phase with the voltage vector (component I_a), one in quadrature (lagging by 90 degrees) with the voltage vector (component I_r). See Fig. L1.

I_a is called the active component of the current.

I_r is called the reactive component of the current.

Fig. L1 – Current vector diagram

The previous diagram drawn up for currents also applies to powers, by multiplying each current by the common voltage V. See Fig. L2.

We thus define:

• Apparent power: S = V x I (kVA)
• Active power: P = V x Ia (kW)
• Reactive power: Q = V x Ir (kvar)

Fig. L2 – Power vector diagram

In this diagram, we can see that:

• Power Factor: P/S = cos φ

This formula is applicable for sinusoidal voltage and current. This is why the Power Factor is then designated as "Displacement Power Factor".

• Q/S = sinφ
• Q/P = tanφ

A simple formula is obtained, linking apparent, active and reactive power:

• S² = P² + Q²

A power factor close to unity means that the apparent power S is minimal. This means that the electrical equipment rating is minimal for the transmission of a given active power P to the load. The reactive power is then small compared with the active power.

A low value of power factor indicates the opposite condition.

Useful formulae (for balanced and near-balanced loads on 4-wire systems):

• Active power P (in kW)
  • Single phase (1 phase and neutral): P = V.I.cos φ
  • Single phase (phase to phase): P = U.I.cos φ
  • Three phase (3 wires or 3 wires + neutral): P = √3.U.I.cos φ
• Reactive power Q (in kvar)
  • Single phase (1 phase and neutral): Q = V.I.sin φ
  • Single phase (phase to phase): Q = U.I.sin φ
  • Three phase (3 wires or 3 wires + neutral): Q = √3.U.I.sin φ
• Apparent power S (in kVA)
  • Single phase (1 phase and neutral): S = V.I
  • Single phase (phase to phase): S = U.I
  • Three phase (3 wires or 3 wires + neutral): S = √3.U.I

where:

V = Voltage between phase and neutral
U = Voltage between phases
I = Line current
φ = Phase angle between vectors V and I.

An example of power calculations (see Fig. L3)

Type of circuit		Apparent power S (kVA)	Active power P (kW)	Reactive power Q (kvar)
Single-phase (phase and neutral)		S = VI	P = VI cos φ	Q = VI sin φ
Single-phase (phase to phase)		S = UI	P = UI cos φ	Q = UI sin φ
Example	5 kW of load	10 kVA	5 kW	8.7 kvar
	cos φ = 0.5
Three phase 3-wires or 3-wires + neutral		S = [math]\displaystyle{ \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\sqrt 3 }[/math] UI	P = [math]\displaystyle{ \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\sqrt 3 }[/math] UI cos φ	Q = [math]\displaystyle{ \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\sqrt 3 }[/math] UI sin φ
Example	Motor Pn = 51 kW	65 kVA	56 kW	33 kvar
	cos φ= 0.86
	ρ= 0.91 (motor efficiency)

Fig. L3 – Example in the calculation of active and reactive power

The calculations for the three-phase example above are as follows:

delivered shaft power = 51 kW = P
active power consumed =

[math]\displaystyle{ P=\frac {Pn}{\rho}=\frac{51}{0.91}=56\, kW }[/math]

S = apparent power

[math]\displaystyle{ S=\frac{P}{cos \varphi}=\frac {56}{0.86}= 65\, kVA }[/math]

So that, on referring to diagram Figure L3 or using a pocket calculator, the value of tan φ corresponding to a cos φ of 0.86 is found to be 0.59

Q = P tan φ = 56 x 0.59 = 33 kvar (see Figure L15).

Alternatively:

[math]\displaystyle{ Q=\sqrt{S^2 - P^2}=\sqrt{65^2 - 56^2}=33\, kvar }[/math]

Fig. L2b – Calculation power diagram