Definition of reactive power

Reactive power is a component of electrical power that oscillates between the source and the load in an alternating current (AC) circuit without being consumed by the load to perform useful work. Unlike real power, which is responsible for doing useful work such as driving motors or lighting bulbs, reactive power facilitates the transfer of energy between the source and the load without actually being converted into useful work.

Reactive power arises primarily due to the presence of inductive or capacitive elements in the circuit. Inductive loads, such as electric motors and transformers, consume reactive power by storing energy in magnetic fields during one part of the AC cycle and returning it to the system during another part. Capacitive loads, on the other hand, supply reactive power by storing energy in electric fields.

Reactive power is measured in volt-amperes reactive (VAR) and is essential for maintaining the voltage levels and stability of AC power systems. While reactive power does not contribute to the real work performed by the load, it is necessary for maintaining the electromagnetic fields in motors and transformers and ensuring the efficient transmission of electrical energy.

Utilities and industries often need to manage reactive power to maintain the stability and efficiency of the power system. Power factor correction techniques, such as installing capacitors or inductors, are used to minimize reactive power and improve the overall efficiency of the system.

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): P = V.I.sin φ
  -  Single phase (phase to phase): Q = U.I.sin φ
  -  Three phase (3 wires or 3 wires + neutral): P = √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): P = √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 =  UI	P =  UI cos φ	Q =  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:

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

S = apparent power

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:

Fig. L2b : Calculation power diagram