# 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.

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)

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:

[math]\displaystyle{ S^2 = P^2 + Q^2 }[/math]

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 cos φ = 0.5 | 10 kVA | 5 kW | 8.7 kvar | |

Three phase 3-wires or 3-wires + neutral | S = [math]\displaystyle{ \sqrt 3 }[/math] UI | P = [math]\displaystyle{ \sqrt 3 }[/math] UI cos φ | Q = [math]\displaystyle{ \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) |

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

**Pn** = 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 **Figure** L16 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** L4).

Alternatively:

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