Reactive power

Prof. Dr. Thorsten Jungmann (last updated 2025-11-16)

Reactive power: it only oscillates back and forth – it does no work, but it loads the grid!

Reactive power (German: Blindleistung) describes the part of the electrical power in AC circuits that is not converted into useful work but oscillates back and forth between source and load. It results from the phase difference between voltage and current and is needed to build up magnetic and electric fields in inductive and capacitive components such as inductors and capacitors.

In the fundamentals of electrical engineering (EE1) only for completeness

This quantity appears in the EE1 script only for the sake of completeness, because it belongs to the components inductor and capacitor. Calculations and a deeper understanding are part of AC circuit theory (EE2).

1 Quantity symbol and unit

The quantity symbol of reactive power is $Q$.
The unit of reactive power is the volt-ampere reactive, abbreviated with the unit symbol var:

$$[Q]=\text{var}$$

The units kvar (kilovar) and Mvar (megavar) are also frequently used.

Sign convention (German: Vorzeichenkonvention):

• $Q>0$: inductive reactive power (e.g. electric motors, transformers)
• $Q<0$: capacitive reactive power (e.g. capacitors for compensation)

2 Practical and worked examples

Reactive power is defined as:

$$Q=U\cdot I\cdot \sin \phi$$

where:

• $U$ the voltage in volts (V)
• $I$ the current in amperes (A)
• $\phi$ the phase angle between voltage and current

From apparent power and active power:

$$Q=\sqrt{S^2-P^2}$$

where:

• $S$ the apparent power in volt-amperes (VA)
• $P$ the active power in watts (W)

Via the power factor:

$$Q=P\cdot \tan \phi =S\cdot \sin \phi$$

2.1 Asynchronous motor

At $U=400\mathrm{V}$, an electric motor draws a current of $I=12\mathrm{A}$ and has a power factor of $\cos \phi =0.8$.

Apparent power:

$$S=U\cdot I=400\mathrm{V}\cdot 12\mathrm{A}=4.8\mathrm{kVA}$$

Active power:

$$P=S\cdot \cos \phi =4.8\mathrm{kVA}\cdot 0.8=3.84\mathrm{kW}$$

Phase angle:

$$\phi =\arccos (0.8)=36.9°$$

Reactive power:

$$Q=S\cdot \sin \phi =4.8\mathrm{kVA}\cdot \sin (36.9°)=4.8\mathrm{kVA}\cdot 0.6=\underline{2.88\mathrm{kvar}}$$

The electric motor requires 2.88 kvar of inductive reactive power for its magnetic field.

2.2 Compensating capacitor

To compensate the electric motor from the previous example from $\cos {\phi }_1=0.8$ to $\cos {\phi }_2=0.95$, a capacitor is connected in parallel.

Required capacitive reactive power:

$$Q_C=P\cdot (\tan {\phi }_1-\tan {\phi }_2)=3.84\mathrm{kW}\cdot (\tan (36.9°)-\tan (18.2°))$$ $$Q_C=3.84\mathrm{kW}\cdot (0.75-0.33)=3.84\mathrm{kW}\cdot 0.42=\underline{1.61\mathrm{kvar}}$$

Required capacitance at 50 Hz:

$$C=\frac{Q_C}{2\pi f\cdot U^2}=\frac{1610\mathrm{var}}{2\pi \cdot 50\mathrm{Hz}\cdot (400\mathrm{V}{)}^2}=\frac{1610}{50265000}\mathrm{F}=\underline{32.0\mu \mathrm{F}}$$

After compensation:

• New reactive power: $Q_2=Q_1-Q_C=2.88\mathrm{kvar}-1.61\mathrm{kvar}=1.27\mathrm{kvar}$
• New apparent power: $S_2=\frac{P}{\cos {\phi }_2}=\frac{3.84\mathrm{kW}}{0.95}=4.04\mathrm{kVA}$
• New current: $I_2=\frac{S_2}{U}=\frac{4.04\mathrm{kVA}}{400\mathrm{V}}=10.1\mathrm{A}$

2.3 Transformer in idle operation

A transformer with $S_{\mathrm{rated}}=100\mathrm{kVA}$ draws, in idle operation at $U=400\mathrm{V}$, a current of $I_0=2.5\mathrm{A}$. The power factor in idle operation is $\cos {\phi }_0=0.15$ (strongly inductive).

Reactive power in idle operation:

$$Q_0=U\cdot I_0\cdot \sin {\phi }_0=400\mathrm{V}\cdot 2.5\mathrm{A}\cdot \sin (\arccos (0.15))$$ $$Q_0=1000\mathrm{VA}\cdot 0.989=\underline{989\mathrm{var}}$$

This reactive power is required to magnetize the transformer core.

2.4 Industrial plant

A plant with many electric motors has:

• Total active power: $P=500\mathrm{kW}$
• Total reactive power: $Q=400\mathrm{kvar}$ (inductive)

Apparent power:

$$S=\sqrt{P^2+Q^2}=\sqrt{500^2+400^2}\mathrm{kVA}=\sqrt{410000}\mathrm{kVA}=\underline{640.3\mathrm{kVA}}$$

Power factor:

$$\cos \phi =\frac{P}{S}=\frac{500\mathrm{kW}}{640.3\mathrm{kVA}}=\underline{0.78}$$

To improve it to $\cos \phi =0.95$, a compensation system with $Q_C=240\mathrm{kvar}$ is installed.

3 Typical orders of magnitude

Inductive reactive power (positive):

• Asynchronous motors (small): $0.5$ to $5\mathrm{kvar}$
• Asynchronous motors (large): $50$ to $500\mathrm{kvar}$
• Transformers (idle operation): $1$ to $10\%$ of the rated apparent power
• Fluorescent lamps (uncompensated): $0.02$ to $0.05\mathrm{kvar}$ per lamp
• Welding transformers: $10$ to $100\mathrm{kvar}$

Capacitive reactive power (negative, for compensation):

• Compensating capacitor (small consumers): $1$ to $10\mathrm{kvar}$
• Compensation system (industry): $50$ to $500\mathrm{kvar}$
• High-voltage cables (natural capacitance): $0.1$ to $1Mvar/km$

Typical ratios Q/P:

• Asynchronous motor (full load): $Q\approx 0.4$ to $0.6\cdot P$
• Asynchronous motor (idle operation): $Q\approx 3$ to $5\cdot P$
• Transformer (full load): $Q\approx 0.02$ to $0.05\cdot P$
• Transformer (idle operation): $Q\approx 5$ to $10\cdot P$

4 Example phrasings

• ‘The electric motor requires an inductive reactive power of $Q=15.5\mathrm{kvar}$.’
• ‘The compensation system provides $Q_C=-120\mathrm{kvar}$ of capacitive reactive power.’
• ‘Reactive power compensation reduces the reactive power from $80\mathrm{kvar}$ to $25\mathrm{kvar}$.’
• ‘The energy supplier charges surcharges once the reactive power drawn reaches 50 % of the active power.’
• ‘The reactive power is $Q=S\cdot \sin \phi =42.3\mathrm{kvar}$.’

5 Related quantities

• Active power:

$$P=S\cdot \cos \phi =\sqrt{S^2-Q^2}$$

• Apparent power:

$$S=\sqrt{P^2+Q^2}$$

• Power factor:

$$\cos \phi =\frac{P}{S}=\frac{P}{\sqrt{P^2+Q^2}}$$

• Reactive factor:

$$\sin \phi =\frac{Q}{S}$$

• Phase angle:

$$\phi =\arctan \left(\frac{Q}{P}\right)$$

• Voltage and current:

$$Q=U\cdot I\cdot \sin \phi$$

• Reactance:

$$Q=I^2\cdot X=\frac{U^2}{X}$$

• Inductance:

$$Q_L=\omega L\cdot I^2=\frac{U^2}{\omega L}$$

• Capacitance:

$$Q_C=-\omega C\cdot U^2=-\frac{I^2}{\omega C}$$

6 Related components

• Inductor: causes inductive reactive power
• Capacitor: supplies capacitive reactive power for compensation
• Electric motor: main cause of inductive reactive power
• Transformer: requires magnetizing reactive power
• Inductor: in fluorescent lamps and switch-mode power supplies
• Synchronous generator: can generate or absorb reactive power