Apparent power

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

Apparent power: what the grid has to supply – not all of it becomes useful output power!

Apparent power (German: Scheinleistung) is the product of voltage and current in AC circuits. It describes the total power provided by the grid and is made up of active power and reactive power (German: Blindleistung). Apparent power is the decisive quantity for dimensioning electric lines, transformers and generators.

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 with the components inductor and capacitor. Calculations and a deeper understanding are part of AC theory (EE2).

1 Quantity symbol and unit

The quantity symbol of apparent power is $S$.
The unit of apparent power is the volt-ampere, abbreviated with the unit symbol $\mathrm{VA}$:

$$[S]=\text{VA}$$

To distinguish it from active power (in watts), the unit volt-ampere is deliberately used instead of the watt, although both are dimensionally identical.

2 Practical and worked examples

Apparent power is defined as:

$$S=U\cdot I$$

where:

• $U$ the voltage in volts (V)
• $I$ the current in amperes (A)

The relationship with active and reactive power:

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

where:

• $P$ the active power in watts (W)
• $Q$ the reactive power in volt-amperes reactive (var)

2.1 Transformer

A transformer has the following data on its rating plate:

• Rated voltage: $U_{\mathrm{rated}}=400\mathrm{V}$
• Rated current: $I_{\mathrm{rated}}=50\mathrm{A}$

Apparent power:

$$S=U_{\mathrm{rated}}\cdot I_{\mathrm{rated}}=400\mathrm{V}\cdot 50\mathrm{A}=\underline{20\mathrm{kVA}}$$

The transformer is designed for an apparent power of 20 kVA.

2.2 Electric motor with reactive power

At $U=400\mathrm{V}$, an electric motor draws a current of $I=15\mathrm{A}$ and delivers an active power of $P=4.5\mathrm{kW}$.

Apparent power:

$$S=U\cdot I=400\mathrm{V}\cdot 15\mathrm{A}=\underline{6.0\mathrm{kVA}}$$

Reactive power:

$$Q=\sqrt{S^2-P^2}=\sqrt{6.0^2-4.5^2}\mathrm{kvar}=\sqrt{36-20.25}\mathrm{kvar}=\underline{3.97\mathrm{kvar}}$$

Power factor:

$$\cos \phi =\frac{P}{S}=\frac{4.5\mathrm{kW}}{6.0\mathrm{kVA}}=\underline{0.75}$$

2.3 Dimensioning of a domestic installation

A detached house has a house connection with 3 phases × 400 V and fuse protection of 3 × 25 A.

Maximum apparent power per phase:

$$S_{phase}=U\cdot I=230\mathrm{V}\cdot 25\mathrm{A}=5.75\mathrm{kVA}$$

Maximum total apparent power (with a symmetrical load):

$$S_{tot}=3\cdot S_{phase}=3\cdot 5.75\mathrm{kVA}=\underline{17.25\mathrm{kVA}}$$

With a typical power factor of $\cos \phi =0.95$, the maximum active power is:

$$P_{max}=S_{tot}\cdot \cos \phi =17.25\mathrm{kVA}\cdot 0.95=\underline{16.4\mathrm{kW}}$$

2.4 Industrial plant before and after compensation

Before compensation, a plant has:

• Active power: $P=100\mathrm{kW}$
• Power factor: $\cos {\phi }_1=0.70$

Apparent power before compensation:

$$S_1=\frac{P}{\cos {\phi }_1}=\frac{100\mathrm{kW}}{0.70}=\underline{142.9\mathrm{kVA}}$$

After improvement to $\cos {\phi }_2=0.95$:

$$S_2=\frac{P}{\cos {\phi }_2}=\frac{100\mathrm{kW}}{0.95}=\underline{105.3\mathrm{kVA}}$$

Reduction in apparent power:

$$\Delta S=142.9\mathrm{kVA}-105.3\mathrm{kVA}=\underline{37.6\mathrm{kVA}}$$

(26 % reduction)

This reduction relieves the grid and allows additional loads to be connected without expanding the grid.

3 Typical orders of magnitude

Low power (< 1 kVA):

• Smartphone charger: $10$ to $30\mathrm{VA}$
• Notebook power supply: $65$ to $120\mathrm{VA}$
• Router/modem: $10$ to $20\mathrm{VA}$

Household appliances (1 – 10 kVA):

• Refrigerator: $100$ to $200\mathrm{VA}$
• Washing machine: $2$ to $3\mathrm{kVA}$
• Electric cooker: $5$ to $8\mathrm{kVA}$
• Instantaneous water heater: $18$ to $27\mathrm{kVA}$

Commercial installations (10 – 1000 kVA):

• House connection of a detached house: $15$ to $30\mathrm{kVA}$
• Small businesses: $30$ to $100\mathrm{kVA}$
• Distribution transformer: $100$ to $630\mathrm{kVA}$
• Medium-sized industrial plants: $400$ to $1000\mathrm{kVA}$

Large installations (> 1 MVA):

• Production halls: $1$ to $10\mathrm{MVA}$
• Substation: $10$ to $100\mathrm{MVA}$
• Power station unit: $500$ to $1500\mathrm{MVA}$

4 Example phrasings

• ‘The transformer has a rated apparent power of $S_{\mathrm{rated}}=630\mathrm{kVA}$.’
• ‘The apparent power is $S=U\cdot I=4.8\mathrm{kVA}$.’
• ‘The generator supplies a maximum of $500\mathrm{MVA}$ apparent power.’
• ‘With an apparent power of 20 kVA and cos φ = 0.9, the active power is 18 kW.’
• ‘The electric line must be dimensioned for an apparent power of 100 kVA.’

5 Related quantities

• Active power:

$$P=S\cdot \cos \phi =U\cdot I\cdot \cos \phi$$

• Reactive power:

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

• Power triangle:

$$S^2=P^2+Q^2$$

• Power factor:

$$\cos \phi =\frac{P}{S}$$

• Voltage and current:

$$S=U\cdot I$$

• Impedance:

$$S=I^2\cdot |Z|=\frac{U^2}{|Z|}$$

• Complex apparent power:

$$\underline{S}=P+jQ=U\cdot I^{\ast }$$

(with $I^{\ast }$ as the conjugate of the current)

6 Related components

• Transformer: dimensioned according to apparent power
• Generator: rated power specified as apparent power
• Electric motor: apparent power drawn is relevant for the grid load
• Fuse: dimensioning according to the current that results from the apparent power
• Electric line: cross-section depends on the current load due to the apparent power
• Inverter: rated power usually specified as apparent power