AC power
Single-phase apparent power
- where
- apparent power (VA)
- voltage (V)
- current (A)
Explain this equation
Gives the apparent power, in volt-amperes, of a single-phase load. Use it to size supplies, transformers, UPSs and generators, which are rated by the current they carry rather than by watts.
V is the RMS voltage and I the RMS current. S is their plain product in VA. It counts both the current that does work and the current that just circulates because of inductance or capacitance.
Cables and windings heat according to the current they carry, whatever its phase. So equipment has to be sized on V × I, regardless of how much of it ends up as useful power.
230 V at 10 A gives 2300 VA, or 2.3 kVA. At power factor 0.8 the load uses only 1840 W but still loads the supply by 2.3 kVA.
Watch out: VA and W are equal only at unity power factor, so don't label a VA result as watts.
Apparent power is useful when discussing transformer and supply loading. It includes the effect of current that does not contribute to active power.
Keep VA separate from W. A 2 kVA load uses 2 kW only at unity power factor; current is found from I = S/V.
Enter your values to see the result.
Common mistake
Do not call a VA result watts without accounting for power factor.
Limits
Use RMS voltage and current. The cos φ form assumes sinusoidal voltage and current; for distorted waveforms use measured true power factor, P/S.
Sources
The tools calculate the displayed equation. They do not automatically choose cable sizes, approve installations, supply current regulatory tables, certify equipment or replace inspection and testing.