Single-phase power
One phase. Not a cable size.
Method
P = VI × pf. Worked case: 230 V, 10 A, pf 0.9 is 2.07 kW and 2.30 kVA.
What to write beside the result
A review can only check this figure if the inputs sit next to it. Write the value, the unit and the assumption the form used. A diameter in millimetres and a diameter in metres are not the same reading. A force in kilonewtons divided as if it were newtons is a thousand-fold error.
The result is the arithmetic on this page. It is not a utilisation, not a selected product, and not a clause from a standard. If the number is going into a shared container, name the page, the date and the inputs. The file-name checker does not read the calculation.
Logic: how this figure is built.
How to read single-phase power
This page answers the real power and the apparent power of a single-phase load: P = VI × power factor, and S = VI. At 230 V, 10 A and a power factor of 0.9, real power is 2.07 kW and apparent power is 2.30 kVA. It is a balanced single-phase product. It is not a cable size.
Power factor is the cosine of the phase angle for a sinusoidal load. A power factor of 1 means real and apparent power match. A power factor of 0.9 means the current is higher than the watts alone would suggest. The page does not separate lagging from leading.
Voltage is the supply voltage at the load, not a nominal 230 V if the point you care about is elsewhere. Current is the line current of that load. A diversity factor, a duty cycle and a starting current are not in the product.
Use it to check a nameplate against a measured current. Do not use it to select a cable, a protective device or a distributor. Those need the installation method and BS 7671. Three-phase loads use the other page.
- Related: three-phase power
- electrical logic
Last checked 3 October 2026. See the disclaimer.