Three-phase power
A steady balanced load. Not a cable size, discrimination study or motor-start check.
Method
P = √3 × V × I × pf. Worked case: 400 V, 32 A, pf 0.85 is 18.85 kW and 22.17 kVA.
Not BS 7671. See what this leaves out.
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 three-phase power
For a balanced load, real power is √3 × VL × IL × power factor. Apparent power is √3 × VL × IL. At 400 V, 32 A and a power factor of 0.85, real power is about 18.85 kW. The √3 belongs in the product once, not twice.
Voltage is the line voltage. Current is the line current. A phase voltage of 230 V used as if it were 400 V understates the power. The page assumes the load is balanced. A single-phase load on a three-phase board is the other page.
Power factor is an input. The page does not measure it. Starting current and a diversity factor are not applied.
Use it to check a board schedule line against a nameplate. It is not a cable size and not a fault-level calculation.
- Related: single-phase power
- electrical logic
Last checked 3 October 2026. A scenario, not a design. See methodology, standards and the disclaimer.