AC power
Balanced three-phase active power
- where
- active power (W)
- line-to-line RMS voltage (V)
- line RMS current (A)
- power factor
Explain this equation
Gives the total real power of a balanced three-phase load from the line voltage, line current and power factor. Use it for motors, heaters and boards on a 400 V supply.
V_L is the line-to-line RMS voltage, such as 400 V. I_L is the current in each line. PF is the power factor. The √3 appears because, in a star system, the line voltage is √3 times the phase voltage, so √3 × V_L × I_L equals 3 × V_ph × I_ph.
Each of the three phases delivers V_ph × I_ph × PF. Writing the total in terms of the line voltage you actually measure swaps the 3 and the phase voltage for √3 and V_L.
400 V × 10 A × 0.8 × 1.732 is about 5.5 kW. A handy figure on 400 V: each amp per line at unity power factor is about 0.69 kW.
Watch out: the result is already the total, so don't multiply by three again. Use line-to-line voltage, not 230 V. A badly unbalanced load needs each phase working out separately.
For a balanced sinusoidal load, one line current and the line-to-line voltage describe all three phases. The factor √3 accounts for the three-phase relationship.
The result is already the total power; do not multiply it by three again. To find current, use I_L = P/(√3 × V_L × PF).
Enter your values to see the result.
Common mistake
Use line-to-line voltage here, not line-to-neutral voltage.
Limits
Balanced sinusoidal load. For unbalanced loads calculate or measure each phase appropriately and sum active powers. Do not use a simple average current for a badly unbalanced installation.
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.