Case 1
- Line-to-line voltage V_L (V)
- 400V
- Line current I_L (A)
- 10A
- Power factor (cos φ)
- 0.8
5,542.56W
Open with these valuesThree-Phase Power Calculator
5,542.56W
Result: 5,542.56 WReal power in a balanced three-phase system is √3 × line voltage × line current × power factor. At 400 V, 10 A and a power factor of 0.8 that is 5542.56 W. Use the line-to-line voltage (400 V in Europe), not the line-to-neutral 230 V — the two differ by exactly the √3 in the formula.
The numbers at a glance
Held fixed: Line-to-line voltage V_L (V) 400.0 V, Line current I_L (A) 10.00 A.
| Power factor (cos φ) | Result (W) |
|---|---|
| 0.60 | 4,156.92 |
| 0.70 | 4,849.74 |
| 0.80Your value | 5,542.56 |
| 0.90 | 6,235.38 |
| 1.00 | 6,928.20 |
Worked examples
Case 2
- Line-to-line voltage V_L (V)
- 480V
- Line current I_L (A)
- 20A
- Power factor (cos φ)
- 0.9
14,964.92W
Open with these valuesCase 3
- Line-to-line voltage V_L (V)
- 690V
- Line current I_L (A)
- 100A
- Power factor (cos φ)
- 0.85
101,584.78W
Open with these valuesHow it's calculated
P = √3 × V_L × I_L × pf
- StepEnter the line-to-line voltage, the value between any two of the three lines.
- StepEnter the current in one line conductor, from the nameplate or a clamp meter.
- StepEnter the power factor as cos φ between 0 and 1; a typical motor runs near 0.8.
- ResultRead the real power in watts — the part that does useful work.
Reference table
| V_L, I_L, pf | Apparent power | Real power in W |
|---|---|---|
| 230, 5, 1 | 1991.86 VA | 1991.86 |
| 400, 10, 0.8 | 6928.20 VA | 5542.56 |
| 400, 16, 0.95 | 11085.13 VA | 10530.87 |
| 480, 20, 0.9 | 16627.69 VA | 14964.92 |
| 690, 100, 0.85 | 119511.51 VA | 101584.78 |
Questions
How do I calculate three-phase power?
Multiply √3, about 1.732, by the line voltage, the line current and the power factor. For 400 V, 10 A and a power factor of 0.8 that is 5542.56 W. Leave the power factor out and you get the apparent power in volt-amperes instead.
Should I enter line voltage or phase voltage?
Enter the line-to-line voltage, measured between any two of the three lines, which is what nameplates quote as 400 V or 480 V. If you only have the phase voltage to neutral, multiply it by √3 first. Getting this wrong scales the answer by 1.73 in either direction.
Why is there a √3 in the formula?
In a balanced three-phase system the line-to-line voltage is √3 times the line-to-neutral voltage. Expressing the power in line quantities carries that √3 through into the formula.
What is the difference between real, apparent and reactive power?
Apparent power is the total the system carries, √3 × V_L × I_L in volt-amperes. Real power is the part that does useful work, apparent power times the power factor, in watts. Reactive power is the rest, √(S² − P²) in VAR, and it flows back and forth without doing net work.
Does star or delta wiring change the result?
No, as long as the load is balanced. The formula is written in line quantities, and those are the same in both connections — they are what a meter on the supply cable sees.
Sources and last check
Information, not professional advice.