Library / Electrical / ELE-21
Electrical
Real power, three-phase AC
Total power delivered by a balanced three-phase supply.
- Ref
- ELE-21
- Rev
- v1
- Field
- Electrical
- Result
- P3 [kW]
P3 total real power, VLL line-to-line RMS voltage, Iline RMS line current and φ the per-phase voltage–current angle. Balanced load.
Three-phase supplies deliver power through three conductors, with √3 times the line voltage and current combining for the total. It is the standard way to move large power efficiently.
Calculator
- Apparent power √3 V I13.86 kVA
- Phase voltage VLL/√3230.9 V
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 400 V three-phase supply carries the same power as a 230 V single-phase supply with less than half the current per conductor.
- The √3 comes from the 120° spacing between phases, not from a loss.
- The formula applies to star and delta connections as long as line quantities are used.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kW) | Local step | Change in result |
|---|---|---|---|---|---|---|
| VLL Line-to-line voltage | 200 to 11,000 V | 5.889 | 323.9 | 318 | +10% | +10 % |
| Iline Line current | 1 to 500 A | 0.5889 | 294.4 | 293.9 | +10% | +10 % |
| cos φ Power factor | 0.5 to 1 | 6.928 | 13.86 | 6.928 | +undefined% | — |
Sweep of Line current
| Line current (A) | Total real power (kW) |
|---|---|
| 1 | 0.5889 |
| 2.175 | 1.281 |
| 4.729 | 2.785 |
| 10.28 | 6.056 |
| 22.36 | 13.17 |
| 48.62 | 28.63 |
| 105.7 | 62.27 |
| 229.9 | 135.4 |
| 500 | 294.4 |
Sweep one variable
Variables and how each one contributes
Total real power
What the equation gives you. Everything else feeds it.
Line-to-line voltage
RMS voltage between any two lines (400 V is common). Directly proportional.
Line current
RMS current in each line. Directly proportional.
Power factor
Per-phase power factor. Directly proportional.
Worked example with the values above
- Apparent power: S = √3 × 400 × 20 = 13,860 VA.
- Real power: P = S cos φ = 13,860 × 0.85 = 11,780 W.
- That is 11.78 kW.
Assumptions and limits
- Balanced three-phase load and supply.
- Line-to-line voltage and line current (not phase values).
- Sinusoidal steady state.
Common mistakes
- Using phase voltage in place of line-to-line voltage.
- Forgetting the √3.
- Using peak values.
Statistics and uncertainty
Relative errors combine as δP/P ≈ √[(δV/V)² + (δI/I)² + (δpf/pf)²]. Unbalanced loads break the formula; the total must then be found phase by phase.
Where it is used
- Generator and motor sizing.
- Ship and shelter power distribution.
- Radar and transmitter supplies.
Maths you need
- Single-phase power: P = VI cos φ
- Phasors: three phases 120° apart