Work out active, reactive and apparent power for a balanced three-phase load from line voltage, line current and power factor.
Solid lines are voltage, dashed are current.
Active power, total
70.67kW
Check the values to see a result.
Balanced three-phase load.
S = √3 × VLL × I ÷ 1000P = S × PFQ = S × √(1 − PF²)
A balanced 480 V three-phase load drawing 100 A in each line at a power factor of 0.85 uses 70.67 kW (83.14 kVA).
Tip: Enter the line-to-line voltage (400 V, not 230 V) and the current in one line, not the total of three.
Three-phase power is √3 times the line-to-line voltage times the line current, which gives apparent power in kVA. Multiply by the power factor for the real power in kW.
The formula assumes the three phases carry equal current at the same power factor. Enter the current in one line, not the total of three.
S = √3 × VLL × I ÷ 1000
Use the voltage between two lines (400 V, 480 V) and the current in one line. Cables, breakers and transformers heat with current, whatever the power factor, so they are rated on this figure.
P = √3 × VLL × I × PF ÷ 1000
The power that does work and that an energy meter integrates into kWh. The power factor is the ratio P ÷ S.
Q = √(S2 − P2)
Motors and transformers draw it lagging to magnetise their cores; capacitor banks supply it leading. The formula holds for sinusoidal current. With harmonic current, √(S2 − P2) also includes distortion power, so it overstates the kvar a capacitor bank can correct.
Pphase = VLN × I × PF ÷ 1000 = P ÷ 3
Use the line-to-neutral voltage (400 ÷ √3 = 231 V, nominally 230 V). The multiplier is then 3 rather than √3, and the total is the same.
For an unbalanced load, measure voltage, current and power factor on each phase and add the three phase powers. The balanced formula with the average current gives the correct total only when all three power factors are equal.
A machine draws the full 32 A rating at a power factor of 0.95: 1.732 × 400 × 32 × 0.95 ÷ 1000 = 21.06 kW. The apparent power is 22.17 kVA and the reactive power 6.92 kvar. The cable and the breaker carry 5% more current than a 21.06 kW load at unity power factor, so size the circuit on amps, not kW.
Active power, total 21.06 kW Open in the calculator
On a North American 480 V system, 100 A in each line at a power factor of 0.85 is 1.732 × 480 × 100 × 0.85 ÷ 1000 = 70.67 kW, with 83.14 kVA and 43.8 kvar lagging. At a power factor of 0.95 the same 70.67 kW needs 70.67 × tan(cos−1 0.95) = 23.2 kvar. A capacitor bank of about 20.6 kvar closes the gap. The power factor correction calculator works this through for other targets.
Active power, total 70.67 kW Open in the calculator
A spot reading of 180 A on each line of a 400 V board, with a power factor of 0.92 from the meter, gives 114.7 kW. That is the load at one moment. A maximum-demand tariff bills the highest 15 or 30 minute average, so log the board over a full working week before you use the figure to size a supply or a transformer. The interval demand calculator turns logged kWh into that demand figure.
Active power, total 114.7 kW Open in the calculator
Active power for a balanced load at a power factor of 0.9, with the current in each line. For kVA, divide a figure by 0.9.
| Current (A) | 208 V (kW) | 400 V (kW) | 480 V (kW) | 690 V (kW) |
|---|---|---|---|---|
| 10 | 3.2 | 6.2 | 7.5 | 10.8 |
| 16 | 5.2 | 10.0 | 12.0 | 17.2 |
| 20 | 6.5 | 12.5 | 15.0 | 21.5 |
| 25 | 8.1 | 15.6 | 18.7 | 26.9 |
| 32 | 10.4 | 20.0 | 23.9 | 34.4 |
| 40 | 13.0 | 24.9 | 29.9 | 43.0 |
| 50 | 16.2 | 31.2 | 37.4 | 53.8 |
| 63 | 20.4 | 39.3 | 47.1 | 67.8 |
| 80 | 25.9 | 49.9 | 59.9 | 86.0 |
| 100 | 32.4 | 62.4 | 74.8 | 107.6 |
| 125 | 40.5 | 77.9 | 93.5 | 134.5 |
| 160 | 51.9 | 99.8 | 119.7 | 172.1 |
| 200 | 64.8 | 124.7 | 149.6 | 215.1 |
| 250 | 81.1 | 155.9 | 187.1 | 268.9 |
| 315 | 102.1 | 196.4 | 235.7 | 338.8 |
| 400 | 129.7 | 249.4 | 299.3 | 430.2 |
| 500 | 162.1 | 311.8 | 374.1 | 537.8 |
| 630 | 204.3 | 392.8 | 471.4 | 677.6 |
| 800 | 259.4 | 498.8 | 598.6 | 860.5 |
| 1,000 | 324.2 | 623.5 | 748.2 | 1,075.6 |
Both are right, with different voltages. Each phase carries VLN × I, so the total is 3 × VLN × I. Because VLL = √3 × VLN, the same total is √3 × VLL × I. Mixing the two is a common mistake: 3 × 400 V overstates the power by 73%, and √3 × 230 V understates it by 42%.
No. √3 × VLL × I gives the total for a star or a delta load, because it uses line quantities. The error comes from where the current is measured. Each winding of a delta carries I ÷ √3. After a star-delta starter, the six leads to the motor carry winding current, so a CT on one of them reads 58% of the line current and understates the power by 42%. Put the CTs on the three supply conductors upstream of the starter.
Enter the true power factor, P ÷ S, to get kW. Displacement power factor, cos φ1, is the cosine of the angle between the fundamental voltage and current only. The two are equal when the current is sinusoidal. With harmonic current and a sinusoidal voltage, true power factor = displacement power factor ÷ √(1 + THDI2). A drive at a displacement power factor of 0.95 with 40% current THD has a true power factor of 0.88, so entering 0.95 overstates its kW by 8%. Many meters label both values PF, so check which one the display shows.
Only when the three power factors are about equal. Take a 230/400 V board with 150 A at 0.95 on L1, 100 A at 0.80 on L2 and 50 A at 0.60 on L3. Phase by phase, that is 32.78 + 18.40 + 6.90 = 58.1 kW. The average current of 100 A with the L1 power factor of 0.95 gives 65.8 kW, which is 13% high. Measure each phase and add the three powers.
A ZEM measures voltage, current and power factor on each phase, so an unbalanced load is measured phase by phase rather than estimated from an average.
