From line voltage, current and power factor, this 3 phase power calculator gives real, apparent and reactive power. Start from a load in kW or kVA instead and it finds the line current. See the voltage and current on each element for star or delta loads, and what the same load draws on common system voltages.
RMS, between two phases. On a 120/208 V system enter 208.
Above 0 up to 1.
Total real power P
70.67 kW
Assumes a balanced load and sinusoidal voltages. For an unbalanced load, work out each phase separately and add the powers.
For the same line voltage, line current and power factor, total power is identical in star and delta. The connection only changes what each element sees.
The same 70.7 kW at PF 0.85 on common systems
| Line voltage | Line current | Star: V per element | Delta: I per element |
|---|---|---|---|
| 208 V | 230.8 A | 120.1 V | 133.2 A |
| 240 V | 200 A | 138.6 V | 115.5 A |
| 400 V | 120 A | 230.9 V | 69.28 A |
| 415 V | 115.7 A | 239.6 V | 66.78 A |
| 480 V | 100 A | 277.1 V | 57.74 A |
| 600 V | 80 A | 346.4 V | 46.19 A |
| 690 V | 69.57 A | 398.4 V | 40.16 A |
Use P = √3 × VL × IL × PF, with line-to-line voltage, line current and power factor. At 480 V, 100 A and PF 0.85, real power is 70.7 kW, apparent power is 83.1 kVA and reactive power is 43.8 kvar. To go the other way, line current is I = P ÷ (√3 × VL × PF). The result is the same whether the load is connected in star or delta.
These hold for a balanced load with sinusoidal voltages, using RMS line-to-line voltage and line current. The results are totals for all three phases.
Solved for the line current:
What each element of the load sees depends on how the three are connected:
From the impedance Z = R + jX of each element, per-phase Ohm’s law gives everything else:
S = √3 × 480 × 100 = 83.1 kVA. P = S × 0.85 = 70.7 kW. Q = √(S² − P²) = 43.8 kvar. In a star load each element sees 277.1 V and carries the full 100 A. In delta each would see 480 V and carry 57.74 A.
At PF 0.7, I = 170,000 ÷ (√3 × 400 × 0.7) = 350.5 A. Correcting to PF 0.9 brings it to 272.6 A for the same real power. ABB uses this case to show the cable dropping from 120 mm² to 70 mm².
Three 2 kW elements. In star each sees 120.1 V, so R = 120² ÷ 2000 = 7.2 Ω and the heater draws 6.01 kW at 16.68 A per line. In delta each sees 208 V, so R = 208² ÷ 2000 = 21.6 Ω, three times larger, for 6.01 kW and the same 16.68 A line current.
Wire the 7.2 Ω elements in delta by mistake and each sees 208 V: the heater draws 18 kW at 50.04 A, three times the design.
Line current I = S ÷ (√3 × VL) for common kVA sizes and system voltages. Power factor does not affect this figure. For sizes not listed, use the 3 phase calculator at the top of the page.
| kVA | 208 V | 240 V | 400 V | 415 V | 480 V | 600 V |
|---|---|---|---|---|---|---|
| 15 | 41.64 A | 36.08 A | 21.65 A | 20.87 A | 18.04 A | 14.43 A |
| 30 | 83.27 A | 72.17 A | 43.3 A | 41.74 A | 36.08 A | 28.87 A |
| 45 | 124.9 A | 108.3 A | 64.95 A | 62.6 A | 54.13 A | 43.3 A |
| 75 | 208.2 A | 180.4 A | 108.3 A | 104.3 A | 90.21 A | 72.17 A |
| 112.5 | 312.3 A | 270.6 A | 162.4 A | 156.5 A | 135.3 A | 108.3 A |
| 150 | 416.4 A | 360.8 A | 216.5 A | 208.7 A | 180.4 A | 144.3 A |
| 225 | 624.5 A | 541.3 A | 324.8 A | 313 A | 270.6 A | 216.5 A |
| 300 | 832.7 A | 721.7 A | 433 A | 417.4 A | 360.8 A | 288.7 A |
| 500 | 1388 A | 1203 A | 721.7 A | 695.6 A | 601.4 A | 481.1 A |
| 750 | 2082 A | 1804 A | 1083 A | 1043 A | 902.1 A | 721.7 A |
| 1000 | 2776 A | 2406 A | 1443 A | 1391 A | 1203 A | 962.3 A |
Amps per kW: multiply by the load in kW. A lower power factor needs more current for the same real power. The three-phase current calculator at the top takes kW or kVA directly.
| Voltage | PF 1.00 | PF 0.95 | PF 0.90 | PF 0.85 | PF 0.80 | PF 0.70 |
|---|---|---|---|---|---|---|
| 208 V | 2.78 | 2.92 | 3.08 | 3.27 | 3.47 | 3.97 |
| 240 V | 2.41 | 2.53 | 2.67 | 2.83 | 3.01 | 3.44 |
| 400 V | 1.44 | 1.52 | 1.6 | 1.7 | 1.8 | 2.06 |
| 415 V | 1.39 | 1.46 | 1.55 | 1.64 | 1.74 | 1.99 |
| 480 V | 1.2 | 1.27 | 1.34 | 1.42 | 1.5 | 1.72 |
| 600 V | 0.962 | 1.01 | 1.07 | 1.13 | 1.2 | 1.37 |
Three 10 Ω resistive elements on a 400 V supply, computed both ways.
| Quantity | Star | Delta |
|---|---|---|
| Voltage across each element | 230.9 V | 400 V |
| Current through each element | 23.09 A | 40 A |
| Line current | 23.09 A | 69.28 A |
| Total power | 16 kW | 48 kW |
Nominal values from IEC 60038 Table 1, written line-to-neutral / line-to-line. The nominal pairs are rounded, so dividing VL by √3 lands close to the first number. Three-wire 240 V, 480 V and 600 V systems at 60 Hz are listed in the same table.
| System | VL ÷ √3 | IEC 60038 |
|---|---|---|
| 120/208 V | 120.1 V | 60 Hz |
| 230/400 V | 230.9 V | 50 Hz and 60 Hz |
| 240/415 V | 239.6 V | Older value; IEC 60038 notes these systems still exist |
| 277/480 V | 277.1 V | 60 Hz |
| 347/600 V | 346.4 V | 60 Hz |
| 400/690 V | 398.4 V | 50 Hz |
P = √3 × VL × IL × PF, with VL the line-to-line voltage, IL the line current and PF the power factor. Apparent power is S = √3 × VL × IL and reactive power is Q = √(S² − P²). All three are totals for the three phases.
I = P ÷ (√3 × VL × PF). A 10 kW load at 400 V and PF 0.85 draws 16.98 A per line. Motors also need the efficiency: divide shaft kW by efficiency first to get the electrical input.
I = S ÷ (√3 × VL). A 75 kVA transformer at 480 V is rated 90.21 A per line. Power factor does not enter, because kVA already includes the reactive part.
kW = √3 × VL × IL × PF ÷ 1000, with VL the line-to-line voltage and IL the line current. At 480 V, 100 A and PF 0.85 that is 70.7 kW.
The voltage between two lines is the difference of two sine waves 120° apart, which is √3 times one line-to-neutral voltage. Total power is three times the per-phase power, and 3 × (VL ÷ √3) × IL simplifies to √3 × VL × IL.
In star (wye) each element connects from a line to the neutral point, so it sees VL ÷ √3 and carries the line current. In delta each element connects between two lines, so it sees the full VL and carries IL ÷ √3.
With the same three elements on the same supply, delta draws three times the power and three times the line current. Three 10 Ω heaters at 400 V give 16 kW in star and 48 kW in delta. For a stated load with given line values, the total power is the same either way.
Divide by √3: 208 V gives 120.1 V and 480 V gives 277.1 V, which is why these systems are written 120/208 V and 277/480 V.
The √3 formulas in this three-phase power calculator assume all three phases carry the same load. For an unbalanced load, calculate V × I × PF for each phase using its own line-to-neutral voltage and current, and add the three results. The neutral then carries the imbalance current.
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