AC Power Calculator (Single & Three-Phase)
Work out real power (kW), apparent power (kVA), reactive power (kvar) and current for single-phase and three-phase AC loads, from the voltage, current and power factor, or find the amps a load of known kW or kVA will draw.
How to Use
- Pick Three-phase or Single-phase.
- Choose what to find: the power from the voltage, current and power factor, or the current a load of known kW or kVA draws.
- Enter the voltage. For three-phase say whether it is line-to-line (480 V, 400 V, 208 V) or line-to-neutral (277 V, 230 V, 120 V), and whether the load is wye or delta.
- Enter the power factor: 1 for heaters and lamps, about 0.8 to 0.9 for loaded motors.
- Read kW, kVA, kvar and the current in the readouts and the highlighted field, and check Show Work for each step. Presets load common examples.
Worked Example
A three-phase motor. A clamp meter reads 14 A on each line of a 480 V supply and the power factor is 0.85. The apparent power is S = √3 × 480 × 14 = 11,639 VA (11.64 kVA). The real power is 11,639 × 0.85 = 9,893 W, and the reactive power is 11,639 × sin(arccos 0.85) = 11,639 × 0.5268 = 6,131 var.
Amps for a known load. A 75 kW machine on 400 V three-phase at PF 0.9 draws I = 75,000 ÷ (√3 × 400 × 0.9) = 120.3 A per line. A 3.5 kW single-phase heater on 240 V draws 3,500 ÷ 240 = 14.58 A.
The common mistake: the wrong factor for three phases. Treating the 75 kW load as single-phase, 75,000 ÷ (400 × 0.9), gives 208.3 A, 73 % too high. Dividing by 3 instead of √3 gives 69.44 A, 42 % too low. The √3 belongs with the line-to-line voltage. If you only know the line-to-neutral voltage (230 V on a 400 V system), either multiply it by √3 first or use P = 3 × V_LN × I × PF, which gives the same answer.
Show Work
Formulas
Why Three Phases, and Where √3 Comes From
Three-phase power was developed in the late 1880s, when Nikola Tesla in the United States, Galileo Ferraris in Italy and Mikhail Dolivo-Dobrovolsky at AEG in Germany all worked on polyphase motors. In 1891 Dolivo-Dobrovolsky’s three-phase system carried power about 175 km from Lauffen to the Frankfurt electrical exhibition, which showed it could work over long distances. Within a few decades almost all large generation and distribution was three-phase.
Three voltages 120° apart add to zero at every instant, so a balanced load needs no return current in the neutral and draws constant power, not the pulsing power of a single phase. That is why three-phase motors start on their own and run smoothly. The √3 comes from the same geometry: two phase voltages 120° apart differ by √3 times either one, so a 230 V line-to-neutral system has 230 × √3 = 398.4 V, nominally 400 V, between lines. The formula P = √3 × VLL × IL × cos φ is the one used in ABB’s Technical Application Paper No. 8 on power factor correction, which works an example of a 300 kW plant at 400 V: at cos φ 0.8 it draws about 540 A.
Apparent power in kVA matters because cables, switchgear, transformers and generators heat up with current whatever the power factor. That is why their ratings are in kVA or amps, while meters charge for kWh of real energy.
About This Calculator
Reference: World Plug & Socket Types.
This calculator handles both single-phase and balanced three-phase AC. Give it the voltage, current and power factor and it returns real, apparent and reactive power and the phase angle. Or give it a load in kW or kVA and it returns the current. For three-phase you can enter either the line-to-line or the line-to-neutral voltage, and choose a wye or delta load to see the voltage across and the current through each phase. The drawing shows the circuit and the power triangle to scale.
The three-phase results assume a balanced load, with the same current on every line. For a badly unbalanced system, measure each phase and add the per-phase powers. For DC circuits, where there is no power factor, use the Power Calculator. Everything runs in your browser; nothing you enter is sent anywhere.
Related tools: Inrush Current Calculator, Power Calculator, and Ohm’s Law Calculator.
Frequently Asked Questions
How do I calculate three-phase power?
Multiply √3 by the line-to-line voltage, the line current and the power factor: P = √3 × V_LL × I_L × PF. A 480 V motor drawing 14 A at PF 0.85 takes 1.732 × 480 × 14 × 0.85 = 9,893 W, or 9.893 kW. The apparent power, without the power factor, is 11.64 kVA.
How many amps does a three-phase load draw?
Divide the real power by √3, the line-to-line voltage and the power factor: I = P ÷ (√3 × V_LL × PF). A 75 kW load on 400 V at PF 0.9 draws 75,000 ÷ (1.732 × 400 × 0.9) = 120.3 A on each line. From kVA it is simpler: 500 kVA on 480 V draws 500,000 ÷ (1.732 × 480) = 601.4 A.
Does the formula change for wye and delta loads?
Not if you use line values: P = √3 × V_LL × I_L × PF gives the total for both. Only the values inside the load differ. In a wye each phase sees the line-to-neutral voltage, 480 ÷ √3 = 277.1 V, and carries the line current. In a delta each phase sees the full 480 V and carries the line current divided by √3, so 14 A on the lines is 8.083 A in each winding.
What is the difference between kW, kVA and kvar?
kW is real power, the part that does work or makes heat. kvar is reactive power, which flows back and forth to charge motor and transformer magnetic fields. kVA is apparent power, the voltage times the current, and it is what cables, transformers and generators must carry. They form a right triangle: kVA² = kW² + kvar². At PF 0.85, 11.64 kVA is 9.893 kW and 6.131 kvar.
How do I calculate single-phase power?
Multiply the voltage, the current and the power factor: P = V × I × PF. A 120 V circuit carrying 15 A into a heater (PF 1) delivers 1,800 W. On 230 V, 10 A at PF 0.95 is 2,300 VA apparent and 2,185 W real. To go the other way, a 3.5 kW heater on 240 V draws 3,500 ÷ 240 = 14.58 A.
How do I use the AC Power Calculator (Single & Three-Phase)?
Simply type your numbers and read the result, which refreshes the instant you change something. There is nothing to submit and nothing to wait for.
Is it free? Does it work without internet?
Yes to both. It is free with no sign-up, and once the page has loaded it keeps working even with no internet.
Where does my data go?
Nowhere — every calculation runs on your own device. Nothing you enter is uploaded, logged, or stored.
Common Use Cases
Motor feeders
A 480 V motor on a clamp meter at 14 A with PF 0.85 is taking 9.893 kW and 11.64 kVA. The cable and breaker must carry the 14 A, not just the kW.
Sizing a supply for a kW load
A 75 kW, 400 V machine at PF 0.9 needs 120.3 A per line. Leaving out the √3 gives 208.3 A, which would oversize everything by 73 %.
Generators and transformers
A 500 kVA, 480 V generator can supply 601.4 A per line. At PF 0.8 that is only 400 kW of real load.
European three-phase
On 230/400 V, a 10 A load at PF 0.9 is 6.928 kVA and 6.235 kW, or 2.078 kW per phase.
Household circuits
A 3.5 kW water heater on 240 V draws 14.58 A; a 1.5 kVA UPS on 120 V can draw up to 12.5 A from the outlet.
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