Calculator
How it's calculated
V = I × RP = V × IP = I² × RP = V² / RS = V × I, P = S × PFS = √3 × V × I, P = S × PFS² = P² + Q², Q = S × sin φ, φ = acos PFI = P / (k × V × PF), k = 1 or √3P in = P out / ηV is voltage, I current, R resistance, P real power in watts, S apparent power in volt-amperes, Q reactive power in var and PF the power factor. In Ohm's law mode, give any two of V, I, R and P and the other two follow from the three relations above: for example from I and P, V = P / I and R = P / I². For three-phase loads V is the voltage between two lines (400 V on a 400/230 V system) and the line-to-neutral voltage is V / √3. A motor's electrical input is its shaft power divided by its efficiency. One mechanical horsepower is 745.7 W and one metric horsepower (PS) is 735.5 W.
Worked examples
| Case | Working | Result |
|---|---|---|
| 12 V across 4 Ω | I = 12 / 4, P = 12² / 4 | 3 A, 36 W |
| Single-phase 230 V, 10 A, PF 0.8 | S = 230 × 10, P = S × 0.8 | 2.3 kVA, 1.84 kW |
| Its reactive power | Q = 2300 × √(1 − 0.8²) | 1.38 kvar, φ = 36.87° |
| Three-phase 400 V, 10 kW, PF 0.85 | S = 10 / 0.85 | 11.76 kVA |
| Its current | I = 11 764.7 / (√3 × 400) | 16.98 A |
| 7.5 kW motor, η 90%, PF 0.85, 400 V three-phase | P in = 7500 / 0.9 = 8333 W | 8.33 kW input |
| Its current | I = 8333 / 0.85 / (√3 × 400) | 14.15 A |
Assumptions
- Three-phase loads are balanced and the waveforms are sinusoidal. Unbalanced or distorted loads (harmonics) need a different calculation.
- Three-phase voltage is line-to-line. Entering the line-to-neutral voltage (for example 230 V) gives a current about 1.73 times too high.
- The power factor is the displacement power factor. It ignores harmonics, which lower the true power factor.
- Reactive power is shown without a sign, so lagging (inductive) and leading (capacitive) loads give the same figure.
- Motor efficiency and power factor default to typical values for a mid-size induction motor. Use the nameplate values when you have them, because both vary with size and load.
- Ohm's law assumes a constant resistance. A real resistor's resistance changes with temperature, so a hot element draws less current than its cold resistance suggests.
- The result is a calculated current. It is not a cable or protective-device rating. Those need allowances for starting current, grouping, temperature and the rules of your wiring code. To check the voltage drop of a cable, use the voltage drop calculator, or the cable size calculator to size a cable for voltage drop and short circuit.
Frequently asked questions
How many amps is 10 kW at 400 V three-phase?
With a power factor of 0.85 it is 16.98 A: I = 10 000 / (√3 × 400 × 0.85). At a power factor of 1 (resistive heating) it is 14.43 A. The lower the power factor, the more current the same kW needs.
How do I convert kW to kVA?
Divide by the power factor: kVA = kW / PF. A 10 kW load at PF 0.8 is 12.5 kVA. Choose "Real power" under "I know" and the calculator shows kVA, kvar and amps together.
How do I convert kW or hp to amps?
Pick the system and voltage, set the power factor, then choose "Real power" or "Motor output" and enter the kW or hp. For a motor, the shaft power is divided by the efficiency first, because the supply must deliver more than the shaft receives.
Why is a motor's current in the NEC table different from this result?
NEC (NFPA 70) Article 430 tells you to size motor branch-circuit conductors and protection from the full-load current tables, such as Table 430.250 for three-phase induction motors, not from a calculated or nameplate value. The table gives 14 A for a 10 hp, 460 V motor, while this calculator gives about 12 A for the same motor at 90% efficiency and PF 0.85. The table figure is deliberately conservative, so use it for code work.
Why does three-phase need less current than single-phase for the same power?
At the same voltage and power factor, three-phase current is 1/√3 (about 58%) of the single-phase current for the same real power.
What power factor should I assume if I do not know it?
Heaters and incandescent lamps are close to 1. Induction motors at full load are typically 0.8 to 0.9 and drop at light load. Check the nameplate or the supplier's data sheet before relying on a number.
Standards and references
- IEC 60038: standard voltages (for example 230/400 V).
- IEC 60034-1: rotating electrical machines, rating and performance.
- NFPA 70 (NEC) Article 430 and Table 430.250: motor circuits and full-load currents.
- IEEE Std 1459 and IEC 60050 (International Electrotechnical Vocabulary): definitions of real, reactive and apparent power.
- Ohm's law is the basic relation between voltage, current and resistance for a constant resistance.
This calculator estimates electrical quantities. It does not size cables, breakers or transformers, and it does not check any code requirement. Have the design checked by a qualified engineer.