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3-Phase Power Calculator — kW to Amps

Convert three-phase kW, kVA or motor hp to line current, or amps back to power, with power factor, kVAR and star or delta phase voltage and current.

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A motor nameplate gives kilowatts or horsepower, but the breaker, the cable and the clamp meter all deal in amps. On a three-phase supply the link between the two is P = √3 × V × I × PF, and this 3-Phase Power Calculator runs it in either direction.

Enter a load in kW, kVA, W or hp to get the line current. Or enter a measured current to get real, apparent and reactive power. Choose star or delta and the table shows what each winding sees as well as the line values.

Worked Example: A 15 kW Load on 400 V

The calculator opens on 15 kW at 400 V with a power factor of 0.85, star connected:

  • Apparent power: 15 ÷ 0.85 = 17.647 kVA
  • Line current: 17,647 ÷ (√3 × 400) = 25.47 A
  • Reactive power: 9.296 kVAR, at a phase angle of 31.79°
  • Per winding: 230.9 V and 25.47 A, carrying 5.000 kW each

Click Amps → find power and 25.47 A is carried into the current box, so the same load is described from the other side. Load Sample continues with a 480 V delta system, a 20 hp motor at 92% efficiency and an 11 kV feeder.

Formula: Where the √3 Comes From

The three phases are 120° apart, so the voltage between two lines is √3 times the voltage across one star winding: 400 V line is 230.9 V phase. Apparent power is S = √3 × VL × IL, real power is S × PF, and reactive power is S × sin φ. In delta the winding sees the full line voltage and the line current splits, so phase current is IL ÷ √3.

Horsepower is the mechanical 745.7 W, and it is a shaft rating. With an efficiency filled in, the tool divides by it first, because the supply also carries the motor’s losses.

What This Rounds and What It Assumes

Currents show two decimals and powers three. Past 1,000 kVA the unit switches to MVA, so the 11 kV sample reads 1.905 MVA instead of 1,905.256 kVA. At a power factor of exactly 1 the reactive power is 0.

The maths assumes a balanced load and a sinusoidal supply. Harmonics, unbalanced phases and starting current are outside it, and it does not size cables or protection. Power factors outside the valid range are refused: “Power factor must be between 0.01 and 1. A resistive heater is 1; a loaded induction motor is usually 0.8 to 0.9.”

I know the

Connection

V

Common voltages

Line current

25.47 A

17.647 kVA apparent · 15.000 kW real · φ = 31.79°

Line current I_L

25.47 A

Real power P

15.000 kW

Apparent power S

17.647 kVA

Reactive power Q

9.296 kVAR

Star (Y)LinePhase (per winding)
Voltage400 V230.9 V
Current25.47 A25.47 A
Real power15.000 kW5.000 kW

Star: phase voltage = line voltage ÷ √3, phase current = line current.

S = √3 × V_L × I_L · P = S × PF · Q = S × sin φ. Assumes a balanced load and sinusoidal supply; it is not a cable-sizing or protection calculation.