Three-Phase Power

Calculate current, kW, kVA, kvar and power factor with worked equations.

Before you calculate

Understanding AC power: P, Q and S

P (kW) is active power that does useful work. Q (kvar) describes reactive power exchanged with electrical and magnetic fields. S (kVA) is apparent power, linked to the total RMS current. Power factor and phase angle connect these quantities.

Learn more — illustrated guide →

Step 1

Input parameters

Enter load and supply characteristics. Results update instantly.

System

Known power

Line to Line Voltage (L-L) (V)

Output power (kW)

Power factor (cos φ)Range: greater than 0, up to 1

Efficiency (%)100% for electrical input kW; lower for mechanical output kW.

Three-phase uses Line to Line Voltage (L-L), 400 V by default; single-phase uses Line to Neutral Voltage (L-N), 230 V by default. Assumes a balanced, sinusoidal, steady-state AC load. Reactive power is displayed as magnitude; leading versus lagging is not distinguished. This is an educational calculation, not a complete installation design.

Step 2

Calculated results

Line current72.17 A

Apparent power50 kVA

Electrical active power45 kW

Reactive power magnitude21.79 kvar

Phase angle: 25.84° · cos φ = 0.9

Step 3

Mathematical explanation

Values are rounded for readability; calculations retain full precision.

1. Electrical active power

If the entered kW is a mechanical output rating, account for equipment efficiency to obtain electrical input power. Use 100% for an electrical input power rating.

P_{\mathrm{in}}=\frac{P_{\mathrm{out}}}{\eta}

P_{\mathrm{in}}=\frac{45}{1}\approx 45\,\mathrm{kW}

2. Apparent power

S=\frac{P}{\cos\varphi}

S=\frac{45}{0.9}\approx 50\,\mathrm{kVA}

3. Line current

For a balanced three-phase circuit, use line-to-line voltage and the square-root-of-three factor.

I=\frac{S\times 1000}{\sqrt{3}\times U_{LL}}

I=\frac{50\,\times\,1000}{\sqrt{3}\,\times\,400}\approx\,72.17\,\mathrm{A}

4. Reactive power magnitude

Use the power triangle. For readability, the numerical example uses the power factor.

|Q|=\sqrt{S^{2}-P^{2}}

|Q|=S\times\sqrt{1-\cos^{2}\varphi}

|Q|=50\,\times\,\sqrt{1-0.9^{2}}

|Q|\approx 21.79\,\mathrm{kvar}

5. Phase angle

\varphi=\cos^{-1}(\mathrm{PF})

\varphi=\cos^{-1}(0.9)\approx 25.84^{\circ}