Three-Phase Power

For a balanced system, regardless of star or delta:

$P = \sqrt{3}\,V_L I_L \cos\phi \qquad Q = \sqrt{3}\,V_L I_L \sin\phi \qquad S = \sqrt{3}\,V_L I_L$

Example: a 415 V motor drawing 20 A at PF 0.85 uses $P = 1.732 \times 415 \times 20 \times 0.85 \approx 12.2$ kW.

Two-wattmeter method measures total power in a three-wire system: $P = W_1 + W_2$ and $\tan\phi = \sqrt{3}\dfrac{W_1 - W_2}{W_1 + W_2}$.