The voltage and current phasors (RMS) of a load are V = 100∠0° V and I = 5∠−30° A. Which of the following is the magnitude of the reactive power of this load?

Answer: A. 250 var

Method 1 (from the power-factor angle): the current lags the voltage by 30°, so the active power is P = VI cos 30° ≈ 433 W and the reactive power is Q = VI sin 30° = 100 × 5 × 0.5 = 250 var. Method 2 (complex power): the complex power is S = V × (conjugate of I) = 100∠0° × 5∠30° = 500∠30° ≈ 433 + j250 VA; the real part is the active power and the imaginary part is the reactive power. The imaginary part is positive, so the load takes 250 var of lagging (inductive) reactive power. The apparent power is |S| = 500 VA.

• A. Correct. Q = VI sin 30° = 250 var.
• B. VI cos 30° ≈ 433 is the active power [W], not the reactive power.
• C. 500 is the apparent power VI [VA], which ignores the phase difference.