A circuit has real power P = 500 W and apparent power S = 600 VA. What is the power factor?

A circuit has real power P = 500 W and apparent power S = 600 VA. What is the power factor?

• A. PF = 0.667
• B. PF = 0.833
• C. PF = 1.0
• D. PF = 0.5

Answer: (B)

Power factor is the ratio of real power to apparent power, showing how effectively the circuit uses power. It’s PF = P/S. With P = 500 W and S = 600 VA, PF = 500/600 ≈ 0.833. This means about 83% of the supplied power does useful work, while the rest is reactive power. The corresponding phase angle satisfies cos φ = PF, so φ ≈ arccos(0.833) ≈ 33 degrees. The reactive power magnitude would be Q = sqrt(S^2 − P^2) ≈ sqrt(360000 − 250000) ≈ 331 VAR. The sign of Q (inductive or capacitive) isn’t specified here, but the ratio confirms the PF around 0.83.

Power factor is the ratio of real power to apparent power, showing how effectively the circuit uses power. It’s PF = P/S. With P = 500 W and S = 600 VA, PF = 500/600 ≈ 0.833.

This means about 83% of the supplied power does useful work, while the rest is reactive power. The corresponding phase angle satisfies cos φ = PF, so φ ≈ arccos(0.833) ≈ 33 degrees. The reactive power magnitude would be Q = sqrt(S^2 − P^2) ≈ sqrt(360000 − 250000) ≈ 331 VAR. The sign of Q (inductive or capacitive) isn’t specified here, but the ratio confirms the PF around 0.83.