A wattmeter connected to an AC load reads 30 kW. A voltmeter reads 440 V RMS and an ammeter reads 90 A RMS. What is the power factor of this load?

Why B is correct

Apparent power = 440 V × 90 A = 39,600 VA = 39.6 kVA. Power factor = true power / apparent power = 30 kW / 39.6 kVA ≈ 0.76. NEETS Mod. 2 §4-2 defines power factor as the ratio of true to apparent power (cos θ). Because shipboard loads are predominantly inductive motors, the factor is lagging.