A circuit having 120 VAC at 5 amps with 20 degrees phase shift would equal what?

• A. P = 120 x 5 x sin 20
• B. P = 120 x 5 x cos 60
• C. P = 120 x 5 x cosine of 20
• D. P = 120 x 5

Answer: (C)

In AC circuits, real (usable) power is found with P = V × I × cos(phi), where phi is the phase angle between the voltage and the current. Here V = 120 V, I = 5 A, and the phase difference is 20 degrees, so the power factor is cos(20°).

Compute: P = 120 × 5 × cos(20°) ≈ 600 × 0.9397 ≈ 564 W.

The sine form would be wrong because power factor uses cosine, not sine. The angle isn’t 60 degrees, so cos(60°) would misrepresent the phase. Dropping the cos term would give apparent power (600 VA) rather than real power, which isn’t correct for the actual power delivered to the load.