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

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.

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.