A load draws 8 A at 120 V with a phase angle of 30 degrees between voltage and current. What are the real power P, apparent power S, and power factor PF?

A load draws 8 A at 120 V with a phase angle of 30 degrees between voltage and current. What are the real power P, apparent power S, and power factor PF?

• A. P ≈ 831 W, S = 960 VA, PF ≈ 0.87
• B. P ≈ 960 W, S = 831 VA, PF ≈ 0.87
• C. P ≈ 831 W, S = 900 VA, PF ≈ 0.90
• D. P ≈ 831 W, S = 960 VA, PF ≈ 0.70

Answer: (A)

Real power is the work actually done and equals P = VI cos(phi). With V = 120 V, I = 8 A, and phi = 30°, cos(30°) ≈ 0.866. So P = 120 × 8 × 0.866 ≈ 831 W. Apparent power is the product of voltage and current, regardless of phase: S = VI = 120 × 8 = 960 VA. Power factor is the ratio of real power to apparent power, PF = P/S = cos(phi) ≈ 0.866 ≈ 0.87. Thus P ≈ 831 W, S = 960 VA, PF ≈ 0.87. The other options either swap P and S, or use an incorrect PF value that doesn’t match cos(30°).

Real power is the work actually done and equals P = VI cos(phi). With V = 120 V, I = 8 A, and phi = 30°, cos(30°) ≈ 0.866. So P = 120 × 8 × 0.866 ≈ 831 W.

Apparent power is the product of voltage and current, regardless of phase: S = VI = 120 × 8 = 960 VA.

Power factor is the ratio of real power to apparent power, PF = P/S = cos(phi) ≈ 0.866 ≈ 0.87.

Thus P ≈ 831 W, S = 960 VA, PF ≈ 0.87. The other options either swap P and S, or use an incorrect PF value that doesn’t match cos(30°).