For a circuit with 120 VAC, 5 A and 20-degree phase shift, the real power is approximately what?

For a circuit with 120 VAC, 5 A and 20-degree phase shift, the real power is approximately what?

• A. 600 W
• B. 564 W
• C. 0 W
• D. A negative value

Answer: (B)

Real power is the part of the power that does useful work, and it equals P = V I cos(phi), where phi is the phase angle between the voltage and current. Here, V = 120 V and I = 5 A, so the apparent power is 600 VA. The circuit has a 20° phase shift, so the real power is 600 × cos(20°). Since cos(20°) is about 0.94, P ≈ 600 × 0.94 ≈ 564 W. This is why the real power is roughly 564 watts. The phase shift means some power is reactive (not doing work) and the real portion is reduced from the maximum 600 W.

Real power is the part of the power that does useful work, and it equals P = V I cos(phi), where phi is the phase angle between the voltage and current. Here, V = 120 V and I = 5 A, so the apparent power is 600 VA. The circuit has a 20° phase shift, so the real power is 600 × cos(20°). Since cos(20°) is about 0.94, P ≈ 600 × 0.94 ≈ 564 W. This is why the real power is roughly 564 watts. The phase shift means some power is reactive (not doing work) and the real portion is reduced from the maximum 600 W.