A circuit having 120 VAC at 5 A with 0 degrees phase shift, the power would be what?

A circuit having 120 VAC at 5 A with 0 degrees phase shift, the power would be what?

• A. P = 120 x 5 x cos(20°)
• B. P = 120 x 5 x cos(0°)
• C. P = 120 x 5 x cos(60°)
• D. P = 120 x 5 x cos(180°)

Answer: (B)

Real power in an AC circuit is found by multiplying voltage, current, and the cosine of the phase angle between them. Since the voltage and current are in phase, the phase angle is 0°, and cos(0°) equals 1. With 120 V (RMS) and 5 A (RMS), P = 120 × 5 × 1 = 600 watts. If there were any phase difference, cos(phi) would be less than 1, reducing real power; a 180° difference would make cos(180°) = -1, reversing the sign of the instantaneous power, but with the given in-phase condition you get the full product.

Real power in an AC circuit is found by multiplying voltage, current, and the cosine of the phase angle between them. Since the voltage and current are in phase, the phase angle is 0°, and cos(0°) equals 1. With 120 V (RMS) and 5 A (RMS), P = 120 × 5 × 1 = 600 watts. If there were any phase difference, cos(phi) would be less than 1, reducing real power; a 180° difference would make cos(180°) = -1, reversing the sign of the instantaneous power, but with the given in-phase condition you get the full product.