Real power in an AC circuit is defined as the product of the applied voltage and which current?

Real power in an AC circuit is defined as the product of the applied voltage and which current?

• A. The current in phase with the voltage
• B. The RMS current
• C. The current leads the voltage by 90 degrees
• D. The current lags the voltage by 90 degrees

Answer: (A)

Real power is the average power delivered to a load, and it depends on the portion of the current that is in phase with the voltage. In AC circuits, the voltage and current can be out of phase by some angle φ, and the real power is given by P = V_rms I_rms cos(φ). The part of the current that is in phase with the voltage does the actual work each cycle, while the current that leads or lags by 90 degrees is reactive and stores energy temporarily rather than delivering it as useful power. If the current is entirely in phase (φ = 0), the real power equals V_rms times I_rms. The RMS current alone isn’t the real power without considering the phase relationship. For example, with a phase angle of 60 degrees, real power is reduced by cos(60°) = 0.5, even though the RMS values are still 120 V and 10 A. The in-phase component of the current is 5 A, and P = 120 V × 5 A = 600 W.

Real power is the average power delivered to a load, and it depends on the portion of the current that is in phase with the voltage. In AC circuits, the voltage and current can be out of phase by some angle φ, and the real power is given by P = V_rms I_rms cos(φ). The part of the current that is in phase with the voltage does the actual work each cycle, while the current that leads or lags by 90 degrees is reactive and stores energy temporarily rather than delivering it as useful power. If the current is entirely in phase (φ = 0), the real power equals V_rms times I_rms. The RMS current alone isn’t the real power without considering the phase relationship. For example, with a phase angle of 60 degrees, real power is reduced by cos(60°) = 0.5, even though the RMS values are still 120 V and 10 A. The in-phase component of the current is 5 A, and P = 120 V × 5 A = 600 W.