If the peak current in a standard light bulb running on 110 V AC power is 0.72 A,what is the average power drawn by the bulb?
A)56 W
B)80 W
C)110 W
D)40 W

Question

Quizplus · Accepted Answer

The average power drawn by the bulb can be calculated using the formula P = Vrms * Irms, where Vrms is the root mean square voltage and Irms is the root mean square current. For an AC circuit, Vrms = Vpeak / sqrt(2) and Irms = Ipeak / sqrt(2). Given Vpeak = 110 V and Ipeak = 0.72 A, Vrms = 110 / sqrt(2) and Irms = 0.72 / sqrt(2). Therefore, P = (110 / sqrt(2)) * (0.72 / sqrt(2)) = 55.2 W, which rounds to 56 W.

If the Peak Current in a Standard Light Bulb Running