An AC voltage source,with a peak output of 200 V,is connected to a 50-W resistor.What is the rate of energy dissipated due to heat in the resistor?
A)200 W
B)400 W
C)566 W
D)800 W

Question

Quizplus · Accepted Answer

The power dissipated in a resistor in an AC circuit is given by P = V_rms^2 / R. The RMS voltage V_rms = V_peak / √2 = 200 / √2 = 141.42 V. Thus, P = (141.42)^2 / 50 = 400 W.

An AC Voltage Source,with a Peak Output of 200 V,is