An AC voltage source,with a peak output of 200 V,is connected to a 25-Ω resistor.What is the rate of energy dissipated due to heat in the resistor?
A) 566 W
B) 400 W
C) 800 W
D) 200 W

Question

Quizplus · Accepted Answer

The rate of energy dissipated in a resistor can be calculated using the formula P = V^2/R, where P is power, V is voltage, and R is resistance.

In this case, the peak voltage of the AC source is given as 200 V. However, we need to convert this to the RMS voltage in order to use the formula above. The RMS voltage of an AC wave is given by V_RMS = V_peak/sqrt(2). Therefore, the RMS voltage in this case is:

V_RMS = 200/sqrt(2) = 141.4 V

Using this value and the resistance of 25 Ω in the formula above, we get:

P = (141.4^2)/25 = 800 W

Therefore, the rate of energy dissipated in the resistor is 800 W. The correct answer is C.

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