AC Power Generator Part A An AC generator supplies an rms voltage of 110 V at 60

Question

AC Power Generator Part A An AC generator supplies an rms voltage of 110 V at 60.0 Hz. It is connected in series with a 0.650 H inductor, a 4.50 F capacitor and a 261 resistor. What is the impedance of the circuit? 532.14 ohm You have to calculate the impedance for a series RLC-combination. Submit Answer Incorrect. Tries 1/15 Previous Tries Part B What is the rms current through the resistor? Submit Answer Tries 0/15 Part C What is the average power dissipated in the circuit? Submit Answer Tries o/15 Part D What is the peak current through the resistor? Submit Answer Tries 0/15

Explanation / Answer

Part A

X(L) = 2*pi*f*L = 2*3.141*60*0.650 = 245 Ohm

X(C) = 1 / (2*pi*f*C) = 1 / (2*3.141*60*4.50*10^-6) = 590 Ohm

now Z = sqrt[R^2 + (X(L) - X(C))^2] = sqrt[261^2 + (245 - 590)^2] = 432.6 Ohm

Part B-

Irms = 110/432.6 = 0.254 A

Part C -

Average power dissipated P = I^2*R = 0.254^2*261 = 16.9 W

Part D

Peak current through resistor = sqrt[2]*0.254 = 0.36 A.

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