An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in

Question

An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.550 H inductor, a 5.90 ?F capacitor and a 401 ? resistor. What is the impedance of the circuit? What is the rms current through the resistor? What is the average power dissipated in the circuit? What is the peak current through the resistor? What is the peak voltage across the inductor? What is the peak voltage across the capacitor? The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?

Explanation / Answer

a) the impedance of each element is: R = 401 ohms Xl = 2 * pi * freq * inductance = = 2 * pi * 50 * 0.550 = j172.787 ohms Xc = -1/ (2 * pi * freq * capacitance) =-1/ (2 * pi * 50* 5.9*10^-6) -j539.50 ohms Since resistances add in series the total resistance is 291 - j727.1145 z= sqrt [401^2 + ( 172.87- 539.50)^2] = sqrt[84681 + (-366.63)^2] =sqrt[ 160801 + 134423.62] =Sqrt[295224.62] =543.34? or z= 543.34 ? b) the current thru the resistor is the 120V RMS voltage divided by 401 ohms = 0.2992 amps or 299.2 milliamps c) the average power dissipated in the circuit is P = I^2 * Rtotal = (0.2992)^2 * 543.34 = 48.64 watts d) the peak current thru the resistor is the peak voltage thru the resistor, where Vrms / 0.7071 = Vpeak, so 120 / 0.7071 = 169.7056 volts = Vpeak. The peak current is 169.7056 volts / 401 ohms = 0.423 amps or 423.19 milliamps. g) the resonance is 1/ (2* pi* sqrt (L*C) ) = 1/ (2* pi* sqrt (0.55*5.9*10^-6) ) = 88.35 Hertz

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