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 0.750 H inductor, a 5.80 HF capacitor and a 246 omega resistor. What is the impedance of the circuit? Tries 0/20 What is the rms current through the resistor? Tries 0/20 What is the average power dissipated in the circuit? Tries 0/20 What is the peak current through the resistor? Tries 0/20 What is the peak voltage across the inductor? Tries 0/20 What is the peak voltage across the capacitor? Tries 0/20 The generator frequency is now changed so that the circuit is in resonance. What is that new resonance) frequency? Tries 0/20 

Explanation / Answer

Given that R = 246 ohm

C = 5.8 uF and L = 0.75 H

IMpedence is Z = sqrt(R^2+(XL-XC)^2)

XL = 2*pi*f*L = 2*3.142*50*0.75 = 235.65 ohm

XC = 1/(2*pi*f*C) = 1/(2*3.142*50*5.8*10^-6) = 548.73 ohm

R = 246 ohm

Z = sqrt(246^2+(235.65^2-548.73)^2) = 54.98*10^3 ohm = 54.98 kohm

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Irms = Erms/Z = 120/(54.98*10^3) = 2.18*10^-3 A

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Pavg = Irms^2*R = (2.18*10^-3)^2*246 = 1.16*10^-3 W

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Imax = sqrt(2)*Irms = 1.414*2.18*10^-3 = 3.08*10^-3 A

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