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 mu F capacitor and a 246 Ohm resistor. What is the impedance of the circuit? What is the rms current through the resistor? You can use a similar equation of Ohm's law for an RLC circuit. What is the average power dissipated in the circuit? Power is only dissipated in a resistor. Thus, you have to multiply the rms current times the rms voltage for the resistor. 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

(1)

lmpedence is,

Z = sqrt (R^2 + (wL -1 / wC)^2 )

Z = sqrt [246^2 + ((2*pi*50*0.075 - 1) / 2*pi*50*5.8*10^(-6))^2]

Z = 398.25 ohm

(2)

rms voltage across resitor = Vrms*cos(theta)

tan(theta) = (wL-1/wC) / R = (2*pi*50*0.75 - 1 / (2*pi*50*5.8*10^(-6)) / 246 = -1.27

theta = -0.904 rad

V = 120*cos(-0.904) = 74.216 V

lrms = Vrms / R = 0.301 A

(3)

Pavg = Irms^2*R

Pavg = (0.301)^2*246 = 22.39 W

(4)

Peak current through resistor,

Ipeak = Irms * sqrt(2) = 0.301*sqrt(2)

lpeak = 0.426 A

(5)

Peak voltage across inductor,

Vpeak = Ipeak * XL = lpeak*2*pi*50*0.75

Vpeak = 100.47 V

(6)

Peak voltage across cap.

Vpeak = Ipeak*Xc = 0.426 / 2*pi*50*5.8*10^(-6)

Vpeak = 234.27 V

(7)

f = 1/(2*pi*sqrt(LC))

f = 1 / (2*3.14*sqrt(0.75*5.8*10^(-6)

f = 76.34 Hz

answer

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