An AC generator supplies an rms voltage of 120 V at 60.0 Hz. It is connected in

Question

An AC generator supplies an rms voltage of 120 V at 60.0 Hz. It is connected in series with a 0.350 H inductor, a 5.10 F capacitor and a 306 resistor.

a. What is the impedance of the circuit?

b. What is the rms current through the resistor?

c. What is the average power dissipated in the circuit?

d. What is the peak current through the resistor?

e. What is the peak voltage across the inductor?

f. What is the peak voltage across the capacitor?

g. The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?

Explanation / Answer

a)

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

Z = sqrt(306^2 + (2*pi*60*0.35 - 1/(2*pi*60*5.1*10^-6))^2)

Z = sqrt(306^2 + (-388)^2)

Z = 494 ohm

b)

tan(theta) = (wL-1/wC)/R = (2*pi*60*0.35 - 1/(2*pi*60*5.1*10^-6))/306

tan(theta) = -2.365

theta = tan^-1(-2.365) = -1.171 rad

Thus rms voltage across resistor is

= Vrms*cos(theta) = 120 * cos(-1.171) = 46.73 V

Thus rms current = 46.73/306 = 0.1527 amp

c)

Pavg = Irms^2*R

Pavg = (0.1527)^2 x 306

Pavg = 7.14 W

d)

Ipeak = Irms*sqrt(2)

Ipeak = 0.1527 x sqrt(2)

Ipeak = 0.216 amp.

e)

Vpeak = Ipeak*X

Vpeak = 0.216 x (2pi x 60 x 0.35)

Vpeak = 28.5 V

f)

Vpeak = Ipeak*X

Vpeak = 0.216 /(2*pi*60*5.1*10^-6)

Vpeak = 112.34 V

g)

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

f = 1/(2*pi*sqrt(0.35*5.1*10^-6))

f = 119.12 Hz

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