An AC generator supplies an rms voltage of 240 V at 60.0 Hz. It is connected in
ID: 1437716 • Letter: A
Question
An AC generator supplies an rms voltage of 240 V at 60.0 Hz. It is connected in series with a 0.700 H inductor, a 4.30 F capacitor and a 366 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
Impedance of circuit, Z = R + j(XL - Xc)
XL = 2pi fL = 2 x pi x 0.7 x 60 = 263.9 ohm
Xc = 1 / 2pifC = 1 / (2pi x 60 x 4.30 x 10^-6) = 616.9 ohm
R = 366 ohm
Z = 366 + j(263.9 - 616.9) = 366 - j353
|Z| = sqrt(366^2 + 353^2) = 508.5 ohm
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Irms = Vrms / Z = 240 / 508.5 =0.47 A
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P = Vrms Irms = 240 x 0.47 = 113.3 W
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Ipeal = Irms sqrt(2) = 1.414 x 0.47 = 0.665 A
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Vpeak _ inductor = 0.665 x 263.9 = 175.5 Volt
Vprak_capacitor = 0.665 x 616.9 = 410.2m Volt
at resonance freuency,
XL - Xc = 0
XL = Xc
2 pi f L = 1 / (2pi f C)
f^2 = 1 / (4 pi^2 L C) = 1 / (4 x pi^2 x 0.7 x 4.30 x 10^-6)
f = 91.74 Hz
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