An AC generator supplies an rms voltage of 240 V at 50.0 Hz. It is connected in

Question

An AC generator supplies an rms voltage of 240 V at 50.0 Hz. It is connected in series with a 0.600 H inductor, a 4.10 ?F capacitor and a 341 ? resistor.
1) What is the impedance of the circuit?_____________________

2) What is the rms current through the resistor?________________

3)   What is the average power dissipated in the circuit?____________

4) What is the peak current through the resistor? _______________

5) What is the peak voltage across the inductor?_____________

6)   What is the peak voltage across the capacitor?______________

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

Explanation / Answer

1) XC = 1/(2*pi*f*C) = 1/(2*pi*50*4.1*10^-6) = 776.8 ohms

XL = 2*pi*f*L = 2*pi*50*0.6 = 188.4 ohms

impedance, Z = sqrt( R^2 + (XL - Xc)^2)

= sqrt(341^2 + (188.4-776.8)^2)

= 680 ohms

2) Irms = Vrms/z

= 240/680

= 0.353 A

3) Pavg = Irms*Vrms

= 0.353*240

= 84.7 Watts

4) Ipeak = Irms*sqrt(2)

= 0.353*sqrt(2)

= 0.5 A

5) VPeak_L = Ipeak*XL

= 0.5*188.4

= 94.2 volts

6) VPeak_C = Ipeak*XC

= 0.5*776.8

= 388.4 volts

7) f = 1/(2*pi*sqrt(L*C))

= 1/(2*pi*sqrt(0.6*4.1*10^-6))

= 101.5 Hz

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