Hey guys, so I\'ve figured out 4 out of 7 parts of this problem and I\'m just a
ID: 1402777 • Letter: H
Question
Hey guys, so I've figured out 4 out of 7 parts of this problem and I'm just a little lost.. hopefully you can help!
An AC generator supplies an rms voltage of 230 V at 60.0 Hz. It is connected in series with a 0.450 H inductor, a 5.50 F capacitor and a 301 resistor.
I figured out the impedance of the circuit is 4.34×102 ohm.
the rms current through the resistor is 5.30×10-1 A.
the average power dissipated in the circuit is 8.45×101 W.
the peak current through the resistor is 7.49×10-1 A.
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I need help finding out:
1) What is the peak voltage across the inductor?
2) What is the peak voltage across the capacitor?
3) The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?
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Thank you, thank you, thank you!!!
Explanation / Answer
w=2*3.14*60=376.8 rad/sec
XL=wL
=376.8*0.450
XL=169.56 ohm
Xc=1/wC=1/376.8*5.50*10^-6
Xc=482.5ohm
Peak current=230/434
=0.529 amp
1) VL=I peak*XL
=0.529*169.56 ohm
=89.85V
2)VC=I peak*XC
=0.529*482.5
Vc=255.2 V
3) Xc=XL (incase of resonance)
1/wc=wL
1/LC=w^2
w=undrt(1/LC)
=undrt(1/0.450*5.50*10^-6)
=0.6356*10^3
w=635.6 rad/sec
w=2*pie*f
635.6=2*3.14*f
f=635.6/2*3.14
f=101.2 Hz--answer