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Consider an $R L C$ circuit with $R=105.2, L=518 \mathrm{mH},$ and $C=0.200 \mu F .($ a) At what frequency is this circuit in resonance?(b) Find the impedance of this circuit if the frequency has thevalue found in part (a), but the capacitance is increased to0.220$\mu \mathrm{F} .$ (c) What is the power factor for the situation describedin part (b)?
a) 494 $\mathrm{Hz}$b) 180 $\mathrm{ohms}$c) 0.58
Physics 102 Electricity and Magnetism
Chapter 24
Alternating-Current Circuits
Current, Resistance, and Electromotive Force
Direct-Current Circuits
Electromagnetic Induction
Alternating Current
University of Washington
University of Winnipeg
McMaster University
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in the first item, we have to couple it. What is the resonant frequency off that circuit? The resident frequency is given by one divided by two pi times the square it off the lieutenant's times like fascinators, then putting the given values by the problem. We got one divided by two pi times the square root off 518 times 10 to the minus treat because there is a merely perfect here times 0.2 times 10 to the minus six. Because there is a me Crowe perfect here, and these is approximately 494 hurts in the second night and we have the following situation. We have almost the same circuit, but now the capacitance is It goes to 0.22 Negro fires, and I remember that the circle it is running out 494 hearts. We have to complete what is impedance off that circuit. Impedance is the square it off r squared. Plus, the difference between the reactors is squared. No, we will calculate. What are you react? Ince's The capacity reactors is equals to one divided by two pi times the frequency times the capacitance So it is, of course, true to buy times 494 times zero points, too. 22 times 10 to the minus six. Because there is a me crow here and this is all under one. Then a couple e the value off the inductive reactors Inducted reactors is equal to pi. Thank the frequency times in look tense, which is, of course, to in the following situation to pie times 494 times 5 1/8 time to stand to the minor street because there is Emilie prefix here. Then we plug in all these expressions on the expression for didn't see this to get the following result. The impedance is the course, too. The square root off 105.2 squared plus to buy times 494 times 518 times 10 to the minus street minus to five times 494 times zero points 22 times 10 to the minus six under one on This difference is squared and these is approximately 180 homes. Finally, we have to calculate what is the power factor off that circuit in the situation given on the second item the power factor is it goes to the co sign off the face, which is equal to the resistance divided by the impudence. So these equals to 105 12 divided by 180 on. These is approximately 0.58.
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