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$\bullet$ In an $R-L-C$ series circuit, $R=150 \Omega, L=0.750 \mathrm{H},$ and$C=0.0180 \mu \mathrm{F}$ . The source has voltage amplitude $V=150 \mathrm{V}$and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replacedby one with $C=0.0360 \mu \mathrm{F}$ and the source frequency is adjusted to the new resonance value. Then what is the averagepower delivered by the source?

a) 1.00b) 75.0 $\mathrm{W}$c) 75.0 $\mathrm{W}$

Physics 102 Electricity and Magnetism

Chapter 22

Alternating Current

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

University of Winnipeg

Lectures

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So here report A We have residence so we know that then the inductive reactant is equal in the capacity of reactions. And we know that the power is sent the equal in 1/2 times the current multiplied by the voltage times co sign of the face angle Phi. So we can say that the formula for the phase angle Phi would be equaling are 10 of X sub out minus except C divided by our We know that X the again the inductive reactant equals the capacitive reactor. It's which means that this value zero Which means that fives equaling arc Tanne 00 This would be equaling 20 degrees, which means that CO sign of Fei is equaling co sign of zero degrees, which is equaling one. And so that would be the power factor of the circuit. And so, for part B, we then know that the current I will be equaling to the voltage over. The impotence is easy. And so and I residents, we have that the impotence Z is equaling are and so I z equaling simply the over our and we can say that then the power is equal in 1/2 multiplied by I he co sign of fine. And we know that CO sign of five is actually one. And so this would be equaling to 1/2 V squared over R At this point, we can solve. And so the power with the 1/2 multiplied by 150 volts quantity squared, divided by 150 arms. And this is giving us 75.0 watts. This would be our answer for part B um four part See, We know that the formula for the average power delivered by the source p equaling again 1/2 times v squared over r multiplied by co sign of the phase angle. If I hear the power factor here is again one. And so the equation becomes again V squared over to our, which is exactly what we had in part B. So the average power delivered by the source will be equaling two again 75.0. What's this would be our answer for part C. That is the end of the solution. Thank you for watching

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