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A capacitance $C$ and an inductance $L$ are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If $L=5.00 \mathrm{mH}$ and $C=3.50 \mu \mathrm{F},$ what is the numerical value of the angular frequency in part (a), and what is the reactance of each element?

a) $\omega=\frac{1}{\sqrt{L C}}$b) $=7560 \mathrm{rad} / \mathrm{s}$$=37.8 \Omega$

Physics 102 Electricity and Magnetism

Chapter 22

Alternating Current

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

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So here even l c circuit. We have a maximum current I equaling 0.34 amps. And this would also be our route means squared current. Yeah, So this would be our answer for part a. And we have then a voltage V at 120 forts and our frequency f it's 60 hertz. So we can say that for part B. The current amplitude I is given by radical too multiplied by the roof needs squared current. And at this point, this would be radical two multiplied by 20.34 amps. And this is giving us point from Oh, I fucking hate you. So mental. So here even l c circuit. We have a maximum current I equaling 0.34 amps, and this would also be our route means squared current or 80 amps. So this would be your answer for part B. And for part c, we know that the average of any sign of soda yoke alternating current over a whole number of cycles of zero so we can say average, uh, sign. And so d'oh alternating current over any. Mmm. So this would be our answer for part A and we have then a voltage V at 120 forts and our frequency f at 60 hertz. So we can say that for part B, the current amplitude I is given by radical too multiplied by the root means squared current. And at this point, this would be radical two multiplied by 20.34 amps and this is giving us point. Full number of cycles is equal to zero. And finally, for part deep. We know that. Then the average square of the current we know our ass equaling the square root of the average square of the current. And this would be equaling two lower case I squared. So average equaling the root means squared current or 80 amps. So this would be your answer for part B. And for part C, we know that the average of any sign of soda yoke alternating current over a whole number of cycles of zero so we can say average, uh, sign. And so d'oh alternating current over any, you know, squared. This is gonna be equaling 2.34 amps squared, and we find that the average square of the current is equaling 2.1156 Amperes squared. This would be our final answer for party. That is the end of the solution. Thank you for watching.

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