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Problem

$\bullet$ (a) Compute the reactance of a 0.450 $\…

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Averell H.
Carnegie Mellon University

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46

Problem 3 Easy Difficulty

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?

Answer

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

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Video Transcript

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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Hugh D. Young

College Physics

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