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An $L C$ circuit consists of a 3.00 $\mathrm{mH}$ inductor and a 5.00$\mu F$ capacitor. (a) Find its impedance at 60.0 $\mathrm{Hz}$ and 10.0 $\mathrm{kHz}$ . (b) Compare these values of $Z$ with those found in Example 23.12 in which there was also a resistor.
$52^{n}$
Physics 102 Electricity and Magnetism
Chapter 23
Electromagnetic Induction, AC Circuits, and Electrical Technologies
Electromagnetic Induction
Simon Fraser University
University of Sheffield
University of Winnipeg
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So here we know that first for part A the inductive reactions would be equaling two pi times the frequency times the induct in CE. So this would be to pi times the frequency of 60 hertz times the inductive of three Miller Henry's were three times 10 to the negative Third Henry's and this is equally 1.131 comes then for the capacitive reactant. So this would be equaling one over to pi times the frequency 60 hertz times the capacitance of five micro Farid So five times 10 to the native six marrieds And this is equaling 530 0.52 homes. And so we can then say that here the impotence would be equaling r squared plus x of l minus except see quantity squared, all raised to the 1/2 power. We know that here this is gonna go to zero and we can say that Then the impotence would be equaling the difference between the inductive capacitance. Sorry, the inductive reactant ce My apologies. 1.131 comes minus the capacitive reactant CE 530.52 comes quantity squared raised to the 1/2 power and so essentially Z is equaling. The impotence is equaling. 529. We can save point 39 homes. So this would be at 60 hertz and then continued on for party. No, we have 10,000 hurts. So ex of El Equaling two pi times This will be 10 kilohertz, or 10,000 hurts well supplied by the induct in CE of three times 10 to the negative Third Henry's and this is equal of 188.5 arms. And then the capacitive reactor ants would be won over two pi times the frequency of 10,000 hertz multiplied by again five Micro fair. It's so five times 10th of negative six marrieds and we find that the capacitive reactor is here is 3.183 thumbs and the infants would then be equaling again R r squared to zero. So here we're simply taking the inductive react in ce mai esti capacitive reactor. It's quantity squared, all raised to the 1/2 power and this is gonna be equaling 1 88.5 minus 3.183 quantity squared raised to the 1/2 power and the impotence is equaling. 185 points. 317 thumbs. This would be our impotence essentially at 10,000 hertz for part A or 10 kilohertz for part B. Ah, why're the compared these values and compared these values of Z in which there is also resistor? Well, there we can say that here the resistance are is there's not contribute greatly to the total impotence. And we can say that here the capacitor dominates at low frequencies and here the induct er dominates at high frequencies. Is that essentially comparing to example 20 to 12? That is the end of the solution. Thank you for walking.
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