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The capacitive reactance of a capacitor at 60.0 $\mathrm{Hz}$ is 105$\Omega . \mathrm{At}$what frequency is its capacitive reactance 72.5$\Omega ?$
$\therefore f_{2}=86.9 \mathrm{Hz}$
Physics 102 Electricity and Magnetism
Chapter 24
Alternating-Current Circuits
Current, Resistance, and Electromotive Force
Direct-Current Circuits
Electromagnetic Induction
Alternating Current
Cornell University
Rutgers, The State University of New Jersey
Hope College
University of Sheffield
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we begin this question by solving these relations for the angular frequency to get that the angular frequency is equals to one divided by the reactor's times, the capacitance. And now remember that the angular frequency is he goes to to pi times the frequency. So to pi times, the frequency is because, to one divided by the reactor's times, the capacitance and finally the frequency is equals to one divided by to pie things the reactant times, the capacitance plugging in the values that we are. Even so, these value for their reactions we get that the frequency is equals to one divided by two pi time. 72 0.5 times the capacitance. And now we have to complicate the capacitance and to call click the capacitance. We apply these relations with these data, and then we get that 105 his equals Truth one divided by two pi time. 60 times like capacitance. Finally, we get that the capacitance is equals true, uh, one divided by two pi times 60 times 105. And finally we get that the frequency easy goes to to pie times 72.5 times, one divided by Jew pie time. 60 times 105 and this is all under one. Simplify the two vice, and you're and we get that. The frequency is he goes to, uh, 60 times 105 divided by 72 15 which is approximately 86.9 birds.
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