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$\bullet$ In an $R-L-C$ series circuit the magnitude of the phaseangle is $54.0^{\circ},$ with the source voltage lagging the current. Thereactance of the capacitor is $350 \Omega,$ and the resistor resistance is 180$\Omega .$ The average power delivered by the source is 140 $\mathrm{W}$ . Find (a) the reactance of the inductor; (b) the rms current; (c) the rms voltage of the source.

a.) $102 . \Omega,$ b.) $0.882 \mathrm{A},$ c.) $270 . \mathrm{V}$

Physics 102 Electricity and Magnetism

Chapter 22

Alternating Current

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

University of Sheffield

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you have the magnitude of the phase angle Phi equaling 54.0 degrees. We have the capacitive reactor. Since equaling 350 homes, we have the resistance of the resister. 180 comes and we have the average power delivered by the source equal in 140 watts. Four part, eh, The formula for the face angle. We can say that tangent, uh, fi equaling the inductive reactions minus the capacitive reactant CE divided by our their assistance week and then say that solving for the inductive reactant so this would be equaling 350 homes minus 180 homes multiplied by tangent of 54 0.0 degrees. And so we can say that then the inductive react Ince's equaling 102.251 comes for part B. The average power is equaling to the R. M s current squared multiplied by the resistance are so the R. M s current would be equal in the square root of the average power divided by the resistance are so this would be the square root of 140 watts divided by 180 arms, giving us 0.882 amps For part c, The R M s voltage is equaling the army's current multiplied by the impotence Z and this would be equaling to the are at mass current multiplied by the square root of the resistance squared plus the inductive reacted. It's minus the capacitive reactor is quantity squared and we can say that then the root mean vote rooming squared voltage would be 0.882 amps multiplied by the square root of 180 bums quantity squared plus 102.251 minus 350 quantity squared. I'm squared. Extend the square root and we find that the roots spread voltages equaling 270 0.1 volts. This would be our final answer for part C. That is the end of the solution. Thank you for watching

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