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(a) Sketch the phasor diagram for an ac circuit with a $105-\Omega$ resistor in series with a $22.5-\mathrm{mH}$ inductor and a 32.2 -\muF capacitor.The frequencylof the generator is 60.0 $\mathrm{Hz}$ . (b) If the rms voltage of the generator is $120 \mathrm{V},$ what is the average power consumed bythe circuit?

a) $-35.14^{\circ}$b) 91.7 $\mathrm{W}$

Physics 102 Electricity and Magnetism

Chapter 24

Alternating-Current Circuits

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

Alternating Current

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to solve the first item we have to complete. What is the phase and only then we can draw a face or diagram and to complete the face we have to do the following. Remember that the tangent off the phase is equal to the difference between the inductive and capacity. Reactant is divided by the resistance. So before completing the phase, we have to copulate. What are the reactors is We will do this when you order site beginning by. The reactor's off the doctor. We have the following. So the reacting suffering doctor is a course to two pi times The frequency times they looked at us put me in the values that were given by the problem. We have to pie times 60 times 22 0.5 times 10 to the minus three and uses approximately eight point 482 owns and now we complete The reactor's off the capacitor, which is equals to one divided by two pi times the frequency times, the capacitance bleeding, the values given by the problem. We have one divided by to pie times 60 times authority to 0.2 times 10 to the minus six and these is approximately eight. Two point 38 owns, and then we can complete one. Is the tangent off the face. So the tangent off the fees is equals True heat point 482 minors eight to point 38 divided by 105. And then to phase. Is it close to the inverse tangent off 8.4 H two minors 82.38. I do find it by 105 which is approximately minus 35.14 degrees. Now we can proceed to draw your face or diagram. These will be our reference access. We will have three phasers, one fund a resister, one for the doctor and one for the capacitor, and also another one for the servers. So there is a total of four phasers. Note that the resistance is bigger than both reactors is so our biggest phaser. We don't need their resistance face work. So the resist or phase were we will be these one and no, we can draw either they looked or phaser or the capacitor phaser. We will choose to draw the doctor favor first. So then doctor phaser leads the current phase or by 90 degrees on it is smaller than the current phaser because the reactive look, this is much smaller than the resistance they're for It is smaller than the reason your phaser No, we can draw work a passable phaser which legs behind the resist on favor by 90 degrees. And it is bigger than the doctor phaser. And finally we can draw a service phaser, which is the biggest off every other finger, because it is at least as big as the resistance phaser, and we can see that the react insists will not console each other. So our search things there will be bigger than any of those other three phasers. And you too, like behind the resistance phaser by 35.14 degrees. So it will be something like that. And then we have our face or diagram on the next item. We have to copulate. What is the average power that is considered bar circuit? In order to do that, we have to know what is impedance off our circuit. So we will begin by doing that impedance off our circle. It is because to describe it off r squared, plus the difference between the react tenses squared. That's Newt. And then plugging the values that were given by the problem and calculated by us, we get it square root off 105 square plus 8.4 eight to minus H 2.38 squared And them keep that. Now we can proceed and calculate What is the average power? Consummate their circuits with the following equation The average power consummate by our circuit. Is it close to the R. M s current? Multiply it by the r a mass voltage Multiply it by the cool sign off the feeds. But remember that the coastline of the fees he's equals to the following equation. So the co sign off the phase is it goes to the resistance divided by impedance. And then I also remember that by using poems War we have V A romance is he goes to Why are a mess times the impudence, then the current Is it close to the vault ation divided by the impudence Blogging It is a result now our equation for the average power Consume it. We get that the average power Easy coast too V are a mass divided by the impudence squared times your resistance. Then we plaguing their values. That won't even buy the problem. So for the rotation we have 120 votes for the impedance. We have these expression. So just copy that that expression here, then it is squared times the resistance and the resistance is 105 O's. And this is approximately 91 0.7. What?

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