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In an ac circuit, a 500 , resistor is connected to a 16 volt acsource. A) What is the current? b) What is the power dissipated inthe resistor? C) what is the...

Question

In an ac circuit, a 500 , resistor is connected to a 16 volt acsource. A) What is the current? b) What is the power dissipated inthe resistor? C) what is the RMS voltage? D) what is the rmscurrent?

In an ac circuit, a 500 , resistor is connected to a 16 volt ac source. A) What is the current? b) What is the power dissipated in the resistor? C) what is the RMS voltage? D) what is the rms current?



Answers

A 1.5 $-k \Omega$ resistor is connected to an AC voltage source with an rms voltage of 120 V. (a) What is the maximum voltage across the resistor? (b) What is the maximum current through the resistor? (c) What is the rms current through the resistor? (d) What is the average power dissipated by the resistor?

But a off the even problem. Um, we know from this relation that re maximum physical Thio We are a mess times they're swearing off it too. From here we can write, but a that is a re max will be equal to our mess. Given is 1 20 times where route off too. This gives us remax off 1 70 year olds. I want to be, um Wi Max weekend right is from all those law are times I'm ex and from here I Max will be cool too. You called to 1 70 17 to do right by resistance, which is 1.5 times. Don't our three year old homes. Um, excuse us there. I'm excellent. 0.11 in Pierre, look. See? Oh, we can find the I are in this by dividing I Max, I'm expe I screwed over to this. Is there a 0.11 in Pierre, If I buy skirt off to this gives us Ah, the current off our famous eight times dental baller minus to appear body we confined. Average power is equal to, um the We are in this square. Our miss were divided by the resistance so substituting values 1 20 Um, we have got a power. The re arm s off 1 20 So 1 20 square. Divided by resistance. 1.5 times. Don't power three. This uses the power off 9.6 wars.

This's Chapter 21. Problem number 66 In part A. We are asked to calculate the army's current, and we are given the frequency and 60 hertz and the Aramis potential as 120 holes, and the resistance is 2.8 cure homes and the induct Ince's 350 of merely Henry's. So, in order to get to the Army's current, we need Thio. Divide the Aramis voltage by the impedance right where we know that the impedance equals two R squared plus in the inductive react. Ince's square, right? So we know that the inductive reactant is to pie half induct in ce hell. So we put everything back in the Army's current equation. So the R M pass over then horse word wass hoop i f. L squared. So our Emma's current is you've given to us is 120 walls and the resistance is 2.8. Kill a home, so let's convert that to homes Square Wass two times 3.14 pi Write thie frequency 60 hertz and we have the induct in CE. If we convert this to Henry's, it's going to be 0.35 Henry's right 0.35 and then we're going to square everything. So our Army's current is gonna turn out to be four point 28 times 10 to the power of negative, too EMS now heart be. We're asked the phase angle between Walt Ege and the current right phase tangle. So in order to get these angle by, it's going to be equal to inverse a tangent. The inducted reactive is divided by the resistance. So then we can plug in to pie, have l for the inductive correct ins divided by the resistance. Right, So Converse two times 3.14 times 60 herds times 600.35 Henry's divided by 2.8 times center for three holes. From here The calculations michelle us that the phase angle between the voltage and the current is 2.7 degrees. So in part, see, the power dissipated. So the power disappeared. That is, you know, is proportional to eye the square of the current times the resistance. And in this case, we have the r. M s current, right? So then what? We calculated our answer to part a wass here. No, 4.284 going to eight times tend to grow negative. Two hams were going to take to square and then to going intense them two or three. Those this is going to be equal to 5.13 watts party re asked to calculate their immense potential across the resistor and the in dr So we are and past. Let's start with Dean doctor, right, and that's going to be equal to the inductive reactant times. The Aramis current and again, Let's plug in for the Inductive reactors to Pi have l Kinds by car. Innis, then So two times. 3.14 times 60 hertz times. 600.35 Henry's times for 0.2 eight times. Tempted for negative two amps. We have the enormous potential across Thean Doctor. It has high 0.65 votes now the second part of parties asking a premise potential across the resistor. So then that is going to be the core to the R. M s. Put current times there the resistance, right? So it's going to be 4.2 times tentacle of negative two pants times two point lead times tented our three homes So we have his 100 in 1994. But since there's of a phase difference between the two, then you know that they are a mess. Potentials are not corresponding to the same value.

We have a register connected to an a. c. power supply. Having current equation I equals to 1.20 ampere. Of course 300 T. Okay so for the part of the question we have to calculate the RMS current. So rMS current it will be equals two. I RmS. It will be given by I note by route to where I notice the peak current. And from the equation this is the current value. So 1.20 M. Pierre de whereby under two. So from here after solving we get RmS that is equal to G 0.849 ampere. Okay now 40 part B. We have to calculate for the part B. We have to calculate the resistance value. So we have so we have we are M. S. In the equation. Uh so we can calculate that we RmS that is equal to I RmS manipulated by traditions are so from here we get traditions are equals two RmS developed by RMS. Okay, so we are M. S. Is given as 100 world divided by I. R. M. S. 0.849 mps. So from here after solving we get traditions are that is equal to 117.9 home. Okay so this becomes the answer for the part B. And this becomes the answer for the part A. Now moving to the next part C. In which we have to calculate ah we have to calculate the average power delivered to the register. So average power delivered, it is equals two. We I. RmS manipulated by we R. M. S. Okay, so since equation of the current ease in the cost 300 T. So voltage situation will be in the same because it is the register. So power delivered, it will be equals two RmS which is G 0.849 Ampere. And we are a mess which is 100 holds. So from here, power delivered or ever his power, it is equal to 84.9. What? So this becomes the answer for the party of the problem. Okay.

Belongs to the alternative current in which for the part we have to determine, we have equality as we know, it is equal to 300 world and we have to determine the RMS walt is okay, So we know that we are a mess. It is equal to peak world divided by undertook to soap equals easy here, 300 world divided by undertook through. So from here solving we get our meatball taste. We are a mess. This is equal to 212.1 Volt. Okay, so this becomes the answer for the part of the problem. Now, for the part B, we have RMS value of the current, the I R. M. S. It is equal to 10 mp here. So we have to determine here the P current. So we know that the current. I know it will be under two times of RMS current. So under two replied by I R M. S, which is 10 M. Pierre. So from here, after solving we get the current, this is equal to 14.1 Ampere. So this will be the answer for the part of the of the problem. Okay, so using these two results, we can calculate the relation between peak value and RMS values. Okay?


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