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03: Explain in details the Voltage Standing Wave Ratio. Support your answer with equations:...

Question

03: Explain in details the Voltage Standing Wave Ratio. Support your answer with equations:

03: Explain in details the Voltage Standing Wave Ratio. Support your answer with equations:



Answers

Use the voltage-division principle to calculate $v_{1}, v_{2},$ and $v_{3}$ in Figure $\mathrm{P} 2.36$

Hi they are in this given problem number 12 we have to use the same circuit diagram as we have used in problem number six and seven and eight and nine and 10 and 11. And that circuit diagram is our next combination of various resistors. The first one is our one having a value of 5.0 home which is joined with the parallel combination whose upper branch is having two resistors in series. Arto is equal to 10 0.0 home and our three is equal to 4.0 home. Then the lower branches having a single resistor. Our fight is equal to 20.0 home then the fifth resistor. R. Five which is having a value of 8.0 home. Finally battery having a potential having an M. 415 world. Now in the previous problems we have found the net resistance of this parallel combination in the problem number six we have found it to be we P sorry that was R. P. Is equal to 8.24 home. Then we have found with the help of our urban and art fight to be in series, we have found equivalent resistance of this combination in problem number seven and that was equal to 21.2. Then we have found the total current passing through the circuit in problem number eight and that current was equal to 5.42 ampere. And that was passing through all these trees resistors, R one and R. P. And R. Five. And then we have found the potential drop across this parallel combination also in problem number nine. And that was 44.7. Bold. Then using the concept that potential remain sane in parallel combination, we have found the current passing through the upper branch. If we consider it to be I too and that in the lower branch to be I. Five. So we have find we have found this value of itunes in problem number 10 and that it was 3.19 M. Pierre. Now finally in this problem as current remains the same in series and here this are five. This is not our five. This is actually our for we are making a mistake here. So this is our for that's why this will be I fought our four and iphone. And this guarantee to passing through our two and we can consider it. It'll be I three here. But as currently means same in series, therefore we can say I two is equal to I. Three which is 3.19 and pr which we have found in the problem number 11. Therefore, using arms law, the potential drop, the voltage drop. Using on slaw wall age drop. Atcross the resistance are three. Will be we three. Let it be the three and that is equal to I into our means pi three into our three. This is 3.19 M. Pierre multiplied by our three, which was 4.0 Oh. So finally this voltage drop across our three. Consult to be 12.8 volts, which becomes the answer for this given problem. Thank you.

Well, we are, miss root. Mean square voltage is equal to the square Root off. Uh, we every square right on this can be returnees are integral. Oh, we square, do you, Pete? Up from zero to buy. Divided by two space from zero Took by divided by two divided by divided by by divided by two minus cedo So this is equal to screen Ruto by divided by two oh into integral Oh, uh two divided by why we p times Tetteh hold square on de Tita And limits are from 0 to 2 pi zero to I divided by two. So this one is not hide your money but you look to study lighted by So it's equal to to divide it by pi Right now let's integrate. No, By integrating, we get ah, square root off it Divided by pi Ncube into a weepy square not times Keita Q divided by three. And the limits are Transito too. I divided by two. Right, um so we are a mess is able to weepy divided by fruit tree

Or that given circuit writing. You see it for Notes one and 2, one Nor 1 Week and Right. We went upon pipe Nice. We went -72 upon five Unless. IS. is called zero and for no who it would be. Me too minus deeper and upon fight. That's the beauty of content. My list. Yes, it's far too. But between no one and two We won minus me to his cut of 10 board Solving equations. 1, 2 and three you will get we want to be don't eat by three points. V. two is cooled to minus and by three and parent I. S. Will be my mess and by three appear that. So thanks for watching.

In this question. The first thing we do is we calculate the time constant. Well, I'll call this TC, or in fact, all right out fully. Just say that we don't get confused Later on and his eagle toe, the resistance multiplied by the capacitance, which in our case, is roughly 330 or 33 kilo homes multiplied by 28 Micro Ference. And then this will get us a constant of 0.924 seconds. And then the next thing we realized is that the voltage is 120 volts. Route means squared. So the period of the of the voltage is given by one over 120. And that's roughly equal to zero point. You're 83 seconds. I'm from here. We noticed that the time constant is much greater than the period. All right, so this implies that the voltage across the capacitor can be thought of as constant. And this is important observation to make, because this because this means that the average voltage is equal to the peak voltage. Now that we've done this using homes law, we can say that the average current is equal to the average voltage of the resistance, which is equal to the peak voltage with the resistance on the peak voltages given by route to multiply that by the RMS voltage. And in our case, this would be route to multiplied by 120 they wanted by the resistance, and this is equal to 5.1 millennium's. So the next fall of the question with the different capacitance you will get a different time constant. So the new time constant will be same calculations before just a different value. So it will be are multiplied by C not which is equal to like 0.1 times 10 to the minus six in this case, which is equal to a much smaller time constant. So I can already tell we won't be able to use the same approximation that we did in the first part of the question. So no, because the time constant is smaller than the period. The potential difference across the capacitor is rippled. And I think you can see this in one of the figures in the chapter. I don't recall which one exactly. So now this means of the implication here with the average voltage, is very, very close or roughly the same as the RMS voltage. So the average current using homes or again would SPV arm s over our, which is equal to 100 20 over the resistance. Now we roughly equal to 3.6 million times.


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