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Z0? A msj2 n2 n~j4 nFor the circuit shown; if I = 41, the complex power delivered by the source in VA is:0 A3362.00-j3362.00 0KB 840.50-j840.50 91C.1681.00-j1681,00...

Question

Z0? A msj2 n2 n~j4 nFor the circuit shown; if I = 41, the complex power delivered by the source in VA is:0 A3362.00-j3362.00 0KB 840.50-j840.50 91C.1681.00-j1681,00 420,25-j420.25 E: 560.33-j560.33

Z0? A ms j2 n 2 n ~j4 n For the circuit shown; if I = 41, the complex power delivered by the source in VA is: 0 A3362.00-j3362.00 0KB 840.50-j840.50 91C.1681.00-j1681,00 420,25-j420.25 E: 560.33-j560.33



Answers

Given $I_{o}=2 \mathrm{mA}$ in the circuit in Fig. P2.104, find $I_{A}$.

You knew this problem, we're going to be looking at the face or diagram that they mentioned in the problem. And so from that diagram, I see the co sign of the phase angle fly. So this is the power factor is equal to the voltage across the resistor over the total voltage. And so now I can use owns a law to relate these two voltages to the currents and that can cancel occurrence and keep the resistance is and the Indians and so v r is equal to current times resistance. And then the total B is equal to current times and peons. Then we can cancel the currents when we get that, which is what we want to show in the first place. And probably we're going to start with this formula. Yeah, which power is equal to the Armas voltage times? The army's current times, the power factor. We know what the power factor is, so let's go ahead and substitute that and with them. And then now I'm going to move this impedance over into this term. And so if I just erase this here, that impedance is going to go with the voltage our mess like that. And then now I'm going to use the fact that the voltage armistice over the impudence is equal to the current armistice. And you'll see, whenever I multiply this by this, it's gonna square it. I'll get I squared R and this is what we want show and so

I know yet it is given that we have to justify that, we have to show that so power factor is our way. Get for it and er circuit and be part who showed that average power. It's going to I out amid two square into our no to prove a B will use federal egg up. This is that got in or potential difference across the register. Just a moment I have to read Loyt. This is potentially across in doctor and this is puttin Celik wrote Trappist. This is the next potential in the circuit. On this is the face relation so value off our backs, Of course, of fight. If you major from the battle and Elytis, it will be potato. It crosses register upon not what it did in the circuit, but until they process, it is stories. I got Emmitt into our and potential in the circuited I Adam, it's upon debt, so power factor you will get are upon jet. So this is the prove off part 89 part B average Barbara is defined it the elements in tow I, Aramis and course of fight so we can write the aromas and toe I, Aramis Course off by it are up one jet, but brmsg upon dead. Gamble it in it, I estimate. Would you ever be question we can write? I Aram is in tow. I aram is in tow. Are so it becomes ice where Aramis you are. That's what for this problem. Thanks for watching it.

For this question about the question you want to find the unknown currents. The first that I can do is to look at loops that you've office small number off unknowns as possible. It's one of loops is over here. So in this looked as the only unknown there'll be involved. It's I fall. Okay, writing Now the jobs look rule. You start off with the battery moving from minus two. Positive. So we get that V two. So this is fight falls. Then we bypass are four in the same direction s to current, so we some tree away Deportation drop. So this tool times I fall every create this 20 so you can find I fought to be quite five MPs. All right. Next unit for other equations. To find the father unknowns, I went to high five. So, Defense Secretary, I'm gonna use T Junction Junction rules made ready. Incoming current must be close to the outgoing current. So, for your first junction, all the rights we have eye for the incoming current and your going will be I five and I treat for a second junction. I tree and I one incoming and I two is the only elk away current. The 30 question We're gonna use the different look, Rose. So if we can do you can look at is over here and for this loop, we start off with the best for you again. And so we are moving from positive to negative. So we got a minus away, V two. So that's minus five. Debbie. Bypass artery. Same direction. Nasty current. So minus Sometimes I tree very bypass are one in the opposite direction. So we plus quit is 20 The last equation Internet Yes, over here. This particular group the start off Tibet tree from minus to plus. So we at the fee one this shelf by plus ah, one witness full time. So for my one Demi bypass, I to which is also subject because it's in the same direction is the current maybe three times side too. Question zero now Too soft for this ik regent's from equation too. We're going to from equation to find ways I won, I want Yes, he goes to my two minus all right tree. We substitute that into both equation tree and four so in equation tree off the substituting in you get minus five minus to a tree, us four times off to my nose, for I treat close to zero. Simplifying debts gets minus five minus six tree plus for two because it was substituting debt into equation for my substituting the equation for I one into equation for get shelf minus for two plus for a tree. My next shii to west with zero get shelf plus four tree minus seven I to close to zero. We have two equations and toe unknowns. It's like yourself. So after substituting, uh, we adjusting the equation I left. We get our i to in terms off a tree and the from the equation on the right. All right. We can substitute the I two into question over here, so we get chop. Thus for a tree minus seven, four times six. Tree bus five, close toe zero. So this is the video expression full I tree and then we bring all the constants to the right hand side. So that's trough doing it. I tree to be close to 0.5 MPs From here, we can find any other values quite easy. I to using the above expression we can't two MPs. Then we can start you to into the equation for I one right over here, which is a goes to I to my nose. I tree should be a 1.5. Finally, we can't get five from the first equation. Jessica's toe I four minus. All right, treat. So this by fine minus my fire, which is two MPs s well, no to find a Pulis applied by the different what age is. We just have to multiply the current 50 voltage. So the current passing Trudy respective for teach We use that current to get power and poets goes to feed times I So the public or city Rotich one get few one times I to write nasty current housing. Truitt. So it'll be journey for what's for P two. The power for the second footage that is V two, which is five votes went Bye bye. I five i five is skipper nous to MPs so this will be 10 watts to find the power dissipated by the resistance. We already have t various currents. So we just have to use the current I square. Uh, my current square time. See resistance to get it power. So power for our first. The sister. Uh, one. We take high one square times the resistance off one first night. What's let me do the same thing for the other resistance, right? So I'm gonna do them directly for you. For our tree. The carbon this point finds a boy five square, perhaps two homes. What fight? What's for the current ISS? I for just 2.5 resistances too. You should get show my fight. What's

All right. This problem deals with why delta transformations. And I've redrawn the circuit firstly over here to make it a little more clear. So you can see that there are two bridge circuits now in this case, all the resistance is are equal to 12 kg. So further I can draw that as to delta's. So I have one delta here and one delta there. So now I'm going to move on to the delta Y transformation and that's on the next slide. So here I have the delta, the delta has an R. One and R. Two and R. Three. And then I label my corners A. B and C. That doesn't matter too much. The big thing to remember for any of these resistors over here in the UAE for instance are A then our A. Is equal to the product of the resistance is next to it divided by the sum of all three of the resistance is. So in this case all the resistance is being 12. Then I have for our A R. B R. C and for each of those two deltas, then I have 12 times 12 and that's divided by the sum of all the resistance is in the delta and that's 36. So in each case then my resistance is equal to four kg. So once I've transformed those into two wise then the circuit becomes pretty easy and pretty straightforward. So here's the circuit redrawn and in this case each of these resistance is will be four kg. There we go. So it's four kg. And that means that this branch here, since it's in series is eight killer arms. And so is this one over here. And of course that being the case and then the resistance for that parallel branch is equal to one over 1/8 plus 1/8. And any time you have to equal resistance is then the resistance for that is going to be half of it. So it's going to be four. So I have four, four and four. And so my total resistance is equal to 12 kg. Now, what I need to find is the power that's produced. So firstly I need to use the definition to use arms law rather. And arms law says that the current is equal to voltage divided by resistance. So I have 24 volts divided by 12 kg. And that means my current is two mg. Well, power is current times voltage that's equal to two mila amps times 24 volts. And so my power was equal to 48 Noah wants. So really using those delta y translations, it becomes a fairly simple circuit. It's just a matter of using OM's law and the definition of power and that's all there is to it.


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