SOLVED:Rx=10n M Rz 20 0 R, =200 Av Rs= 30 n If V-40 V and the battery Is Ideal In the figure shown; AJ what Is the equlvalent resistance In Ohms of the clrcult? (Express your answer to two slgnlficant

Video Transcript

uh, we ‘re given a circuit diagram on the leave in the book. They, rather of labeling these he one are one art to our three. Eat you and our four. They put the actual numbers other racing circuit, but I labeled them this room because it ‘s a little spot easier to work with on the values are given above where, ah, the international monetary fund for a dinge, one is sixteen bolts. The inner electric resistance associated with that barrage is one point six homes. The resistor. Our two has a value. Five homes. The internal electric resistance of the battery below are three has one sharpen four homes and the associate id enough. His papal metric ton and the final resistor has a value of nine oats in ah, part a of this question were asked to find the current in the circuit both its order of magnitude and its guidance. so, uh, they ‘re two batteries here, Um, which means that the current could be flying in one of two ways it can not be flowing in the clockwise or counterclockwise commission. Um, there was It ‘s long in a counterclockwise will be flowing from a perplex, eh ? To see what you got A little error s So it ‘s just going around this way of his kind of clockwise. And if it ‘s clockwise, it ‘s the other way around A to B to C. So I ‘m gon na make a guess. And I ‘m going to assume that the current is flowing counterclockwise and it actually does n’t matter which steering I picked because the sign of the current will tell me whether or not I picked the right direction or if it ‘s the face-to-face guidance. And what I mean is, when I solve for the current, say, assume that it ‘s going counterclockwise. If I get a incontrovertible value for stream When I saw for it, then that means this steering that I ‘ve picked is the correct one. And if I get a negative value, then that means the direction that I pick. Uh, their current is actually flowing in the opposite direction. So if I could, if I turned out to be negative, then they will be flowing clockwise alternatively of counterclockwise. So the white to solve her eye is to remember that if you start in a certain point in the circuit and you come back and you make a full round travel so that you end where you started then the total potential dispute. The sum of all the likely remainder is that you encounter is zero. So let ‘s go across. We start here at at the first Battery and then we ‘LL good going to make our way around in a counterclockwise steering. So we encounter in one and then then there ‘s our one with the stream eyes. So there ‘s that folded your problem there and then another vault in patronize of our two high right now as we go across here and then the next resistor. Reese, he ‘s our three. So the electric potential cliff is our three I. And now we ‘re going. We ‘re going across, you know, we ‘re crossing this under this other battery here, but this is the the positive side of the negative side and the way that we determined a venous sinus. What we go from negative to positive that that ‘s a that ‘s a asset gestural where we go from the higher electric potential and to Lord likely. And over here we ‘re going in the opposition way were going from the lower potentials are higher potential in terms of the battery. So we ‘re crossing the battery in the opposition way, so we have to put a mind assigned to account for that. And then we have a boulder speculate here minus four before I and that ‘s equal zero. So that ‘s the total union of all the voltages Target potential differences in in the circuit. So we want to solve her eye, besides. Let ‘s actually collect all the terms of I and rewrite this equation. So minus I And this, uh, here will precisely be the sum of off the resistance that ‘s in the racing circuit. And we can easily solve this equality for eye. So it ‘s just the difference in the IMF. Know where Osama, the immunity. And we actually have all the numbers. Uh, so that ‘s sixteen bowls minus a bolt over to him our wan, which is one decimal point six homes in order to which is five bombs in our three one bespeak four homes. And finally our four, which is nine homes. So we find that the stream, uh, is your sharpen four seven amperes. immediately move on to share B. Okay, thus, separate B, we ‘re asked to Ah, so find the terminal voltage V a barn of the sixteen wax battery. So let me cursorily just give a sketch again of the Stargate. So we can have her points of address. So remember, this is B. This is a This is the one. This is our one and yeah, that ‘s that ‘s all I need for this especial part of the problem. so I know I ‘m not going to label in the rest, but, um, what we ‘re gon na do is we ‘re started, BB, and we ‘re going to go over here. Davey, eh ? indeed have a child and then thank you. Pick up a contribution from the battery and then have ah, voltage drop hera of part one. I in That should be you, Teo. Hey. And then the baby is the deviation between via and V b. But the A is given by all of this stuff over here. So the pamper forcible excessively b-complex vitamin plus you want minus R one ? I, uh, And then, child, So all of this Hey, so be a B is equal to anyone minus our one eye. And again, we have the numbers for all of this stuff. So that ‘s sixteen folds minus one indicate six homes. time zero sharpen four seven amperes, which is a current we ‘d found, uh, in part a. So the pamper, it ‘s, uh, fifteen decimal point two, five. crippled ! Right. So is in a lower a dental, then. Ethan B. Okay, And now we move on to part, See so share. See, um, using the potential sulfur So we have to find the likely deviation. B A c a point adenine with respect to point C. So it ‘s exchangeable to Teo partially be accepted that we ‘re finding the electric potential difference between A and C rather than a bee. thus again, that ‘s Jonah disgusted it. So we know where everything is located. Okay, so that ‘s our one. This is the matchless, and that ‘s our two. And that ‘s our three. This is into that ‘s our four. This point is a In this point, it ‘s C. And actually, you can find that in in one of two ways so you can go from they just see by traveling this way or by travelling this room either direction should n’t give you likely remainder. so, um, that ‘s big. Ah, let ‘s Let ‘s go. Let ‘s go with the black, The black note. Okay, so we have. So we started VC, and then we pick up a contribution from from the IMF and ah, normally, when we go across this resistor, it would be a minus I r three. But we already know that point A is at a higher potential than then point fit. thus when we go across the obstructionist, we actually put a summation sign, like therefore And the lapp matter here when we go across our excessively. And that ‘s peer Teo via so, um v c it is me, eh ? Minus BC, which is that mathematics ? Plus, I are two plus three, and that gives us They ‘re eight votes plus your point four seven amperes and our DOAs five rooms. And our three is one point for own ‘s in that Give this football team, Moz. Right. And you, Khun, repeat the routine by going the direction of the blue line and give me the lapp soul. Okay ? And immediately we come to party of the trouble and party. Ah, they ‘re asking a secondhand figure twenty five twenty dollar bill as a model. Er five hundred Graf, the potential rise and drops in the circumference. So we ‘re gon na model after that name that mention. then fair the figure that they reference here is twenty dollar bill five twenty. Okay, so we use that as a model. Um, so over here we have sixteen bolts, and huh we ‘LL start and point B. So we put that there and point is a little act aways from that. And then point coke is a short further away. So let ‘s see if I could do this department of justice. We ‘re hera and then we move a little spot and then we see that potential difference. The drop in in in electric potential because of the internal resistance, and then move along until we encounter another resistor that are besides resistor. And then we move along in a ceaseless way until we encounter our three and then we ‘re in Constant for a little bite. And Stan See, then we have, like, bad drop and then were constant and then another big drop. then this big job is from the eight Bolt resistor. This drop is from the are for that ‘s from our three. That ‘s from art, besides. In that time, our one

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