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Find the current through the 9.0 0 resistor.6.099.02.0912v0.71 A21A13A0.86...

Question

Find the current through the 9.0 0 resistor.6.099.02.0912v0.71 A21A13A0.86

Find the current through the 9.0 0 resistor. 6.09 9.0 2.09 12v 0.71 A 21A 13A 0.86



Answers

Find the current in the $12-\Omega$ resistor in Figure P 18.15.

In this question we are given that yeah given this uh circuit. Okay. Battery with three resistance. And then uh given that the E. M. F. is 9.0 votes. R. One is 150 homes. R. Two is 250 homes. Our tree is 1000 homes want to find a current in each resistor. Okay so to solve this question first we want to find our equivalent resistance in the circuit. So we see that to an artery in parallel So we can buy our 2. 3 is one of our two plus one of archery and the recipe book of it. Okay and then our equivalent resistance is one plus R. Two tree. Okay so uh of our fast one of our two plus one of archery. The reciprocal substitute the numbers one of 1 15 plus 1/2 50 plus one over 1000. The reciprocal of it and the Cuban resistance you calculate you get 250 once. Okay? And then the current in the circuit current supplied by Petry. If you go to the mm, do I buy the equivalent resistance? So nine by Try and 50 and you get 0.0257 MPS. He then um I won is going to be 0.0257 MPS. Because uh is in series with the battery and then I too Is equal to be to divide by our two and B two is e minus. We won then be one is uh I one or 1 developed by our two. So you can substitute nine minus. So I want 0.257 R one is 115 and then divide by 215. Okay, So you get 0.0206 Mps. And then I three is equal to 1 -2 T. And the answer is 5.14 times 10 to the negative three mps. Okay, so here are the answers for this question and that's all

And this problem we have the circuit here, We're all we're given. Is that the current of this? 13.8 own resistor Is given as .795. Yeah, You can call this current 1/17 13.8 Can be called Resistor one. So let's call 7.17.2 are too The 8.45. These are three, someone at the bottom is our four. And on the left side 15 on will be 5 12.5. All will be R. six. And we want to find the current for each resistor. So to do this we can use the junction war which is the total current of it equals the sum of all the currents in the circuit. But there's two things to notice here. The first is that from the battery we'll have the total current coming out of it. So we have the total current flowing into our five and into Our six. But then when it comes to this place right here it'll split off, it will split off to this current going down into another current going to the right. I went to the right will flow through our three and it will also flow to our four. So R five and R six are in series so they will have the same current flowing through them and on the right side R. Three and R. For they are also in Syria so they'll have the same current flowing to them as well. So now let's look at The current going through our two. So we see that are one and are too are in parallel to each other which means that they will both have the same potential difference or the same voltage across across both of them. So we can find the voltage From resistible one. And then I go as the voltage for resistor to and then we can find the current flowing through them. So again we're looking for the voltage for Is this the one we can use we can do that with arms long which is voltage equals current time resistance. So it will be the one weakness I one times Are one and we know what I won is is given to us is .795 are one is 13.8. So the voltage across was just one is 10 point 97 folks. And this is the same voltage once registered to. So he had V two is also 10.97V. So now we can find the current falling through itunes again from mom's long. We have v. two equals 2 times are too. And we can solve for I too. So the vertical sites are too So we get voltage two divided by R. two. I'm not plugging in our values. We have the top 10.97 Provided by our two 17.2. So the current going through resistor to Is .64. Yeah. And now let's look at R. three and R. four. So we can find the current falling through both of them because they are in serious. So they should have the same one. And again we call this current. That's just how I. Prime. So from um salah we have voltage equals current. I prime times The equivalent resistance. And here here is the sum of the resistors. R. three and R. four. So solving for I. Prime. Okay I'm trying it was the voltage battery by the equivalent resistance. Now for voltage we go back to our circuit diagram Because R. three and R four are in parallel with the resistors inside that means they also have the same voltage. So the three and the form will have the same voltage as Our women are too. Which we found as 1097 folds. And again, this is only because they are in parallel to each other. So now to find the current pipeline, we just need to find the equivalent Resistance for our three or 4 because they are in series. The equation is just R. three. R 4. So plugging in the values eight points. That's why Plus 4.11. So the equivalent resistance is talk on 56 Holmes. So now you can plug this into. I'm fine. So we have the voltage 10.97 divided by the equivalent resistance 12.56. So we get here .87 eps. And so this is a current for I three and I four again because they are in series 87 amps for I three. And my foreign is also 0.87 Yes. So now we just need to find the current flowing through. Is this for five and 6 on the left side? So now we can use the junction rule. We can add up all the currents to get the total current because again, the total current is what's flowing out of the battery. There's green line here so the total current will fill our five in our six. Once we find that we find the current for archive and our six. So the total current With the sum of all the currents. So we have the current one .795 Plus. current two which is .64 plus. So the current is the same for R. Three and R. For which we found to be with 87 So you add this up, you get the total current for the circuit is point There's 2.3 amps. And again, this is the current coming out of the battery. This is the current flowing through R five and R six because they are in series. So we have uh All right, fine, 2.3 Amps and I six. 2.3. Yes.

Okay, so this is a very lengthy problem. Um, first, So we have this circuit here, and Ah, As you can see, I have marked all the currents that's passing through our each resistors. Now, I marked the current with the, uh, value of the resistor correspondent. I'm among the current corresponding to the value of the resistor. So, for example, I six is the one that's passing to six homes. Idol is passing through twelve homes. I six point eight passing to six point agent. So on, so forth on DH. Now we see we have one, two, three, four, five unknowns. So we need five questions Now we can use Junction rule at A and E, and then we can use lope rule at left, middle and right, Luke. So let's first supply junction rule at point and be so that point a we see that top is going away. I six is going away and I five is going into the junction. So that means I five should be called Ice Express. I adopt similarly in this junction. I top is going coming in, and I six hundred it is coming in and then twelve is going over it. So that's what we want here. And if we act, it's all for it. If we actually substitute items from both of the questions, we get a relation between I five, six point eight, twelve and six, right? So that's That's a little attention now for the loop rule. What we're doing here is first. We're starting with the seven bowls here, so these two are in serious. So that's why we add them. And we are taking our direction of current as clock ways, and similarly, we're taking for the middle one. We're taking the direction as club plays again. And for the right Luke, we're taking the direction as anti clockwise, right? So here, as you can see, we're adding twelve and five so way have folded gain on both Ah, marriage. So that's where we have seven holes on. There's a plus sign. Then the kind is in the same direction as the look. So that's why you have a negative sign and then we have I five times five homes, then who have negative sixteen six jobs because of this current. So that's our left. Look for the right look. We have to various connected in Siri's. And then we have a voltage again. So we have ten bowls, then twelve times twelve, and there's a negative because of the same direction. Then again, six fine date is in same direction with the low obstructions, or it's going to be six point eight times Isaac's pointy and is a native. And for the center look, we see that ice sixes in opposite direction with the obstructions or that. So sorry. Ah, it's going to be the opposite. But it doesn't matter, actually, because since there are only two currents, we'LL have the same expression. But anyway, so we see that Ice twelve is in opposite direction with the low traction. So that's why it's positive. And I six is in the same direction of the looks, tricks and direction. So so it's native, but we're not concerning. This ight off current here because there's no resistors. So that's what it is, a double digit drop, so we don't have to worry about that right? So now we have, ah, technically, four questions. So we have marched this equation into one, so that's where four questions Let's try to solve them. So we start with the fourth question. So you see, it's it's the shortest off all. So we get a relation between six and twelve, and then we can use this and substitute I six on the other equation. So that's what we did. So when we substituted all the I six, we got expression like this. So these are our new ah equations. So these are the three new questions that we have. So we go to the bar for decoration. Now, what we can do here is from there we used question one or the first one toe eliminate six point eight. So in our forest ignition is this so we are eliminating six find? Ah, yeah, we're eliminating six point eight from the first equation. So we're writing in terms ofthe tighter than I fight. So write if we substitute back Toa I too and I we get these to falling expressions on da. Okay, so this thing is going to be zero, right? So then this is our third equation. This is our second question. Now we see that we only have five and I dwell on both of the questions. So what we did here is we now found an expression for if I from the second equation s o a cz we see here that we got rid of our first equations. Now, from the second creation, we write I five in terms ofthe twelve, and we substitute that Any question three And if we do so we'll have on ly I dwelt, um, present in this third equation and then we can solve for it. Well, so that's what we did here. And then once we put all that on all the numbers together, we got I twelve as point six seven nine, eight amps. Now, since we know I'd well, we can figure out if I buy this expression. So And we got a fight as one point seven six eight EMS again, we can go back. And since we know I five, we can know. We know I six point eight by this expression over here. So that gave us this this point two seven one for EMS, And then finally, if we use this expression like here, that gives us the expression for our value for six, which is one point two six APS. So, yeah, that's how we're going tow. We got our occurrence, but make sure you check the calculation so that you can do it by yourself. Thank you.

Hi in the given problem, This is a series combination of the two resistors. This is our one having of resistance of 15.0 oh and in series with it there is another resistor having a resistance of our two is equal to 9.0 home. Then a battery is attached with this serious combination which is having an IMF off 18.0 volt. So the equivalent resistance of the city's combination will be given by our one plus R two which will come out to be 15.0 plus 9.0 Whom? Or we can say this is 24.0 home. So using home slope, the current in this circuit, the current passing through the city's combination will be given by I is equal to net mm headlight divided by the equally violent resistance of the city's combination. So here it will be 18.0 world divided by 24.0 home, which finally comes out to be 0.75 M p R. Which is the answer for this given problem. Thank you.


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