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In the circuit of the figure below, the current I1 =2.20 A and the values of € for the ideal battery and R are unknown_24.0 VS.0U 0}G.0U I2(a) What is the cur...

Question

In the circuit of the figure below, the current I1 =2.20 A and the values of € for the ideal battery and R are unknown_24.0 VS.0U 0}G.0U I2(a) What is the current I2? magnitude direction Select -(b) What is the current I3? magnitude direction Select -(c) Can you find the values of € and R? If sO, find thelr values: (If it Is not possible to find these values enter NONE" . )Explain:This answer has not been graded yet:Need Help?Readilt

In the circuit of the figure below, the current I1 =2.20 A and the values of € for the ideal battery and R are unknown_ 24.0 V S.0U 0} G.0U I2 (a) What is the current I2? magnitude direction Select - (b) What is the current I3? magnitude direction Select - (c) Can you find the values of € and R? If sO, find thelr values: (If it Is not possible to find these values enter NONE" . ) Explain: This answer has not been graded yet: Need Help? Readilt



Answers

In the circuit of Figure $\operatorname{P} 28.30,$ the current $I_{1}=$ 3.00 $\mathrm{A}$ and the values of $\varepsilon$ for the ideal battery and $R$ are unknown. What are the currents (a) $I_{2}$ and $(\mathrm{b}) I_{3} ?(\mathrm{c})$ Can you find the values of $\varepsilon$ and $R$ ? If so, find their values. If not, explain.

So we're here we have a network of the resistors and we will need to find out the current flowing through each resistors. So let's say the current flowing is I one in this branch and I took in this branch. So applying ketchups, current law, we will get the current flowing through this five home as I one minus. I do. So similarly here the current flowing will be I too. And we take the current flowing through the six home as I three. So the current flowing through the six home will be I tu minus I three. And here these two currents will combine and the total current flowing through the six volt will be I want minus I do plus I three. And here as usual I even current goes out of the battery. So even current will come out here. So we will apply ketchups voltage law across three loops because we have three different unknown variables. So let us say first we apply in this look. So here we start from this point we take it as zero potential. So potential at this point will be 12 and pay attention at this point will be decreased by this five home resistor. And then we are moving from positive to negative. So there will be document of six volt and then this 10 and three are in series. So there will be decreasing potential due to this two resistors. So 13 Taiwan and this will be zero. So this will be the first equation. So let us simplify it even more. So taking five Ivan and 13 islands will get minus 18 Ivan plus five. I do plus six equals to zero. So this is first situation. Now we consider let's sit this look. So we start at this point with zero potential and we are moving oppose it to the direction of current. So potential here will be right, I want minus I took and then potential here will be minus eight I two. And here the potential will be minus before I do. And similarly now the potential will be minus six I three and it will be zero. So we'll simplify it even more. So we get 551 minus five to minus eight. I do. That is minus 13, I do. And then minus four I two. So that will make it minus 17 I took and minus six I three as you should. But this is the question too, let's put them in the box so that we are sure that these are the equations. So similarly will take third. Look let's say I take it this one. So we start at this 0.0. Okay plus six. So the third question from this loop we use ketchup voltage logging so six minus so here we are going oppose it to the direction of current. So it will be plastic society minus six times I two minus three and this will be zero. So simplifying it six I three plus six I three here. So we get 12 I three minus six I two plus six equals to zero. And this is the equation three. So we saw these three equations so we'll take two equation at the time and get the value of Ivan and I two and then substitute in the third to get in I three. So from the first equation we can write I won has five I two plus six by 18. The second equation it is also in terms of Ivan and I two and I three. So we can take the value of I three now and converted in the form of I two. So we get five instead of Ivan we'll write it as five I two plus six by 18 minus 17. I two minus six. I three equals to zero. So converting the second equation we'll get 25 I two plus 30 by 18 minus 17 I two minus six I three is equal to zero. So here we need to calculate them a bit because it's involving quite bigger numbers. So here we get 25 I two plus 30 minus 18 times 17. That is three not 68 to minus 18 times six. That is 108 Hi three equals to zero. So from here we find out a tree. So we combine 25 I two and 306 itunes. So that is minus 281 I two plus 30 minus 108 I three. So from here we get I three as 30 minus 281 I two. Bye one not eight. So we take this value of Ivan and I three in terms of I two and replacing this 30 questions. So let's replace that we get 12 times I three. So you're right 12 times the value of I three. From here there is 30 minus 281 I two over 108 and then minus six I two. We will keep it as it is and then we have plus six is equal to zero. So from here we can figure out I took so we'll get 3 60 minus 2 81 times. Yeah, so that will be 3372 I do minus 108 times six. That will be 648 I two and then again 108 times six wickets 648 he goes to zero. So from here we figured out the value of I do. So we combine like terms, we get one double 08 minus 337 plus 648 We get 9 85 I 20 So from here we can easily get I to ask for one point 02 amperes. So once we have got this I two has 1.2 Let's put it. The situation I two s 1.2 So Ivan will come out to be five times 1.2 plus 6/18. This will come out to be 0.62 NPR's similarly I three will come out to me two point four amperes. Let's take in terms of two DP So 2.3 eight amperes it's coming as negative. So negative sign just indicates that the value of I three, whatever we have considered in this situation, it will be not the same as we have taken, it will be a positive. So this will be the value the direction of three. So we have to figure out the current across all the resistors. So once we have figured out Ivan I two and I three, we can see that the current across four home and it home. Yes, I took So let's list all the resistors here and the currents. So there are different resistors that we have over here in the diagram that is for our own resistance. We will see the current through this forum, we'll see the resist the current through this. A two six on resistance we have another six armed resistance. We have and then we have five own resistance and then 10 homes and three owned resistance. So across four home and eight we can see this four oh mandate. they are in series So current will be I too. So we figured out it was 1.2 NPR's so current through them will be 1.2 amperes. So six on there are 26 on resistance. So six on. Let us say I call this middle resistance. The current will be I three in the upward direction and I threes 2.38 amperes. So this will be 2.38 amperes across the middle one and this through the sixth on the current is I two minus I three. So we have calculated I two and I think so if we subtract, I do minus I three, so 0.62 plus two point 38 That comes out to be three amperes. And across five own resistance it's Ivan minus I took All right, this is the current across five bombs. So I want minus I do so 0.6 to minus 1.2 So it will be 0.4 amperes. Its direction will be again oppose it, oppose it to the direction what we have taken and then finally comes to 10 home and three um there also in series and the current across them would be I want that is 0.62 amperes. So we have figured out the current flowing through each resistance connected in the circuit.

So for this problem, we're looking forward, the direction and the magnitude of the current going through all these resistors. So first, let's take a guess as to what direction the currents air going in. And then we can write some equations to figure out if our guesses are correct, and so current typically goes out of the positive end of a battery in into the negative end. And so let's take a guess that the current going through our one is going in this direction and we'll call I want, and then we'll make the same guests for our three will say it's going out of the positive end of the battery. Yes, that will be our direction for I three and then for our two. Let's say it's going this way and we'll call it I two and again, these are just guesses. We'll see if they're right when we write our equations out. And so the first equation we're going to write has to do with the junction rule at this junction here. The amount of current going into the junction has to equal the amount of current going out of the junction, So the current going in is going to be I one plus I three and then the current going out is going to be I two. That's going to be our first equation that we're going to use to solve this. And then for the second equation, we're going to look at two of the loops the first loop will look at is going to be this loop, and we're gonna write down all the voltage drops along that loop. And so starting right here we have E m F one and then we're going to subtract the vultures, drop across the resistor r one That's going to be I one times are one and then keep going around clockwise. We're going to subtract the voltage drop across resistor to which is I two times are too and that's all going to equal zero. Then we're gonna do the same thing for this loop here, starting with the first battery battery number two. Then we're going to subtract the voltage drop from resistor three. So we have I three are three, then we're going to subtract the voltage drop from resistor to so I two r two and that completes the loop. And so that's going to equal zero now. There is a lot of different ways to consol of this system of equations. Um, do whichever way works best for you. But if you solve it by plugging in the values of the MF's from the battery and the resisters, you're able to solve it for I one I two and I three. And what we find is that I one is equal 2.133 amps. I to is equal 2.315 AM's and I three is equal 2.181 amps. Now, all of these values are positive, which means that our guesses for the direction of each of these currents is gonna be correct. So for the directions will use the three directions we have in the diagram here and then for the magnitudes. We'll use what we found from the equations

Okay, So when whenever a circuit like this Ah, we see that there are two batteries, they're connected and then we have three resistors. Now we always when way have to use your chops Law here because it's a complicated circuit way have two batteries and that ketchup, ketchup slow has two sections, will start the junction rule and then we'LL do the Lupron. So when we started Junction Rule, what that tells us is first, we need to find out a junction. So, for example, we have two junctions here on let's consider one of the junction. So for this problem, I'm taking this one as ah junction that I'm ah applying my rule Now. What this junction room tells us is if there's a current that is going out of the junction, that should be equal to the currents, that air coming towards the junction. So we see that I do is going out and idea and I won are coming in. So that means our If we want to apply the rule, it's going to be I too in court isto I want plus three. So that's we found that using our intentional now we could have used this junction again because we have to consider all the junctions to use a rule. But as you can see that in this area, the eye too is coming towards the junction and then I ten. I want her going out So it's going to give you the same expression like this so we don't need to do this for, for this example where there might be other examples where we might need to use all the junctions. Ah, okay, so now we'LL use the loops rule. So far, lucre. What we know is, if there's a potential drop, then there were negative sign. And if there is a potential game, then there is a positive sign. So that's one thing to remember and the other one that we should remember or which will simply their problem is if we consider the direction off our loop. So let's consider this look and let's stay clockwise direction as our direction. So if the current goes with the direction that we specified, then there there's been a negative sign, and if that goes against our if the current goes against our specified direction that there's a positive sign. So ah two ways to remember. First of all, if we say that there's a potential drop, so as you can see that in this very we have negative to positive. So that means there's a potential gains. So, um, Ian, if will be positive, then here there's a potential drop. And also since I won is going with the loop. So that's a negative. And then I do is also going with the loot. So that's a negative again. It's going to be I do are two we calls to zero, and this whole thing should be equal to zero. So we got desiccation using loop rule, so we applied it on Loop one. So let's call this Loop one, and let's call this look too similarly if we set the direction as anti clockwise, we see that the direction that we specified is consistent with the direction of the current. So again, Artie, I d will be negative and I who are you with the natives? But then, since there's a potential gain because it's negative to positive, so this guy will be positive and this will be eye to no one. Okay, so that means Teo in minus I'd artery minus. I too are too is equal to zero. So? So we applied Lope rule two seconds. Now let's see what's given here. So we know all the resisters. We know that internal instance Now we don't need it for the first part we need it for the second part. So we'LL come to that then we have the image Eso this one has nine bowls and this one has twelve volts. Okay, so now we have three questions and we need to find out the current. So since we have three unknown so we'll be able to solve this. So I've already written down how to do it and s o that I don't bore you s So what we did here is we wrote I'd won in terms of I to a nightie. And then we have this equations. So we planting all the values off. So we plugged in the value of our one and our two And then we plugged in the value ofthe the IMF and we substituted I won from this expression. So we took I won from this expression. Put it here on DA. That gave us this. So now we have an equation ofthe eye to an eye tree only. And from here, since this equation has a to n igh t only we wrote I too in terms ofthe tree and we substituted thiss in here. So we substituted. I do here and that on ly gave on Lee had one unknown and if we solve for it, we got the value ofthe tree now, since we know it tree, What we did is we put that back in this equation and that gave us the value ofthe eye too. Now, once we know I do and I'm tree, we plug these two guys back in this equation to find out I want on. That's what we got as I want. All right, so now it's all for eye. One eye to an eye tree on Now let's goto part b and see what thing is we have to make now? The problem tells us that way. Now have internal resistance tow these am a weather. Now, since the internal residence is in serious connection with let's ah, let's if we consider this looks so the Internet recent is in serious connection with our one and arto and for this case, internal resistance is in serious connection with arson are too. So we can add the internal resistance it with one off the resistors and that we can do the same thing because for serious connection, we know that the resistance um Adlai nearly So if we have an internal resistance that's around here, we can add that with our one and then we'll just have an increase off our one And then if we do the same thing for this loop, we just add our internal resistance to our three. So we'll just have an increasing alti and then the loop remains the same as part, eh? So we can carry forward the same calculation and we can solve for I when I do a nice tree. So that's what I did hear What I did was I added Internal reasons are So let me write that down here. So what I did was I added the internal resistance wit. Our given are one and that gave us our new are one. And for Artie we did the same thing. We added our internal instance too rc, and that gave us our new resistance. And then we can we did the same calculation. But instead of twenty five and thirty five, we put twenty six and thirty six and then we solve it on DH. If we do that, we again get our I want I do, and I treat. So you should check. You should do the calculation and check if you're getting the same answer as me or not. Thank you.

So what we want to registrants is So basically, Uh, there is a battery like this one. Okay. And one registrants is over here. Wanted to know what happened And one which took over here. This is a hard one. This is the heart. And this is a hard three. This is battery. And the potential difference of the battery is given the difficult to 14 or 15 gold and R1 equals to what, 10, 15, 20. So this is 10, this is 15, this is 20 only activity. Okay, So first of all we have to do and the equal tradition says then the currents of the register and the total total currently just get. So basically we can clearly see that very simple question. What we can do over here is that since the same current passes through them because they are in the streets are equivalent even do nothing. But just that some of the iron plus R plus R three, that one still plus 15 plus 20. That means the 35 39 45 on in the next part. What we need to do is the current we need to find the current restoration. So we already know that the current through its rays changes will be saying because they are in the street. So I will be nothing but Mm 2015.45. So that will be won by three that once a 0.33 NPR. And it seems we already know that the current through the target will be current through a heated exchange. So current through the entire circuit. That will be also said that the 0.33 NPR. Okay. Thank you for watching. Yeah.


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