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Find the currents flowing in the circuit in Figure 21.55. Explicitly show how you follow the steps in the ProblemSolving Strategies for Series and Parallel Resistors.

$I _ { 1 } = - 0.345\mathrm { A }$$I _ { 2 } = 0.379 \mathrm { A }$$I _ { 3 } = 0.034 \mathrm { A }$

Physics 102 Electricity and Magnetism

Chapter 21

Circuits, Bioelectricity, and DC Instruments

Direct-Current Circuits

Cornell University

Simon Fraser University

University of Sheffield

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Hi. In the given problem here, this is Ah, complicated circuit diagram which is having two loops. The upper loop is a k a l e B c b and a again. On the lower loop is a G i h g f e l k and a again no. If we apply junction rule at Junction A, we come to know that this current I tree, which is entering into the junction, is equal toe I one plus i to the algebraic sum off the two currents Ivan and I two now this current Taiwan between same in the upper branch, it will pass through this internal resistance. Arman it will pass through this are five also and will reach to the junction e again. And this I do will pass through this are two and then are 2 14 and will reach to the Junction E. And then this I to will combine with this I wan and will become I three again which will enter into the negative terminal off cell, the three. And then we'll come out off its positive terminal again. No, If we apply for jobs, Lou, Cool in the upper loop line, the jobs loop rule in the closed loop e g l e be c b A. We know that this rule says the algebraic sum off all the potential drops taking place in the loop is equal to algebraic sum off all the IMF's put in the loop on the EMFs off the loop. And here this algebraic means we have to consider the sign convention before adding them. So it says all the potential drops in which the currents are passing clockwise will be taken as positive. For example, this Ivan organ will be positive here. This Ivan into our five will be positive because I wanted to Five will be positive. But this I toe in 20.5 will be negative. As this is counter clockwise and this 40 into, I too will also be negative. As this is also counterclockwise. This Santa convention also saves if the I m f is sending its current in a clockwise direction. It will be taken as positive, so even is positive, while the MF, which will send its current in counterclockwise direction, will be taken as negative. So here this Ito is negative. So using all these sign conventions we come to know that in the left hand side the algebraic sum of potential robs Ivan Urban Plus I one r one minus I to into our to minus I to into our to Plus I want in tow. Our fight is equal toe even minus it'll now plugging in all known values. For example, urban is giving us five home so it becomes five i one plus 0.10 I won minus 0.50 I do minus 40. I do plus 20 I won is equal to 24 minus 48. So finally it becomes 25.1 I won minus 40.5. I do is equal to minus 24 which becomes situation number one No, we apply Sing loop rule in the closed loop in the closed loop a que el e f h i g and a again it becomes I do are too positive. Plus i two r two again positive as it is clockwise plus like three. I've are three rightly we will put later on I one plus I to plus I three are four and at last Plus I three are three is equal to this is E two. There are three batteries e to and the three sending their currents in clockwise direction. So this is a two plus e three, but he for risk sending its current in counterclockwise direction. So negative now, finally plugging in all known values it becomes 0.50 I do plus 40. I do. Plus 78 for our three and four i three. This is Ivan, plus I to less 0.2 for our food. But I I three again, this is I want Plus, I do Yes. 0.0 fight for Arthuis and I one plus I to again for I treat is equal to 48 plus six minus 36 from the IMF's. So, finally the situation comes out. Toby 78.25 off Ivan plus 118. 175 off I to is equal to plus 18, which is equation number two. Now, if we right equation number one again. Below it it is 25.1. Dr. One minus 40.5. I to is equal to minus 24 which was equation about one now, too. Solve these two equations for Ivan and I to we use the rules off equating the coefficients. To do this, we multiply the equation. Number two bye. 14.5 and we multiply in question number one by 118 0.75 So it becomes a question number. Oh, becomes 3169 point 1 to 5 I won plus 4, 809.375 I do is equal to 729 and the situation number one becomes 2980.6 to 5, 451 minus 4809.375 I to is equal to minus two a by zero. So finally, if we add these two equations decide toe we we canceled. So finally we become 6149.75 time So 51 is equal toe minus 21 toe one. So finding the answer for this current Ivan comes out to be minus 0134 5 a.m. p year. This is one off the answer for the given problem and here negative sign shows that the direction given in this figure is actually wrong. The current Ivan is in a direction opposite toe that we have choosing no. In order to find the value of 51 we will put the situation in equation. We will put this value off given in equation bourbon. No putting this value off Ivan Immigration involvement when you get 25.1 multiplied by Ivan Means minus 0.345 minus 40.52 He's going toe minus 24. So when it becomes minus 8.6595 minus 40.5 times off, two is equal to minus 24. So we can say this is 40.5 I to is equal to 24 minus 8.6 595 which is equal to 15.3405 So finally the value off I two comes out to be positive. Zero country, 79 and beer. The second answer for the problem. And finally, we have to find the value off I three. So that is given by algebraic sum off I want and I to means with sign convention. So for Ivan, this is minus 0.345 ampere. Right, so this is plus 0.3, 79 am here. So final answer Variety comes out to be zero and zero 34 ampere. And this is the last answer for the given problem. Thank you.

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