Find the currents flowing in the circuit in Figure 21.52. Explicitly show how you follow the steps in the Problem- Solving Strategies for Series and Parallel Resistors.
<b>Figure 21.52</b>
Figure 21.52
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Final Answer

I1=0.345 AI_1 = -0.345\textrm{ A}
I2=0.379 AI_2 = 0.379\textrm{ A}
I3=0.034 AI_3 = 0.034\textrm{ A}

Solution video

OpenStax College Physics for AP® Courses, Chapter 21, Problem 38 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. Given this circuit from figure [21.52], we are going to figure out what each of the currents are: this is current I 1 and current I 2 and current I 3. So we will use Kirchhoff's Loop rule two different times for two different loops and then we will use Kirchhoff's junction rule as our third equation because there are three things that we don't know—I 1, I 2 and I 3— we need to have three different equations in order to find three unknowns; you always need as many equations as you have unknowns. Okay! So there are a variety of different loops that we could think about and there's no benefit to choosing one versus another so let's just pick loop abcde and then fghij and then back to a. So that's this loop on the outside perimeter that's loop number one and we are going to traverse it in this direction with I 1 and then with I 3 here. Then we will consider loop abcde— just the same as before— so go up from point a and across and then down to point e but this time we will cut across the middle branch and so we are going to letters l, k and then a so we are traversing in this direction like this. So when we go through the middle part, we are traversing in the opposite direction to current I 2 and so we have to pay attention to that when we traverse these resistors because the change in potential will be positive in that case when traversing in the opposite direction to the current. Okay. And these currents by the way are just guesses as to the direction... it doesn't matter if the guess is correct or not; in our final answer, if the current is in fact going the other way than what we have drawn then the answer will be negative. Okay! So equation from this first loop will be negative I 1R 1 as we traverse resistor one in the same direction as the current and then we go from the negative terminal of this emf one ending at the positive terminal so that's an increase in potential there so we have plus ε 1 and then traversing little r 1 in the same direction as the current through it means minus current one times little r 1 and then minus I 1R 5 minus I 3 now because it's current number three that's going between points e and down and up to point a here so this is minus I 3r 4 and we go from the positive terminal ending at the negative terminal of this emf and so that's a drop in potential so that's why there's a minus ε 4 there and then we have minus I 3r 3 as we traverse from right to left in the same direction as the current I 3 through this resistor little r 3 and then we go from negative to positive on ε 3 so that's plus ε 3 and then minus I 3R 3 as we traverse this resistor here and all of that equals zero and that's the Loop rule, which says the total change in potential as you go in any loop and return to your starting point has to be zero. And then we will consider loop two and so we have all the same terms up to point e here so we have negative I 1R 1 plus ε 1 minus I 1r 1 minus I 1R 5 and then plus I 2R 2 as we traverse this resistor in the opposite direction to the current—the current is going to the right but our direction of traversal is to the left— and so when that happens, we have a gain in potential and so we are multiplying I 2 by R 2 but we are adding that change in potential and then plus I 2r 2 as we traverse this internal resistance of the battery number two and then minus ε 2 as we go from the positive to the negative terminal of this emf and so all of that equals zero as we reach our starting point. And then equation three says that the total currents going into any particular junction have to equal the total currents coming out of that junction— charge can be neither created nor destroyed, in other words— and so whatever goes in has to come out somewhere else and so that's conservation of charge. So we have the currents going in or just one this I 3 going into the junction at point a and then out of that junction comes currents I 1 and I 2 so I 3 equals I 1 plus I 2. There! So that's the physics taken care of and from this point on, it's algebra and solving this system of simultaneous equations. So there are a variety of strategies you could use and I will show you what I have done here. First of all, let's plug in some numbers and to save some writing since there is so much writing to begin with, we are not gonna bother writing in units nor am I gonna write in precisions, I am just gonna write the least amount necessary. So we have I 1... is traversing this 5.0 ohm resistor to begin with but I am just gonna write negative 5I 1 plus 24 because that's emf one minus 0.1 times I 1 minus 20I 1 minus 0.2 times I 3 and that comes from traversing little r 4 and then minus 36 as go from f to g then minus 0.05 times I 3 and then plus 6 and then minus 78 times I 3 so all that equals zero and then we'll collect the like terms to write our more simplified version of equation one where we have substituted for all the resistances and the emf's. So we end up with this I 1 term, this I 1 term and this I 1 term all put together make negative 25.1I 1 minus this I 3 term, this I 3 term and this I 3 term which together make negative 78.25 times I 3 and then we consider all of the constant terms— 24, negative 36 and plus 6— all that makes minus 6 and that equals zero and we can double check that we have accounted for every single type of term here and notice I am using different symbols to organize the work here— check marks for all the I 1 terms, x's for all the I 3 terms and circles for the all constant terms— you can use what strategy works for you but to me that's what makes sense I'll just double check that everything has some symbol on it to show that I have accounted for it. Then we consider equation two and make substitutions for all the resistances there. The first four terms will be identical as it was for equation one as we go across this top branch here and then we have plus 40 times I 2 as we traverse this resistor here and then plus 0.5 times I 2 as we traverse this internal resistance and then minus 48 as we go here and then all that equals zero and combining the I 1 terms gives us minus 25.1I 1, combining the I 2 terms gives us 40.5I 2 and then all the constant terms work out to negative 24 and all that equals zero. So we have these three equations now— this one, this one and this one to work with— and we have to eliminate the unknowns in order to solve for one of them. So let's consider equation one and we will eliminate the variable I 3 by substituting in its place I 1 plus I 2 and so here is that done we are substituting equation three into equation one so I am rewriting equation one but putting I 1 plus I 2 instead of I 3 and doing this eliminates the I 3 variable between these two equations; now we have two equations and two unknowns so that's a step in the right direction. And then this minus 78.25 gets multiplied into the bracket and so we have minus 78.25I 1 minus 78.25I 2 and then combining the I 1 terms together, we end up with minus 103.35I 1 minus 78.25I 2 minus 6 equals zero. So it's this equation and this equation that we have to work with now. And the strategy I will use this point is to make the coefficient of one of the variables the same in the two equations and then subtract the equations. So in order to make this coefficient minus 25.1, I am gonna multiply both sides of this equation by 25.1 divided by 103.35 because doing that will cancel the coefficient that I 1 currently has and replace it with 25.1 and so that's useful because now it has the same coefficient as it does in equation two and then we'll subtract the equations; I like to write subtraction as adding the negative— I find it easier to think of it that way— so we are gonna add the opposite of equation two so I am adding positive 25.1I 1 and then opposite again so this becomes a minus 40.5I 2 and this becomes a plus 24 but I am getting little bit ahead of myself there though but first, we have to continue multiplying every term in equation 1b by this fraction and so we have negative 25.1 over 103.35 times 78.25 times I 2 minus 6 times that fraction and we multiply the right hand side by it as well because you know, the only reason we are allowed to do this multiplication on the left is because we are doing the same thing to the right hand side but the right hand side is zero so it doesn't matter. Okay! So we get out our calculator and work through what these results are and we end up with well, the I 1 term disappears, which is by design, that's why we did this and so we don't have to think about that term but because you know, it's a positive 25.1I 1 here and a negative 25.1I 1 here so it adds to make zero. So we have negative 19.00411I 2 minus 1.457184 minus 40.5I 2 plus 24. Okay and all that equals zero. So we combine the I 2 terms together and the constant terms together to get this line here and then subtract 22.542816 from both sides and then after doing so, divide both sides by this coefficient for I 2 divide by negative 59.50411 and then we get our answer of 0.379 amps. There! We have one of the three answers— that's awesome— and I wrote the unrounded number here because we are going to use this number in our subsequent calculations to avoid intermediate rounding error; we don't wanna do rounding until we get our final answer. So but this is the final answer for I 2, 0.379 amps with three significant figures. So let's solve for I 1. We will use equation two in its sort of original form here and solve for I 1 in terms of I 2 and so I have copied it here and then add 24 to both sides, subtract 40.5I 2 from both sides and then you have this line here and then divide both sides by negative 25.1 and you get this line here and then we can substitute for I 2 based on what we calculated before so that gets put here and we end up with negative 0.345 amps and this negative sign indicates that the current in fact is going this way for I 1. Okay! Then we have two of our currents now and the third one's gonna be easy to find because equation three is just directly adding these two currents together. So current I 3 is negative 0.344892 amps plus 0.378845 amps which is 0.034 amps. And there we go! If you made it this far, give yourself a pat on the back that was a lot to get through and congratulations!