How much current flows through the bottom wire in FIGURE P28.6...

Question

How much current flows through the bottom wire in FIGURE P28.66, and in which direction?

Accepted Answer

Welcome back, everyone in this problem, we want to find the amount of current that flows in the two ohm resistor of the circuit as shown in the diagram below. For our answer choices A is two amps. B is one AMP C is 0.9 S and D is 0.8 amps. Now, we have our circuit and you notice that it has two resistors connected in parallel on either side. So we may have to use Kirk's laws to help us to figure out the currents in each of these resistors and at different points in the circuit, once we can find how those currents are flowing, then we can create probably a system of equations that will help us to find the current that's flowing through the two ohm resistor. So let's go ahead and do that first. Let's label our diagram. So we can first apply Kerk off's junction rule. OK. And recall that what this rule is telling us is that all the current that is flowing through the circuit is equal, it's the same. So in other words, if I were to label my junctions in the circuit as ABC and D then Kirk's junction rule is telling us that the current flowing from A to B is the same. So we could say that starting from our voltage source, 12 volts, remember we're assuming that the circuit is ideal, meaning there's no internal resistance in the batteries or other sources of energy loss, that the resistors have no temperature dependence. That is that they are omic and that the batteries are ideal voltage sources with constant electromotive forces. So assuming that then by Krakow's junction rule, we know that we'll have a current from our battery source, our 12 volt battery source. That's I one. When it comes to junction A, it splits into two currents. Let's call those currents I two and I three, no I two and I three come back together at junction B. So we know that that must be I one. OK? Because I two plus I three would have been I one, it flows to the next resistor at junction C. So that current I one or same current I one is now split into two currents, one current across the three ohm resistor. Let's call, let's give that current a name. Let's call that current I five and one across the two ohm resistor. The current that we're really interested in, let's call that current I six. They combine back into I one and go to our 12 vote source. So what does that tell us then? Well, by Kerkow junction rule, this is telling us that I, one equals I two plus I three. And it's also telling us that I one equals I five plus I six. Now, if I were to rewrite this equation to get all the variables on the same side of our equation, then we could write it as I one minus I two minus I three equals zero and I one minus I five minus is six also equals zero. So that's what Kirk's junction rule tells us. Recall that Ko's rules also apply to our loops. And by Kirkoff loop rule, what that rule tells us is that the sum of electromotive forces that's moving through a circuit equals zero. So if we were to find the sum for the loops that is, that are in the circuit, then we would be applying kerk off's loop rule. So first, let's start with our outer loop. OK. Let's start with our outer loop here. So let me highlight that. OK. So starting from our 12 volt source, we focus on the resistors that are on the outside. Now, if we're looking at our loop, I remember now we're talking about voltages, then starting from our 12 volt source, we'll have 12 volts. OK, minus or voltage at the three ohm resistor which is three multiplied by I two minus or a six volt battery source minus two multiplied by I six. And all of that will be equal to zero. We can rewrite this, we can simplify this even further because notice that we have two voltage sources here, two constants as it were and 12 minus six equals six. So I can rewrite this as six minus. Let me put that in black just to show that we're doing some extra work here. So we'll have six minus three I two minus two I six equals zero. And we can carry our constant to the next side and multiply our equation by negative one. So that will leave us with three I two plus two I six equals a positive six. Now let's focus on our inner loop. So let's take that one out and no, let's do our inner loop. So we're starting again at a 12 volt source. But now we're focusing on our forearm resistor and our three own resistor respectively. Now focusing on that loop. What do we have? Well, we start at our 12 volt source. We go to our 44 ohm resistor source. So that's minus four I three. We get out, go to our six volt battery source minus six volts and then we go to our three ohm resistor. So that's minus three I five and that also equals zero to make it simpler. Again, we have our 12 volt and our negative six volts. So we can simplify that to say six minus four I three minus three multiplied by I five equals zero. And we can bring the, the, the variables to the other side of the equation. So that will leave us with four I three plus three I five equals six. There's one more source that we can or one more loop rather that we can represent. And that's the loop between our three and four arm resistors. OK. So if I oops, I shouldn't have done that. If I were to focus on that loop between our three ohm and four ohm resistors, remember their sum must be equal to zero. So by applying Ko's rule, notice that we have negative three I two, right plus four, multiplied by I three. And that would be equal to zero. No, we could figure out more loops if we wanted to. But guess what we're trying to solve for I six, the current through the two ohm resistor. And by my count, we already have, let's see, we have 123 or five equations. So we have five equations, six different currents. So it's six unknowns and we want to solve for that sixth unknown I six. So we have enough equations to help us to solve for that unknown. So we can set up our Gaussian elimination equation. That's that our, our Gaussian elimination matrix because that's the easiest way for us to be able to solve for that sixth current. It's, it's best used when we're dealing with an equation that has tons of variables. So let's go ahead, set up our equations in row echelon form and see if we can simplify to solve for I six. So let me come down here. OK. And using Gaussian elimination, OK. Very handy tool. Using Gaussian elimination. Let's set up our equations. So what do we, what do we already know? Well, let's make a note of it. So here, let's let me call, we have I 1 to 61 I, two, I, three I four I five and I six. OK. And we're going to have an equal sign here. OK? And then we're going to have our constant on this side. So what do we, what do we know? Well, for our first equation, we said that I one minus I two minus I three equals zero. So I one minus I two because we are writing our coefficients minus I three equals zero. Let me fill the rest with our zeros here. For our second equation I one minus I five minus I six equals zero. So I one is one I five is negative one I six is negative one and that will be zero. Next we, oh let me put in the zeros here. Next we're going to have uh three I two plus two I six. So three I two plus two I six equals six filling our zeros for our fourth equation. We'll have four I three plus three I five equals six. So four I three, right? Oops four I three. Yeah, four I three plus three I five equals six as well. Just fill in our zeros. And then for our last equation, so we have our first row, our second row, our third row and our fourth row. For our fifth rule, it will be zeros the empty, empty row. And then for our sixth row, we know that negative three I two plus four I three equals zero. So we can fill in our zeros as well. No, guess what guys. Let me draw my line a bit longer here. Now we have our matrix. OK. And this is our equation in raw echelon form. Now, I'm going to be working mainly with this equation, this matrix because we don't have much space to work with. So let me just copy this matrix here. OK? And I'm going to make it smaller because we're going to be doing a lot of work. So it would be good for us to remember what our original matrix look like. And we're also going to keep note of the equations that we're forming from our, from our calculations. OK. The changes that we're making to the roles rather. So now that we have all of our values, remember that the purpose of Gaussian illumination is to help us to solve in one variable and to do that, we can try to get all of the values below the leading diagonal. OK. So our leading diagonal would be what I'm highlighting here in yellow. We're trying to get all of the values below our leading diagonal to be zero. And that makes sense because if all of those are zero, then we can solve for our unknown value that we want Isaac. So let's try and see if we can figure out how we're going to do that. Oops, I should have included this one as well. So let's see how we can get all of those values to be equal to zero. So let's start with our second rule. Let me keep that here. So we want to make our second rule, we want to make one be equal to zero. How could we do that? Well, we could make, so this is the first one. We could make our first row negative and add it to our second row and then make that the new row too. So what we're saying here is that we could multiply four I one, it would be negative one plus one and the negative one plus one equals zero. So you see, you see why we did that because no, we can make that coefficient be equal to zero. Likewise, we have to do the same for the rest. So it's going to be negative negative one plus zero, which is going to just be one. OK? Then we have negative one multiply sorry, negative negative one, which is positive one plus zero. Again, that's going to be a positive one. And then we have negative zero plus zero. That's still zero. But I'll, I'll rewrite it in red just to show that we changed it because it's good to keep track of that. Then we have negative zero plus negative one. That's still negative one. And then for I six, it's, it's the same thing. Negative zero plus negative one, it's still negative one. And then we'd still have zero. So now we've made the entry in row two or first entry zero. So we still have some more work because we need to make the rest of those variables or the rest of those coefficients rather be equal to zero. So since you got the hang of it for the first one, I'm going to go through each and I'm going to write it ahead of time and then we'll talk about it a little bit. So now let's move on to our next row. We want to make the three in row three be equal to zero. So how can we do that? Well, we could multiply row two by negative three and add it to row three. Let me put that in black just to show that's our work that we're doing. So that's negative three multiplied by row two plus row three. And that is going to be the new row three. So what does that look like when I'm finished? Well, no, notice that zero multiplied by negative three plus zero equals zero. Let's focus on our coefficients that have changed though. So notice that negative three multiplied by one is negative three and negative three plus zero would give us negative three, negative, negative three multiplied by negative one would have been three and three plus zero equals three. Likewise negative three times negative one would be positive three and three plus two equals five. Negative three times zero is zero and zero plus six equals six. What next? Well, no, we want to get row four. When we go to row four, we want to make the four be equal to zero. Ok. So how can we do that? Well, if we look at the previous row row three, notice that if we multiply, let me put that in red, if we multiplied row three by four thirds. Ok. Multiplied row three by four thirds and added it to row four, we could make it the new row four because that will make our value of four be equal to zero. So let's go ahead and do that. How about you give it a try? Is this what you got? Because notice that four thirds of negative three equals negative four and the negative four plus four equals zero. Likewise, we got seven here because four thirds of three is four and four plus three equals seven. And then we got 23rd here because four thirds of five is 23rd and then four thirds of six is eight and eight plus six equals 14. Now, what else do we want to make zero? Let's see. Hm. How about making the three in rows in the negative three in row six being equal to zero? That makes sense. So let's do that. We could multiply row two by three and then add it to row six. So we're gonna multiply row two by three, right? And then we're going to add it to row six and that will be the new row six. No notice that we have a seven in row six at I three, a negative three at I five and a negative three at I six. So we were able to get rid of our negative three, but now we have some more unknowns. So that means we need to get rid of seven. How can we, how can we do that? Well, again, what if we use our third row? Because let's put that in red here. If we multiply our third row by seven thirds and add it to the sixth row, we can make that our new sixth row. Ok. So let's go ahead and do that. No, this would be our result. We're all most there. No, we just need to try and find I sixth from the sixth row. Let's think which other row. Notice that the sixth row has four zeros, which other row has four zeros. Well, row four has four zeros and in row four, there are coefficients at I five and I six. So if we can get rid of I five, then we should be able to find the coefficient of I six and have a zero in row six for I five. So let's think about it. How can we, how can we get rid of I five there? Well, we're talking about seven and four. So here's what we could do if we could multiply rule four by negative 4/7. Ok. And then add it to row six. Guess what? We can make that our new row six. So let's go ahead and do that. No, we've got it. No notice that we have only a six remaining in row six and we have our constant. No, that means we can go ahead and write our equation. So our Gaussian elimination told us that because remember we're solving for Isaac, the current that's flowing through the two or resist. It tells us that 34 7th of is six equals six votes. Therefore, that means I six equals six, multiplied by seven, divided by 34. OK? And when we put that into our calculators, you'll find that we get approximately 1 a.m. pair. So that means the current flowing through the two ohm Resistor is one umpire. When we go back to our answer choices that would have been answer choice. B great job today guys. You really did. Well, thanks a lot for watching. I hope this video helps.

How much current flows through the bottom wire in FIGURE P28.66, and in which direction?

Key Concepts

Ohm's Law

Series and Parallel Circuits

Kirchhoff's Laws

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