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In this exercise, you have two different circuits, and you're supposed to use kirchhoff's rules to find the currents i -1, i -2, and i -3.
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And they've already specified for you, i -1, i -2, and i -3.
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Now, and they've given you all the resistance values and any voltage sources and so on.
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I've labeled the edges, you know, so i can do my loops and you can see what i'm doing.
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Now, you're free to choose your currents any way you want.
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I've chosen i want to be moving from b to c, i2 from d to a, i3 from f to e, you know, going around.
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I could choose them any, i could, does not matter.
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If you might say, well, what happens if i1 isn't going from b to c, but it's really going the other way? c to b.
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What you will find is that i won when you do your analysis will be negative.
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They'll tell you that your assumption was wrong.
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It was not b to c was c to b.
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But the value is exactly the same.
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The value of the absolute value, if you like, of the current is going to be exactly the same.
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The sign tells you the direction.
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If it's plus, you chose correctly.
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If it's minus, you chose the opposite direction.
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Just flip your direction.
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That's it.
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That's all there is to it.
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But you're free to choose anything you want.
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Likewise, when you do loop rule, you're free to go around counterclockwise or clockwise.
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It does not matter.
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Do whatever you want to do.
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Just have everything laid out, and you've got to be consistent as you go around the loop with that choice.
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That's all.
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Just be consistent.
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All right.
01:59
Now, kerchhoff's has loop rule, it also has a junction rule.
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You could do this problem with three loops, the top loop, the bound loop, and the outer loop.
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Three equations three are no.
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Or you could do it with two loops, any two of the three loops i just said, and the junction say at a or d.
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Those are the junction points.
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A and d are the junctions where you have, you basically have three lines coming together.
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Nowhere else, these are just corners.
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There's nothing special.
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A and d are the junctions.
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So, you know, what, let's do junction a.
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Now, what is this? this is just a statement that what comes in has to go out, has to be equal.
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So you're right, whatever is coming in, which is i2, is equal to whatever is going out, which is i1 plus i3.
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It's a steady state situation.
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So you can't have a disparity.
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How could that be? if more comes in that's going out, then that's not a steady state situation.
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Now, is it? so that's equation one.
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Now you might say, what happens if it's that two is going out and three is the one coming in? well, when we do our analysis, those minus signs will flip this equation around, and we'll show you that.
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So don't worry about it.
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Just be consistent, solve your equations, see what you get.
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Now i'm going to do loop a, d, c, b, c, b, c, a.
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And that means i'm going to do it counterclockwise.
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C -c -c -w counter -clockwise.
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I could do it clockwise c -w.
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But in the next example, the next problem, i will do it clockwise.
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Okay.
03:59
Now, a couple of things.
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When, what does really the loop rule represent? think of a bank account.
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A is the beginning of the day.
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You got a certain amount in there.
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When you come back to that, that means you have, so it's beginning of the day, you get a certain amount, at the end of the day, you're supposed to have the same amount.
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You can do whatever you want during the day, but you've got to have the same.
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When you open that bank, you know, whatever that value is, and that bank closes, it's got to be the same.
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That's the rule.
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Somebody put that rule on you.
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You can do whatever you want during the day.
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It's got to be the same at the end of the day.
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So getting back to a is coming back at the end of the day.
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So that means as you traverse one of these loops, as you rise or fall crossing a resistor or rise or fall with a battery, all those things in the end, the rises and falls, you have to add to zero.
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All the deposits and withdrawals must add to zero because you're back to where you started from.
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So you took out 10, you had to put in 10.
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So when you go around a loop, all those rises and falls got to equal zero.
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Have to.
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Because you're back where you started from.
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So, let me, so you can see it.
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So we're starting at a going around counterclockwise, a to d.
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And this is a, that is a d there.
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Out of b, let me fix that.
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Okay.
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Now, when you go with the current, you're dropping in voltage.
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When you go against the current, you're rising.
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Current always flows to lower potential.
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It's positive charge, lower potential.
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So, starting a day, we're going against the current.
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So that's plus, and i'm going to drop all units, plus 5i2.
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Then we reach the battery.
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It's called a battery.
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Whatever it is, it's just a voltage source.
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It's a battery.
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And this point is 12 volts higher than that, so plus 12.
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Then we're going against the current again.
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So plus 2.
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I1 is equal to zero.
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This is equation two.
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Can't do anything yet.
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Two equations, three unknowns.
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Got to have three equations.
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All right.
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Now we're going to do loop, and let me move this up a little bit.
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Loop a, f, e, d, a.
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Okay.
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So we're going to start at a again, going around kind of clockwise again.
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Could go clockwise if i wanted to.
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The next problem i'll show you clockwise.
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That's not matter.
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All right, so we come down.
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We're now going with the current minus 10 i3, minus 9.
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We're dropping.
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This point is 9 volts lower than this.
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Minus 9 going with the current minus 5 i2 equals 0.
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And this is equation 3.
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So we have three equations and three unknowns.
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That's what we need it.
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Now it's the math problem.
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Now it's a math problem.
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Now, on the web, you might be able to find places that will solve these, calculators, computer programs, whatever.
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You know, having access to those is fine.
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You can check yourself, but you still want to get the practice of doing it.
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So let's do some work here.
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Notice in two and three, there's a five plus five i2 minus five or two.
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So we add those, we get rid of the i2.
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So let's do that.
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So here is just a matter of being careful.
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But one thing i should say, if you get in a situation where you have fractions, do not calculate the decimal numbers.
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You're going to be carrying around too much garbage, and it's going to lead to mistakes.
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Just waste your time.
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Keep, if you've got to divide, you know, two divided by five, leave two divided by five.
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If it's a whole number, like you're going to see me do on this first problem, it's a whole number, that's okay.
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But if it's going to have decimal number of digits and you got to round, do not do it.
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Do not do it.
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And even if you don't have to round, if you have to, if it's got many digits, you're going to turn 1 .675, sooner or later it's going to become 1 .657 without question.
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Sooner later, you do these problems.
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They're big enough.
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You're going to, you're going to, something's going to get mixed up.
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So try not to do that.
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If you have to just keep everything as fractions.
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Okay, so adding two and three, five i2 plus 12, plus 2 i1, i's 10, i3, minus 9, minus 5 i2 is equal to 0.
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So what do we get, a 12 minus 9, we get a 3 plus 2 i1 minus 10i3, the 5i2s go away, and it's equal to 0.
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Gives me the 2 i1 is equal to 2 i1 is equal to 10 i3 minus 3 or i 1 is equal to 5 i 3 now this is 3 divided by 2 is 1 .5 so we can live with that we can live with that that's okay if you if you're not counted at all 3 divided by 2 whatever works for you that's equation 4 now place 4 into 1.
10:51
So i2 is equal to 5, i3 minus 1 .5 plus i3 is equal to 6i3 minus 1 .5.
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And that i'll call equation 5.
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Now, place 5 into 3.
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Myself some boom.
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So minus 10.
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I3 minus 9 minus 5 times 6 i3 minus 1 .5 is equal to 0 minus 10 i 3 minus 9 minus 30 i 3 plus 7 .5 is equal to 0 so that gives me minus 40 i3 is equal to 1 .5 you got the minus here you're going to we'd have a minus 1 .5 here but it's going to go on the side so it's plus i saw some room here applies i3 is equal to minus 1 .5 over 40 which is minus now this is exact this is exact minus 0 .0374.
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And i'll put in the units now amps.
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So the minus sign tells me that i3 is in the opposite direction.
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So instead of from f to e, in that bottom, it is moving from e to f.
13:02
So i chose wrong...