0:00:02.540,0:00:06.190
Hello, this is Dr. Cynthia Furse,[br]from the University of Utah.

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Today, we're going to talk[br]about Kirchhoff's Laws.

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Kirchhoff's Current Law says that the sum[br]of the currents entering a node is = 0.

0:00:16.068,0:00:19.604
Or equivalently, the sum of[br]currents leaving a node is 0, or

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the sum of currents leaving is equal[br]to the sum of currents entering a node.

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So we can write the equation[br]four different ways.

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When we're talking about a node, we're[br]talking about something like this right

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here, that's in the center[br]of all four of the wires.

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Now, when we say the sum of[br]the currents entering the node,

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we can see that i1 and i4 are entering[br]the node while i2 and i3 are leaving.

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So we're going to say i1 + i4[br]those are the currents entering.

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And i2 which is leaving, we're gonna[br]put it there with the minus sign and

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i3 we're going to put with a minus sign.

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When we write the equation where we have[br]the sum of the currents leaving the node

0:00:57.453,0:01:00.876
is equal to 0, what we're doing is[br]just switching the minus signs.

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Minus and plus signs here, as you can see.

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When we talk about the sum of the currents[br]leaving is equal to the sum of

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the currents entering.

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Then you could say,[br]here's the sum of the currents entering,

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here's the sum of the currents leaving,[br]and they are equal.

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These four different ways of writing[br]the node equation are all equivalent.

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This is Kirchoff's Current Law, or KCL.

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Kirchhoff's Voltage Law or[br]KVL is the other law that we will use.

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This says that the sum of the voltage[br]around a closed path is = 0.

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The way you do this is you[br]run a loop like this, and

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you just add up all of the voltage[br]as you're going around the loop.

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So for example,[br]as we're going through here,

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we say -VDC +, see there's the + V1 + V2,

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and that's equal to 0.

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So -VDC + V1 + V2 = 0.

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Alternatively, we could say that the sum[br]of all the voltage drops is equal to

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the sum of all the voltage rises.

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So the voltage rise is +VDC and[br]the voltage drop is V1 + V2.

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So there's just a convention[br]that typically,

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we use a systematic clockwise[br]motion around the loop.

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We could go counter clockwise, but

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convention is just that[br]we normally go clockwise.

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We also assign a positive sign to[br]the voltage across an element,

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if the positive sign of that[br]voltage is encountered first.

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So that's why I was kinda circling[br]this thing as I went to your + V1.

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Now, let's talk about a simple circuit[br]where we can apply KCL and KVL.

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So here's our simple circuit,[br]let's first apply the loop equation.

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Right here's our loop and[br]let's apply the equation.

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So we can see minus,[br]see we count the minus sign first,

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-VDC + V1 + V2 = 0,[br]that's the equation right here.

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That is the KVL equation or[br]the loop equation.

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Now, let's examine the circuit,[br]what do we know?

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Well, VDC is defined that's 120 volts,

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R1 is defined that's 10 ohms,[br]R2 is defined that's 50 ohms.

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What's unknown?

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V1 and V2 are both unknown.

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So if you remember from math,[br]if you have two things that are unknown,

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then you're going to need two equations.

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But we only have one equation,[br]so what else can we do?

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Well Ohm's Law will give us two more[br]equations, so why don't we try that?

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So we can see here for example that if we[br]have a current I1 that's running through

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R1, we know that V1 = (I1) times (R1),[br]we also know that V2 is (I2) times (R2).

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Well, look, we got two more equations.

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But we also have two more unknowns,[br]I1 and I2.

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So now, we have four unknowns and[br]three equations.

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So what we always need[br]is one more equation and

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this one's going to come from a node.

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So here's our node equation right here,[br]that's going to be our KCL.

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And remember, that's the sum of the[br]currents coming in to the node is equal to

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the sum of the currents[br]going out of the node.

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So I1, which is all the currents[br]coming into that node is equal to I2,

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where the currents are going out.

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So we now have four unknowns, V1, V2,[br]I1, and I2, but we have four equations.

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So now, we can solve it.

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You're going to say, heavens.

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It looks like we could have done this[br]equation a whole lot simpler and

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absolutely, you could have.

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But what I'm showing you is the systematic[br]way of handling the KCL and the KVL.

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Okay, so now,[br]let's actually solve this equation.

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So here's our first[br]equation where we have VDC.

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Well, that's know.

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We could substitute that in there and[br]V1 and V2.

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We know R1, we know R2,[br]let's substitute those in there.

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And we know that I1 = I2 so[br]let's just call it a value I.

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So just substituting in, what I'm[br]gonna do is substitute I in there,

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and I in here and of course,[br]all of our constants.

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This equation now becomes this right here,[br]and I can solve if for I.

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I = 120 volts, let's just bring[br]in the 120 over to that side,

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divided by the sum of 10 and[br]50, so I = 2 amps.

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Then if we wanted to know V1, we would[br]just bring our 2 amps back up in here.

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Let's substitute it into the equations for[br]V1 and V2,

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and you can see that we have 20 volts[br]across 1 and 100 volt across the other.

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Let's just do a quick check,[br]let's just check our loop equation.

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We have 120 volts, so that's -120

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volts + 20 + 100 is that equal to 0?

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Yes, it is, so[br]we've been able to check our equation.

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Now, what I showed you was[br]an extremely simple circuit, and

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yes, there's an easier[br]way to solve the circuit.

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But let's use it to define a cookbook that[br]will let us do more complicated problems.

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So if I were to set a cookbook for[br]this I would say,

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the first thing I would want to do[br]is write down all known values.

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What was known?

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The 120 volts, the 10 ohms,[br]and the 50 ohms.

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Then I would keep track of all the unknown[br]values as I went throughout my

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circuit because I know that I need as[br]many equations as I have unknowns.

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Now, we would write all of the KVL or[br]loop equations.

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An important thing is that each new loop[br]must pick up at least one new element,

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like a new resistor or[br]a new voltage element.

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Also current sources,

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which we'll talk about in a minute,[br]can't be included in any loops.

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Fourth, I'd write Ohm's Law.

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In fact, I usually prefer to[br]write Ohm's Law as I go just for

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convenience because I know I'm[br]going to have to apply it anyway.

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Ohm's Law is just applying the current and

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the resistor together to get the voltage[br]any time we have a resistor.

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Then I would apply as many KCL or

0:07:04.027,0:07:07.558
node equations as are needed[br]to fill in my unknowns.

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Each KCL equation must also pick[br]up at least one new current,

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then I'd solve for all the unknowns.

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Okay, let's take this cookbook now and[br]see how we can apply it.

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So the first thing we do, and[br]let's apply it to a problem,

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that's just a little bit more[br]complicated than one we just did.

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I'm going to add a current[br]source I3 here which is 6 amps.

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So let's apply this, and as you can see,[br]we're going to have another element.

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So let's just apply each of the steps.

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First of all,[br]let's write all of the known values.

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Well, VDC is known, R1, R2 and[br]I3 are all known values.

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Now, let's keep track of our unknowns.

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I've already drawn two unknowns here,[br]I1 and I2.

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So we have two unknowns, and[br]we need to have at least two equations.

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So let's write, The loop equation's next.

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Well, this same loop equation is the one[br]that we just wrote and it hasn't

0:08:02.665,0:08:06.932
changed at all, even though we've added[br]another element over here on the right.

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And so for the loop equation,[br]it's -VDC + (I1)(R1),

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so here is -VDC + (I1)(R1) + (I2)(R2) = 0.

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This is what I mean by[br]applying Ohm's Law as I go.

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I'm just doing + (I1)(R1)[br]instead of saying + V1, and

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later having to go back in and substitute.

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Well, we could say each loop needed[br]to pick up at least one new element.

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But which elements did this loop pick up?

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It picked up the VDC, the R1 and the R2.

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Now, current sources cannot be[br]counted as new elements, and

0:08:45.243,0:08:47.631
in fact, I can't include them in my loops.

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So there's no other loop that[br]I could do in this picture,

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because any other loop would have to[br]go through the the current source.

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This is a current source right here, and[br]I cannot include that in a KVL loop.

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Now, let's apply Ohm's Law,[br]as I went, that's right there.

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And then let's apply as many node[br]equations as needed to fill in

0:09:06.797,0:09:07.658
the unknowns.

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Well how many unknowns do I have?

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Two, how many equations do I have?

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One, so I need at least one node equation.

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Here's my node right there, and[br]let's apply the node equation.

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Let's say that every current[br]that's coming in, I1 + I3.

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So here's I3, it's coming into that node.

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I1 + I3 = I2.

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Okay, that's my second equation.

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So now, I can solve for[br]all of my unknowns.

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Let's just do the solution,[br]just one time, for fun.

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So here was my VDC, I1, R1, I2,

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R2 = 0 and I1 + 6 amps = I2.

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So what I'm going to do is substitute[br]in order to remove the variables.

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What I'll do here is I'll[br]substitute in for I2,

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so this equation now[br]becomes this right here.

0:10:00.565,0:10:03.683
Let's solve for[br]the remaining variable, that's I1.

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That gives us the equation that you[br]see here, and now let's go back and

0:10:08.724,0:10:12.748
actually do the math that[br]tells us that I1 is -3 amps.

0:10:12.748,0:10:15.917
And then,[br]if I go back to my original equation,

0:10:15.917,0:10:18.466
in order to find the other variables.

0:10:18.466,0:10:24.020
Let's say I want to find I2, I2 = 3 amps.

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Now, let's talk about what[br]it means to have -3 amps and

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what it means to have +3 amps.

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So right here, when I did my analysis,[br]I said that this value I1 was a variable.

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And when I solved for it,[br]I discovered that it is -3 amps.

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What does that mean?

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Well, it means I've guessed the direction[br]of the current incorrectly.

0:10:44.720,0:10:48.660
It means that the current is actually[br]going from right to left instead of from

0:10:48.660,0:10:53.630
left to right, so that I have three[br]amps going from left to right.

0:10:53.630,0:10:56.943
Now, if I had actually put my[br]arrow this way in the first

0:10:56.943,0:11:01.855
place when I did my solution, I would have[br]found out that I1 was equal to 3 amps.

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And I would have had[br]the correct direction.

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It doesn't matter which way[br]you guess your current.

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If you guess it wrong, it's gonna[br]come up with a negative number for

0:11:09.600,0:11:11.430
the current, and that's just fine.

0:11:11.430,0:11:16.761
So go ahead and leave it like this, and[br]just tell me that you have -3 amps for I1.

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Okay, let's go one step further.

0:11:22.310,0:11:23.873
Let's add one more loop.

0:11:23.873,0:11:29.124
So let's apply our cookbook[br]to this system right here.

0:11:29.124,0:11:32.064
First of all,[br]we need to write all known values.

0:11:32.064,0:11:33.460
V1 will be known.

0:11:33.460,0:11:34.940
I1 will be known.

0:11:34.940,0:11:37.836
R1, R2, and R3.

0:11:37.836,0:11:40.963
Then we're going to keep track[br]of all of our unknown values.

0:11:40.963,0:11:47.870
I already have an I1, an I2, and an I3,[br]that I can see are going to be unknown.

0:11:47.870,0:11:49.900
Then we're going to write[br]all the loop equations.

0:11:49.900,0:11:51.820
Let's see how many loops we could do.

0:11:51.820,0:11:54.170
Well, here's our first loop.

0:11:54.170,0:11:56.280
That's a loop we've been doing all along.

0:11:56.280,0:11:57.900
Here is a second loop.

0:11:59.260,0:12:00.920
Is there room for another loop?

0:12:00.920,0:12:03.190
Can I do a third loop here?

0:12:03.190,0:12:07.480
No, I can't, because I have this[br]current source right there.

0:12:07.480,0:12:09.888
Can I do a third loop around this way?

0:12:11.077,0:12:14.685
No, I can't,[br]because that's including a current loop.

0:12:14.685,0:12:17.029
Could I do a third loop here?

0:12:18.835,0:12:21.300
Well, that's a legitimate loop.

0:12:21.300,0:12:25.325
But I can't do that, because it[br]doesn't pick up any new elements.

0:12:25.325,0:12:28.046
My first loop picked up R1 and R2.

0:12:28.046,0:12:30.294
My second loop picked up R3.

0:12:30.294,0:12:32.418
There isn't anything else to be picked up.

0:12:32.418,0:12:37.655
So I am only going to have the two[br]blue loops that are shown here.

0:12:37.655,0:12:39.875
Then I would apply as many KCL or

0:12:39.875,0:12:42.745
node equations as are needed[br]to fill in the unknowns.

0:12:42.745,0:12:47.891
So I'm going to have two loop equations,[br]the blue lines that are shown.

0:12:47.891,0:12:52.478
And I'm going to have one node equation[br]here, where things are orange, and

0:12:52.478,0:12:54.950
that's going to fill in my unknowns.

0:12:54.950,0:12:57.607
And then we would solve for the unknowns.

0:12:57.607,0:13:00.786
Okay, now,[br]let's take a look at the special node.

0:13:00.786,0:13:04.687
So the node remember was right there,[br]but there was a line or

0:13:04.687,0:13:07.090
a short circuit right here.

0:13:07.090,0:13:12.475
Any time you have a short circuit, it just[br]means that you have a really big node.

0:13:12.475,0:13:15.060
You don't need to include[br]two separate nodes.

0:13:15.060,0:13:17.540
This is just one really big node.

0:13:19.610,0:13:23.992
Okay, now, let's imagine even[br]adding another node, sorry,

0:13:23.992,0:13:25.868
even adding another loop.

0:13:25.868,0:13:28.746
So let's talk about how[br]we might do this problem.

0:13:28.746,0:13:33.425
Okay, so[br]write down all the known values, V1,

0:13:33.425,0:13:36.630
R1, R2, R3, R4, and I1.

0:13:36.630,0:13:37.798
Those are all known.

0:13:37.798,0:13:39.455
Keep track of the unknowns.

0:13:39.455,0:13:46.669
They would be I1, I2, and[br]I3, and let's call this IDC.

0:13:46.669,0:13:48.810
I know it says I1, let's call it IDC.

0:13:48.810,0:13:50.473
All right, so those are the unknowns.

0:13:50.473,0:13:56.111
So we know that we're going to need,[br]and there's also going to be an I4.

0:13:56.111,0:13:59.410
So we're going to have four equations and[br]four unknowns.

0:14:01.300,0:14:03.790
Now, let's see how many loops we can do.

0:14:03.790,0:14:06.419
First of all,[br]we've got our first loop right here.

0:14:06.419,0:14:09.808
That pick up V1, R1, and R2.

0:14:09.808,0:14:12.030
That's one loop equation.

0:14:12.030,0:14:13.950
Here's a second loop equation.

0:14:13.950,0:14:16.168
Picking up R1 and R3.

0:14:16.168,0:14:19.030
There's at least one new element, the R3.

0:14:19.030,0:14:21.400
Can we do a loop here for I4?

0:14:21.400,0:14:25.353
No, we can't, because that's going[br]to include the current source, so

0:14:25.353,0:14:26.835
we cannot do that as a loop.

0:14:26.835,0:14:33.896
All right, so we have two loop equations,[br]here's one node equation.

0:14:33.896,0:14:38.795
And in fact, this node is actually going[br]to encompass all of this area where you

0:14:38.795,0:14:41.182
can see it short circuited together.

0:14:41.182,0:14:43.663
I need at least one more equation and

0:14:43.663,0:14:47.610
somehow I need to pick[br]up R4 in that equation.

0:14:47.610,0:14:48.885
So let's see what we would do.

0:14:50.791,0:14:53.813
Okay, here is a very good choice for[br]handling that.

0:14:53.813,0:15:00.334
We would do a loop all the way around[br]the outside like this picking up R4.

0:15:00.334,0:15:04.987
So that would give me[br]three loop equations,

0:15:04.987,0:15:09.779
and one node equation for[br]the four unknowns.

0:15:09.779,0:15:12.489
So my problem would end[br]up looking like this.

0:15:12.489,0:15:19.845
3 loops, 1 node would give me 4 equations,[br]sorry, 4 unknowns.

0:15:19.845,0:15:28.438
3 loop equations, 1 node equation[br]would be 4 unknowns and 4 equations.

0:15:28.438,0:15:34.000
Okay, now let's take a look and just kind[br]of expand our understanding of the nodes.

0:15:34.000,0:15:37.930
Could I have used this as[br]a node equation instead?

0:15:37.930,0:15:41.415
Well, the answer is, no, I can't,[br]because I don't know how much

0:15:41.415,0:15:44.365
current is going to be coming[br]out of this voltage source.

0:15:44.365,0:15:48.925
So just like the fact that I cannot[br]include loops when I have a current

0:15:48.925,0:15:53.965
source, I cannot include nodes that[br]are attaching voltage sources.

0:15:53.965,0:15:56.775
So I'm gonna add that information[br]to my cookbook as well.

0:15:59.099,0:16:00.584
Okay, so let's do this problem.

0:16:00.584,0:16:03.270
Maybe it looks a little complicated,[br]but it really isn't.

0:16:03.270,0:16:05.491
So let's do each of our loops.

0:16:05.491,0:16:11.077
My first loop is the VDC[br]loop right here which

0:16:11.077,0:16:16.967
is going to say minus,[br]this is the minus plus,

0:16:16.967,0:16:21.051
-VDC + I1R1 + I2R2 = 0.

0:16:21.051,0:16:23.577
This is my loop 1.

0:16:23.577,0:16:26.274
Now let's do loop 2.

0:16:26.274,0:16:32.697
And remember, I've got the plus minus[br]going from my current like this.

0:16:32.697,0:16:38.447
So +I3R3, this is I3 + I3R3,

0:16:38.447,0:16:42.525
you hear that's all short

0:16:42.525,0:16:47.357
circuited minus I1R1 = 0.

0:16:47.357,0:16:50.931
Again, let's call this IDC, okay?

0:16:50.931,0:16:55.816
And let's do this outside[br]loop right there.

0:16:55.816,0:17:01.187
So I'm gonna say -VDC

0:17:01.187,0:17:07.416
+ I3R3 + I4R4 = 0.

0:17:07.416,0:17:10.440
Now, let's do the node.

0:17:10.440,0:17:15.381
So right here is the big node[br]that I'm gonna be interested in.

0:17:15.381,0:17:17.582
So the node equation is going to say,

0:17:17.582,0:17:22.065
all of the currents that come in has to[br]equal all of the currents that go out.

0:17:22.065,0:17:24.079
Or all the currents[br]coming in are positive,

0:17:24.079,0:17:27.027
all the currents going out[br]are negative and that's equal to 0.

0:17:27.027,0:17:31.597
So I3 is coming in, that's positive.

0:17:31.597,0:17:34.567
I4 is going out, that's negative.

0:17:34.567,0:17:37.247
I1 is coming in, that's positive.

0:17:37.247,0:17:39.927
I2 is going out, that's negative.

0:17:39.927,0:17:46.969
I also have to include plus IDC,[br]and that whole thing is equal to 0.

0:17:46.969,0:17:50.217
Okay, now let's keep track[br]of how many unknowns I have.

0:17:50.217,0:17:51.367
What are my unknowns?

0:17:52.650,0:17:59.000
My unknowns are I1, I2, I3, and I4.

0:17:59.000,0:18:00.930
How many equations do I have?

0:18:00.930,0:18:05.726
I have th3 ee loop equations and[br]1 node equation.

0:18:05.726,0:18:08.449
So I have four equations and[br]four unknowns.

0:18:11.439,0:18:14.536
Okay let's talk about the math for[br]this for just a minute.

0:18:14.536,0:18:16.463
We can solve this using algebra, and

0:18:16.463,0:18:19.086
I know that you've had[br]some practice doing that.

0:18:19.086,0:18:23.984
But I want to show you one specific[br]method of being able to solve the KCL and

0:18:23.984,0:18:24.720
KVL math.

0:18:24.720,0:18:27.892
It's gonna be very handy[br]when we apply it in MATLAB.

0:18:27.892,0:18:30.722
Any time we have four equations like this,[br]and

0:18:30.722,0:18:35.312
four unknowns, we can write them as[br]a matrix, as I'm gonna do right here.

0:18:39.892,0:18:42.043
So let's take our first equation.

0:18:42.043,0:18:47.938
Here's our first equation right there.

0:18:47.938,0:18:53.385
So the unknowns are gonna be I1,[br]I2, I3, and I4.

0:18:53.385,0:18:57.438
Anything that does not have an unknown[br]moves over to the right-hand side.

0:18:57.438,0:19:00.562
So we have +VDC on the right-hand side.

0:19:00.562,0:19:06.182
Anything that does have an unknown,[br]I've put the multiplier, like R1 times I1.

0:19:06.182,0:19:12.138
See here's R1 times I1 and R2 times I2.

0:19:12.138,0:19:14.191
That's right here, R2 times I2.

0:19:14.191,0:19:17.747
Okay, and then there are no I3's and[br]there are no I4's in that equation.

0:19:17.747,0:19:20.741
So they are 0 and 0.

0:19:20.741,0:19:22.689
Let's go to our next equation.

0:19:22.689,0:19:27.895
Our next equation right here says, there[br]is nothing that doesn't have an unknown.

0:19:27.895,0:19:33.001
So I have 0 over here[br]on the right hand side,

0:19:33.001,0:19:36.324
then I have -R1 times I1.

0:19:36.324,0:19:39.314
I have nothing times R2 and

0:19:39.314,0:19:43.995
I have R3 times I3 and nothing times R4.

0:19:43.995,0:19:46.324
Okay, let's go down to the third equation.

0:19:46.324,0:19:50.501
The third equation has VDC,[br]which has no unknowns.

0:19:50.501,0:19:53.619
So let's bring it over thee there's +VDC.

0:19:53.619,0:19:55.970
Do I have anything times I1?

0:19:55.970,0:19:57.891
No, I2?

0:19:57.891,0:19:59.944
No, I3?

0:19:59.944,0:20:01.349
Yes, R3?

0:20:01.349,0:20:02.437
Yes, I4?

0:20:02.437,0:20:04.589
Yes, R4?

0:20:04.589,0:20:08.056
So there's the matrix, and[br]then this very last one,

0:20:08.056,0:20:12.270
what is over here that doesn't[br]have a unknown in front of it?

0:20:12.270,0:20:17.088
Take that over, that's gonna become -IDC,

0:20:17.088,0:20:24.013
what do we have times I1,[br]a +1, I2-1, I3 + 1, I4-1.

0:20:24.013,0:20:25.780
So here's my matrix.

0:20:27.660,0:20:29.180
Here's my vector of unknowns.

0:20:29.180,0:20:30.920
Here's my vector of constants.

0:20:30.920,0:20:34.866
Then so[br]this is writing the matrix equation for

0:20:34.866,0:20:38.300
this particular circuit element.

0:20:38.300,0:20:42.258
Next, you can use Gaussian elimination or[br]MATLAB matrix solution.

0:20:42.258,0:20:43.706
So I'm going to leave this here.

0:20:43.706,0:20:46.082
We're not actually going to[br]do the algebra to finish it.

0:20:46.082,0:20:49.527
Because in general, I'm going to assume,[br]you'd use MATLAB to solve this problem.

0:20:52.092,0:20:55.300
So in short,[br]we've talked about two important laws.

0:20:55.300,0:20:59.260
One is Kirchhoff's Current Law that[br]says the sum of all the currents

0:20:59.260,0:21:00.730
entering the node = 0.

0:21:00.730,0:21:04.411
The second law is[br]the Kirchhoff's Voltage Law or KVL,

0:21:04.411,0:21:09.144
that says the sum of the voltage is[br]around a closed path or loop is = 0.

0:21:11.310,0:21:15.330
Finally, we wrote down a cookbook for[br]the KCl and kvl.

0:21:15.330,0:21:16.371
The cookbook is this.

0:21:16.371,0:21:18.840
First, write down all the known values.

0:21:18.840,0:21:21.458
Keep track of how many[br]unknown values you have,

0:21:21.458,0:21:24.150
write all the loop equations[br]that you can write.

0:21:24.150,0:21:28.616
Remember that each loop needs to[br]pick up at least one new element and

0:21:28.616,0:21:31.419
that current sources can't be in loops.

0:21:31.419,0:21:35.714
Then apply Ohm's law which I usually[br]choose to do as I go for convenience.

0:21:35.714,0:21:40.100
And that is equal to the voltage is equal[br]to the current times the resistance.

0:21:40.100,0:21:44.790
And then apply as many nodes as you[br]need in order to fill in the unknowns.

0:21:44.790,0:21:48.906
Each KCL equation must pick[br]up at least one new current.

0:21:48.906,0:21:54.079
Shorts as they are here can be used[br]to combine nodes into a single node,

0:21:54.079,0:21:57.161
and nodes can't touch voltage sources.

0:21:57.161,0:22:02.562
Then you solve for[br]the unknowns using matrix math.

0:22:02.562,0:22:05.914
So the picture that's on[br]the front that's in Red Canyon.

0:22:05.914,0:22:09.966
That's near the bottom of[br]Thunder Mountain, near Bryce Canyon, Utah.