[MUSIC] Welcome back to linear circuit. This is Dr. Ferry. This lesson is on Kirchhoff's Current Law. It's a very fundamental concept that will help us tremendously when we're doing analysis net, it's later on. Now, the Kirchhoff's Current Law builds upon two prior concepts. One is Ohm's Law, V=iR and the other R is the idea of Nodes. This particular circuit has four nodes. Remember, a node is a collection of wires that interconnect different circuit elements like here. This is Kirchhoff's Current Law. I'm going to call it KCL for short, just because it's easier to say. It's the sum of the currents leaving a node is the sum of the currents entering the node and a shorthand notation is right here. Now, let's illustrate that. So here, we've got a node. And we've got two currents leaving it, i1 and i3 and one current entering, i2. So the equation is i1+ i3, those leaving is equal to the one entrance since we've only got one. Now, there are different ways of representing the KCL and we can go between them by reversing the direction here on an arrow. For example, right here, if I wanted to reverse the direction of this current of this arrow right here, I would have to indicate the current. So, the equivalent way of writing this particular node expression would be like this. I reversed the direction of the arrow, I've negated the current. Since I haven't changed the circuit actually, I have the same equation. So, I take the original equation and I represent it that way. Now, the different way of representing the KCL is to just call it something else. Instead of -i2, I'm going to call it i4. So, another way of representing the KCL is right here. The sum of the currents leaving a node is equal to zero. Because in the previous case, I had an i2 that was entering. So, I just reverse the direction of i2. Now call it i4, which is actually equal to -i2 and I get this expression. The sum of the currents leaving is equal to zero. Similarly, if I play this same sort of game and I reverse directions on all of my arrows and negate all of my currents. I get another form of the KCL, which is the sum of the currents entering a node is equal to 0. So, it's shown right here. So anytime, I can use any of these three versions of it and it depends on how my circuit looks to determine which one I want to use. Now in this particular case, we're looking at it applied to an actual circuit. So this right here, I've got a node and I've got i1 entering, i1 has to be leaving. This is a very simple node, only one branch enters and one leaves. Now in this particular case, we've got these elements are in series. So the current right here, that's entering the component has to be the same one that's leading the component and the same with right here. So all of these elements are actually in series with one another, because they share the same current. So, the current is the same through series elements. Now if I want to go ahead and apply the KCL at say, this particular node right there, then I would define two more currents. Those that are leaving that node right there. And then in that particular case, I'd have i2 plus i3 is equal to i1. Now, we're going to apply the KCL to this particular example and then I'm going to have you do a quiz. Now in this particular case, we have several nodes and I've color coded them. So we've got the simple node right here and up here, we've got a big node that connects all of these components right here and then we've got a node right there and then up in the corner another node. So, suppose I give some values right here. In this particular case, the node is actually a lot larger, but a KCL also holds for a sub note and this sub note is something that just connects these wires together. So this particular case, I've got two amps leaving here. Two amps leaving here and the sum of them have to go to zero. So I'm using this version of the KCL, the sum of the current leaving the nodes has to equal zero. Now, I want for you to do a quiz on your own. I want you to solve for first of all, this current right there and then this current right there. And to do that, you might use the other versions of the KCL. So, we want to solve for this current right here and I see that I have a current here of two amps. That's going to flow through here and enter the node right there and I've got two amps coming here and there. So we've got two amps and two amps both entering, and that means the sum of the currents entering the node has to be zero. So this actually has to be a -4, so that it all sums to zero. So once I know what that one is I've got a -4 here and this particular one, I've got actually this big node there and I'm going to kind of circle it right here. That's the big node. So if I'm looking at the currents, I've got this current, that current, that current and that current right there. So if I sum those up, I'm going to have. They're all leaving. So I've got a 2 that's leaving, I've got a -5. I've got this one, which I don't know. I've got the -4. So, this one since it's all going to have to go to 0, that's going to be a plus 7=0. So, this has gotta be seven in order for that to all sum to zero. Now, looking at this particular example. I'm going to be combining a KVL and a KCL, because I want to utilize them both to be able to do an entire circuit problem. This particular circuit, I want to solve for my currents. I want i2 and i3. I can use a KCL and I'm going to write that up here as a KCL at this upper ode right here. So at that upper node, I've got one current that's entering the node. And that's 2 and that's got to equal all of the sum of the currents leaving, which is i1+i2+i3. Now on the left side, I can do a KVL. So, if I'm going to be going around this way. That's going to be a plus and minus, because of the direction of the current and that current's also going to be flowing through this component. So now if I write that, I'm going to be doing 50*i3+509*i3 and then this is going to be a minus here, because the current's coming from this direction. So, it's -100i2=0. Now if I do a KCL in the middle, well, these two components are actually in parallel with one another. And so that particular case, KVL middle. In that particular case, I've got 100i2=50i1. So what I now have is three equations and three end nodes, and I can solve for the currents in those particular cases from those equations. And in that case, if I solve them, I would have i1=1, i2=0.5 and i3=0.5. To summarize this lesson, we'll go over the key concepts that we've looked at. The first has to do with manipulating currents right here. These two particular configurations of this node are the same, because all we've done is I've reversed the arrow on this current right there and in doing so. In order to make it equivalent, I had to negate the sign of that current. And in doing that, we were able to derive three different versions of the KCL right here and you use whichever one is most convenient for you. Now, another concept that we covered is that series components have the same current. So this particular case, we've got two resistors. They both have the same current through them and the last thing is we did example where we show that how powerful it is to combine both the KCL, and the KVL to solve the circuit problem. Thank you. [MUSIC]