I'd like to go through this problem that we started in class.
This is an interesting Kirchhoff’s law problem because it has a supernode
here in the middle.
A supernode is any time that you have a current source or a voltage source,
whether it's dependent or independent, that's in between two nodes.
And it doesn't have anything else, no resistors in between these two nodes.
We have to handle that in a special way.
Let's say that we have been given our two sources, our voltage source and
our current source, and all of the resistors that are in this circuit.
Then we can use this analysis technique where we know that we have our sources,
in fact, we also know our current source, and our resistors.
And we can use Kirchhoff’s law in order to determine the currents, so let's begin.
The first thing that I'm going to do is this
loop right here, -Vs + I1R1 + I2R2 = 0.
Now let's do this loop, in this case,
I'm going to put pluses and minuses on my currents.
This would be -I2R2- Vx,
which is 3I2 + I3R3 = 0.
I can't do any more loops because if I did this loop,
it would have to go through a current source, and that's not allowed.
So the next thing I'm going to do is a node.
I'll show you the easy node, and
then we're going to actually do a supernode, so here is one node.
And that's actually the easy node to do in this case.
If I did the pink node, I would have all of the currents coming in
equal all of the currents going out, so what's the current here?
The current and the branch is always the same, so this is I1.
I1 is the same current in that whole branch.
So the currents that are coming in are I2 + I3.
And the currents that are going out are I1 + Is,
that's if I did the pink node.
As an alternative, I could have done this supernode right here.
Now, you'll notice that I actually have two nodes.
But in between them is just a voltage source that has nothing else,
no other resistors.
So if I wanted to be able to do the orange node, for instance,
I would have the current coming in, I1, and the current going to out here.
But what current do I have there?
I don't know.
If I tried to do this node, I'd have the current going out,
the current coming in, and what current is there?
I don't know, so there are two ways of handling the supernode.
The easy way, I think, is to define a current, let's just call it Ix.
That's the same current everyplace in that branch.
And then I can actually do both this node and this node.
Because I've defined this new unknown,
I now need to find also Ix, so I'm going to need an extra equation.
So as an alternative to this, I could do the orange node,
which would be currents coming in, I1 = Ix + I2.
And the other one would be what's coming in,
Ix + Is = I3, going out.
So this is the orange node, and this is the yellow node.
So I have two different ways I could solve this problem.
I can either use equations 1 and 2 for my two loops, plus my pink node.
Or I could use 1 and 2 plus my orange and yellow.
These, I would turn into a matrix equation.
Let's do the one with the pink on this side.
I would be solving for I1, I2, and I3, my variables would be I1, I2, I3.
I would be solving for these equations, like so.
And then there would be a constant on this side.
So it would be Vs coming over to this side.
I1 times R1, I2 times R2, and 0 times I3.
Here we would have 0 times I1, -R2, aha, there's a 3R2 there.
So combine those two, so it's actually,
The I2 is going to have a -R2 and a -3 right there,
and then here is R3, any constants?
No, no constants over on that side.
Right here I would have, let's get these all on the same side.
I'm gonna move these two over to this side.
This was Is, not I3, I had a little trouble keeping track of that one.
So here's I1, here's -I2, and
here is -I3, and then over here we have Is.
So these are the equations that we would solve in order to find our solution.
Okay, let's go over to this side.
If I wanted to do this set instead, I would have actually had four unknowns,
I1, I2, I3, and Ix.
I1, I2, I3, and Ix, and
then over here's my vector of constants.
My equation number 1 would be just the same, R1, R2, 0, 0, and Vs.
My equation 2 would be the same, 0, (-R2, -3).
This would be R3, 0, and 0.
Now, we're replacing the pink with the two orange and yellows.
So this would be, I'm gonna move this over here.
0 is equal to minus I1 plus 1 times Ix plus 1 times I2.
And the yellow one is going to be 1 times Ix.
Let's move Is over to this side and I3 over to that side, like so.