1 kvl to current

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.