In this video,
we introduce the concept of a super mesh.

Here we have a circuit that
involves three meshes.

The left-hand mesh, which has a mesh
current associated with it of I1,

a mesh here with a mesh current defined of
I2, and a mesh current here defined of I3.

The problem arises when
we have a branch that

contains an independent current source.

By definition, an independent current
source produces, in this case,

I0 amps of current ,no matter
what the voltage is across it.

In other words, there is no relationship
between a current source current and

its voltage.

Let's demonstrate the problem by
attempting to write the mesh equation

around this loop right here starting
in the lower left-hand corner.

We've got the voltage drop across R2,

which would be R2 x the current through R2

going in this direction, is I2- I1.

Coming down the right-hand side, we have
the voltage drop across R3 is R3 X I2.

And now we get to this independent source.

We're attempting to
write the voltage drops

around this mesh, but
we have no expression for

the voltage drop across
I0 in terms of I2 and I3.

How do we resolve this?

Well, to resolve it, we take
advantage of the fact that the sum of

the voltage drops around any
closed loop must equal zero.

So instead of going around each of these
meshes separately, we're going to make

a trip around the outside of
both of these loops combined.

Let's show what we mean here
by now analyzing this circuit.

Let's start by writing the loop
equation around this left-hand mesh.

We'll have -V0 + R1I1 + now

coming down this branch here,

voltage drop across R2
will be R2 x I1- I2.

Plus the voltage drop across R5,

which will be R5 x I1- I3.

The sum of those voltage drops
must equal 0, nothing new there.

Now let's write the KDL equation
going around our combined super mesh.

Starting down here in the lower left-hand
corner and going in a clockwise direction,

we'll have voltage drop across R5 will
be R5 times, let's be careful here.

We're going up across R5.

The current going up is I3- I1.

So R5 x I3- I1,

plus the voltage drop across R2

going in this direction will be

I2- I1 + the voltage drop across

R3 is just gonna be R3 x I2.

And finally, we've got the voltage
drop across R6, which will be R6 x I3.

And the sum of those
voltage drops will equal 0.

There, we've got the two equations.

Again, we're gonna get our
third equation by taking

advantage of the fact that
the current in this branch is I0.

And that that current I0,
in terms of the two mesh currents,

is equal to the mesh current
flowing in this direction,

minus the mesh current
flowing in that direction.

Or, I0 will equal I2- I3.

Thus, we have the three equations that we
need to cover the three variables that we

defined.

Again, our first equation came from going
around this mesh, as we have always done.

The second equation came from
going around the super mesh,

effectively avoiding the need
to go across the volt or

to know the voltage across
that current source.

And then the third equation came from
the fact that the branch current in this

sources is in fact I0.

And in terms of the mesh currents,
the current in that branch is I2- I3.

Three equations, three unknowns,
be ready to solve it,

using whatever method for solving
systems equations you choose to use.

In the next video, we'll do
a numerical example of a super mesh.