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.