WEBVTT

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

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Here we have a circuit that
involves three meshes.

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

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

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The problem arises when
we have a branch that

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contains an independent current source.

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

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

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

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its voltage.

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

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

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We've got the voltage drop across R2,

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which would be R2 x the current through R2

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going in this direction, is I2- I1.

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

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And now we get to this independent source.

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We're attempting to
write the voltage drops

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around this mesh, but
we have no expression for

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

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How do we resolve this?

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

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the voltage drops around any
closed loop must equal zero.

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

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

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

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

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We'll have -V0 + R1I1 + now

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coming down this branch here,

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

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Plus the voltage drop across R5,

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which will be R5 x I1- I3.

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

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

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

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

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We're going up across R5.

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The current going up is I3- I1.

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So R5 x I3- I1,

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plus the voltage drop across R2

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going in this direction will be

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I2- I1 + the voltage drop across

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R3 is just gonna be R3 x I2.

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

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And the sum of those
voltage drops will equal 0.

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There, we've got the two equations.

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

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

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

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is equal to the mesh current
flowing in this direction,

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minus the mesh current
flowing in that direction.

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Or, I0 will equal I2- I3.

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

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defined.

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

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The second equation came from
going around the super mesh,

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effectively avoiding the need
to go across the volt or

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to know the voltage across
that current source.

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

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sources is in fact I0.

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

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

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

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