L3 2 2Mesh Analysis Example no numbers

Example analyzing circuits using Mesh Currents

Alright, let's formalize our
approach to mesh analysis.

When analyzing a circuit using mesh
analysis we first of all identify

the meshes.

Once again, a mesh is a closed loop
that contains no other loops within it.

We'll next assign variable names for
each of the mesh currents.

We'll then write KVL equations around
each mesh in terms of the mesh current.

And then at that point, we have a system
of equations in terms of our mesh currents

that simply need to be solved, and
we've got the problem taken care of.

Let's take a look at this circuit here.

First of all, identify the meshes.

We've got a mesh here on the left,
we've got this upper mash, a lower mesh.

And again, this outer loop is not a mesh
because in this case it contains three

different meshes within it.

So we identify then or we define three
different mesh currents, one for

each of the meshes.

We'll define the current
In this mesh to be i1 and

referenced in that direction,
the mesh in the other,

mesh current in this mesh we call i2 and
the mesh current in this mesh we call i3.

Lets go ahead now then and write the three

Mesh equations in terms of I 1 and
I 2 and I 3 starting right here.

Going up this way we cross the voltage
source going from minus to plus

therefore it's a voltage increase and
again that would be a negative V 0.

Plus the votage drop across
R1 is just R1 times I1.

Now coming down across R2,
we need to be carefully.

The current through R2

involves two different mis currents
because we are going down at this point

The current through here is gonna be the
mesh current reference down which is I 1,

minus I 2, which is referenced
in the opposite direction.

So we'll have then plus
R 2 times I 1 minus I 2.

Now continuing on down, across R5.

The current flowing through
R5 is going to be I1- I3.

And those terms must then add to 0.

Let's take a look at the top mesh.

Starting right here and

going up we've got R2 times the current

flowing up which is I2 minus I1.

And again let us just point
out that the current going up

I2 minus I1 In this equation
is the opposite current that

was flowing down which was I1- I2 and
this equation.

All right continuing on
around we have in plus

R3 times the current
through R3 is simply I2.

Coming back to the left through R4
we'll have plus R4 times the current

flowing in R4, which

is I2- I3.

That brings us back to where we started so
the sum of those terms must equal zero.

And finally we do a KVL around this bottom

mesh starting here and
we'll have R five times

the current flowing through R five in the
direction we're going which is I three.

Minus I1

plus coming across here, plus R4 times

the current flowing left to right in R4.

In terms of the mesh
currents is I3 minus i2 and

than finally the current coming down
here through that r6 will be r6 times i3

and than some of those
currents equal zero.

Up here, lets just go ahead and

complete this by factoring out the mesh
currents and combining like terms.

We have, for the top one,
the top equation we've got I1 times R1.

Let's see what we've got.

R1 there, there and there.

So, we've R1 plus R2 plus R5.

+ I2 times, we have one I2 term there,

it's got a negative.

R2 + I3 We've got one I3 term here
with a negative sign in front of it.

So that would be I + I3
times a negative R5.

And the sum of those then equals,
we've got the negative V0 that we

need to bring over to
the other side as a positive.

V 0 combining like terms in the second
equation we factor out the I 2 I'm

sorry the I 1 here and I've got one

I 1 term, it's got a negative on it so
it'd be a negative

R 2 Plus I2 times.

You've got I2 here, here, and here.

All positives will be R2 plus R3 plus R4.

Plus R3 plus R4 and

then our I3 turns There is only one.

It's got a negative, so
it'll be a negative R4.

And the sum of those
terms have to equal 0.

And finally, the bottom equation here,
factoring out the I1s to begin with.

I've got an I1 term here.

Times what is that?

That's R5 isn't it?

So I1's got a negative R5.

Plus I2.

once again there's only one I2
[INAUDIBLE] has a negative R4.

Plus I3.

And there are three I3 terms,

I got R4 plus R5 plus R6,

R4 + R5 + R6.

And the sum of those has to equal 0.

Plug it in, well, of course, with
the values of the resistors and the V0,

plug it in to your matrix solver or
your solve button on your calculator,

and you've got everything you need
to calculate those mesh currents.

Once you know the mesh currents you
can calculate any branch voltage or

branch current that you
might be interested in.