L3 2 3Mesh Example with numbers

Mesh Current circuit analysis

Let's use this as an example
to see how mesh-current

technique can be used to identify or
to calculate, say,

the voltage across that pair
of combination resistors.

In order to do that we are going to need
to know current flowing through the either

the 15 ohm, the 12 ohm and

then the voltage across that will be,
that current times the resistance.

So lets get started here.

Our first step is to identify the meshes.

As you can see we have three meshes,
one here, this one in the center.

And then one there also.

Now let's define mesh-currents
in each of those meshes.

We'll call this I1,
we'll call this one here

I2, and
call this mesh current right there I3.

Now, let's look first of
all at that left hand mesh.

It's interesting here.

We have an independent current source of
100 milliamps flowing in that branch.

Well, that branch current happens to be
the same as this mesh current we have

identified I1.

This we know right now.

Without writing anything else,
I one is equal to 100 million or .1 amps.

That gives us one of our equations, also.

All right.

Let's look at the center mesh, here.

Starting at that point,
We've got 60 Ohms times I2- I1 and

we can certainly say -0.1 because we know
what I1 is but let's just go ahead and

write two additional equations in terms of
I1, I2 and I3 so we'll have a system of

three equations and three unknowns and
we can solve it that way.

So we have 60(I2-I1) +,

coming across here, 30( I2).

Now we have a voltage source here.

We're going from minus to plus.

That represents a voltage increase.

Therefore, it will be a -10 volts.

Coming on down this 15 ohm resistor,
we'll have plus 15,

times the current in that 15 ohm resistor
where the current is referenced going

down cuz that's the direction we're
going The current going down is I2-I3.

And that brings us back
to where we started, so

the sum of those terms equals 0.

And then finally, the going around
the right-hand mesh, we have 15

times I3- I2 + 2

times just I3 = 0.

That gives us three equations
with three unknowns.

I've already pointed out that we know
what I1 is but let's just go ahead and

write them Combing common terms and

factoring out our mesh current variables.

So starting with the top
equation we have I1, equals .1.

The second equation factoring out I1 we're
left with We've only got one i1 term.

It's a negative in front of it with a 60,

it will be a -60 + i2.

We have three i2 terms.

we got 60 +30 +15.

60 +30 +15.

Then for I3 we have just one I3
term it's got a minus sign and

being multiplied by 15 so minus 15 and

then we've got this constant voltage here
of a minus 10 we bring to the other side.

As a plus ten.

So there's our second equation
with the terms combined.

Now the third equation,
the third equation has no I1 terms.

So lets put a I1 with a zero
there to hold the place.

Plus I2 times the negative 15

Plus I3 times 15 plus 12 and

the sum of those terms must equal 0.

There again are three equations and
three unknowns.

Now let's just go ahead and
calculate these.

Let's see this is gonna be 60
plus 30 is 90 plus 15 is 105.

And then this here is 15 plus 12 is 27.

We go ahead and plug that into our
calculator either using the solve button

or the solve facility or

matrix Algebra, and
when we do that, we find that I1

is equal to, we already knew what I1 was,
100 milliamps, or 0.1 amps.

I2 = 0.1655 amps.

And I3 is equal to .092 amps.

But we weren't asked to determine
what the currents were.

We were asked to find what Vout is.

Vout is equal to 12 times The current

flow through that which is
our mesh current I three.

So 12 I three.

12 times .092 and

that equals 1.10 4 volts.

We can determine any other volt for
the current that we want to.

For example, what is the current flowing
down through that 60 ohms resistor?

Let's call it i sub 60.

Well, I sub 60 Is

just going to be I1 minus I2, or

.1 minus 0.1655 Amps and

of course that then equals
negative 0.0655 Amps.