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.