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

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

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the voltage across that pair
of combination resistors.

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

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the 15 ohm, the 12 ohm and

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

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So lets get started here.

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Our first step is to identify the meshes.

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

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And then one there also.

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

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

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I2, and
call this mesh current right there I3.

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

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It's interesting here.

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

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

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identified I1.

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This we know right now.

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

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That gives us one of our equations, also.

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All right.

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Let's look at the center mesh, here.

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

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

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

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

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So we have 60(I2-I1) +,

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coming across here, 30( I2).

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Now we have a voltage source here.

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We're going from minus to plus.

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That represents a voltage increase.

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Therefore, it will be a -10 volts.

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

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

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

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And that brings us back
to where we started, so

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the sum of those terms equals 0.

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

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times I3- I2 + 2

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times just I3 = 0.

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That gives us three equations
with three unknowns.

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

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write them Combing common terms and

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factoring out our mesh current variables.

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

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

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

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it will be a -60 + i2.

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We have three i2 terms.

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we got 60 +30 +15.

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60 +30 +15.

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

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being multiplied by 15 so minus 15 and

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

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As a plus ten.

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

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

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

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Plus I2 times the negative 15

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Plus I3 times 15 plus 12 and

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the sum of those terms must equal 0.

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There again are three equations and
three unknowns.

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Now let's just go ahead and
calculate these.

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

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And then this here is 15 plus 12 is 27.

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

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or the solve facility or

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

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

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I2 = 0.1655 amps.

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And I3 is equal to .092 amps.

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

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We were asked to find what Vout is.

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Vout is equal to 12 times The current

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flow through that which is
our mesh current I three.

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So 12 I three.

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12 times .092 and

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that equals 1.10 4 volts.

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

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

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Let's call it i sub 60.

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Well, I sub 60 Is

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just going to be I1 minus I2, or

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.1 minus 0.1655 Amps and

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of course that then equals
negative 0.0655 Amps.