Let's take a look at our circuit. We want to apply mesh analysis to solve
for this current right here.
[NOISE]
For I sub 0. In mesh analysis, we have to identify our
mesh currents. I've got three loops here that are
non-inclusive, so they don't include any other loop.
And I've got to define my mesh currents, so I'll call this one I1, this one I2,
and this one I3. So, when I solve for this problem I've
got to set up three simultaneous equations and they're going to be in the
form of the unknowns being I1, I2 and I3, a mesh currents.
So, I should have three equations and three unknowns.
Now, sometimes when you set this up, you might find that you can reduce out one of
these pretty easily. But I'll go head and set it up with the
three equations to begin with. Now notice that I got a dependent source
in here this is the dependent current source,, and you often times find circuit
sources in certain examples like amplifier examples.
Amplifiers might become from non-linear elements, and some, sometimes we redraw
those and model those, and when we model them to an equivalent circuit, we
oftentimes find dependent sources like this.
So I just want you to be able to see this.
It does complicate it a little bit, but if you can understand how to handle
dependent sources it'll help you under to be able to solve other problems.
So, I got 3 equations that I need to, to find and I'll find them by doing a KVL
around each of these loops so let me start out with loop 1, that's the I1
mesh, okay. So, this one is pretty straightforward
because I've got this source right here. It's a dependent source.
But the current has to match up with what this mesh current is.
So, I1 It's going to be equal to capital I over 2, alright?
Capital I is a variable I don't want to have in my final equations, because I
only want I1, I2, and I3, so I've got to try to get rid of capital I.
So let me see where else I've got an I. I've got an I going through this loop
right here, this branch. And I also know in terms of my mesh
currents, that I would be equal to capital I sub 2 minus I sub 1.
Because going in this direction, I two is going in this direction.
I one is going in the opposite direction so I have to subtract it off.
So I can now solve for I in terms of I one and I two.
That means plugging back into this equation [SOUND] I had that let's see.
I can reduce this out by looking at yeah, I can plug this back into here.
It's actually easier if I say, 2 times I1 is equal to I.
If I have 2 times I1 is equal to I and then just equate these, I end up with the
expression 3I1 is equal to I2. So now I have an equation that only
involves my mesh currents. Let me look at loop 2.
Around this loops let me start out with the voltage source going around the loop
in a clockwise direction, I'll get to the plus sign first, so I'm going to have a
plus 2 volts, so coming up to here remember that I want to have equations in
terms of my mesh currents so I don't want to write in terms of I, I want to write
in terms of I1 and I2 So, it's plus 5, I 2 minus I 1, going in this direction.
Plus a voltage drop across 2 ohm resistor.
And going into this direction it would be I 2 minus I 3.
Plus the voltage drop across this resistor and the only current going
through that is a mesh current of I 2. Plus I 2 times 1 ohm, that has to equal
0. If I want to clean this up.
Let me re-group my terms. I will have a minus 5I1 plus 8 I2 minus 2
I3 is equal to minus 2 loop 3. So I now have two equations and I need
one more equation, I go to loop 3 and looking at this loop I'm going to have 4
ohms times 3, which is this voltage drop plus this voltage drop which is 2 ohms
times this current which is I3 minus I2 plus this voltage drop, which is going to
be, I 3 minus I 1, is going to equal 0. And if I clean this up by regrouping my
terms, I'll get a minus 10, I1 minus 2I2 plus 16I3 equals 0.
And this is, these are my equations. In this case, I've got three equations
and three unknowns. Now, one of the equations is really
simple. I could substitute this back into here
and solve, reduce this down to two equations and two unknowns.
I'm going to go ahead and keep it as three unknowns right now, because I want
to show you how to do matrix analysis of this.
So, I've got three unknowns I want I2 and I3, that's sum matrix times that is equal
to something. So, taking my first equation this is a
pretty simple one, I've got 3I1 equaled to I2.
Or in other words, I can write this as 3 I 1, minus I 2, equals 0.
So, putting that into the first row of this matrix, I'm going to have a 0,
because it equals 0 and then I've got a 3 times I 1, minus 1 times I 2, and then a
0 plus 0 times I3, that's my first equation.
My second equation, I put in the coefficients here, a minus 5 times I1, 8
times I2, minus 2 times I3 and that equals minus 2, and my last equation I've
got a minus 10 times our 1 minus 2 times I 2 and a 16 times I 3 equals 0.
If I solve this, again, if your calculator does this, use your
calculator. Otherwise, you can go online and find a
linear equation solver. If I solve this, then I would get I 1, I
2, and I 3 equals minus 0 point 118 minus 0 point 353 and minus 0.118.
And of these mesh currents I was interested in I zero to begin with well I
zero is equal to I three minus 0.118 and that's my final answer.
So, to summarize what we did, we were looking for I0, we defined our mesh
currents, which are currents around each of my non inclusive loops, and I want to
make sure you remember I only want equations that only involve these mesh
currents, those are going to be my unknowns.
So I solve to find those equations, I do a [UNKNOWN] around each of my loops and I
make whatever substitutions I make to get rid of any extraneous variables because I
only want the the mesh currents to be in, my only variables in there.
So I do a [UNKNOWN] around each of the loops and in my case, and I came up with
three equations, and three unknowns, and I wrote it into a matrix form, and then I
used a matrix inversion solver. because I invert this matrix to solve for
the I's. Thank you.