[BLANK_AUDIO].
Okay, let's take a look at this circuit. Now this circuit has a volt, a current
source in there and it's got some dependent sources.
This dependent sources is, this is a dependent voltage source.
And I wanted you to be able to see a problem like that.
I might mention that dependent sources sometimes come into use in, in models of
non linear systems. For example, a non linear system with
transistors might have a dependent source in it if you look at a model in the
linear range. So, it's good for you to see some
problems with dependent sources and become familiar with working with them.
So, this is ah, [SOUND] a node analysis approach that we're using here.
So node is, analysis we need to pick a ground node.
I'm going to pick this as my ground node only because I've got a voltage coming
off of it. So, I know this voltage is pretty easy.
That's going to be at 2 volts. So I picked this one.
Its not unique. I could have picked another one.
I typically don't want to pick a node with a current source coming off of it
though, because that's a little hard to, to handle.
I've got two other nodes that I need to identify.
Let me call this one node A and I'll call that voltage V sub a, referenced to the
ground. And this is V sub b, referenced to the
ground. So the two voltages I have, node voltages
V sub a and V sub b, means I'm going to want to write equations with unknowns, V
sub a and V sub b. So I should have two unknowns and I
should come up with two equations for those unknowns.
I need to do a KCL at each of these nodes.
So node A [SOUND]. Let me do a K KCL.
And I always do it where I have, I sum up the currents leaving the node.
So, in this direction, this direction, this direction.
So, leaving the node over here, it's going to be VA minus this voltage.
What this voltage is written a little bit strangely because it says I over 2, where
I is this current right there. But that's okay because it's still a
voltage source, so this is in voltages. So this voltage right here is I over 2
referenced to ground. So V sub a minus [SOUND] I over 2,
divided by that resistance, plus the current going in this direction, V sub a
[SOUND] over 5, plus the current going in this direction.
Well, the voltage drop is V sub a minus V sub b [SOUND] and the resistance is 2.
That has to equal 0. Now as, as I said, I want to only, have
the variables of V sub a and V sub b in my equation.
So I need to get rid of I. How do I get ride of I?
Well, I've got to be able to write I in terms of V sub a or V sub b or some
combination of them. If I look at I as its defined here I can
see that it is related to V sub a. In fact, if I look at this expression
right here, I find that I [SOUND] is equal to V sub a over 5 and then I'm
going to negate that, because I is referenced in this direction, where V sub
A over 5 by itself is the current going in this direction.
So I have to throw the minus sign in because of the direction that I is
defined. Now, let me group my terms here.
I've got V sub a [SOUND], 1 over 10 plus 1 over 100 plus 1 over 5 plus one half,
and then a minus one half V sub b is equal to 0.
And if I add this all together, this is point 71, 0.71 V sub a minus 0.5 [SOUND]
V sub b equals 0. So now I have an equation just in terms
of the Va and Vb. Let's go ahead and do the same thing with
node b. [SOUND].
Node b, I'm going to look at the currents leaving the node.
So that is, this voltage drop is now Vb minus Va [SOUND] divided by 2 ohms, plus
this voltage drop is Vb minus 2 [SOUND] divided by 1 ohm.
And then this current leaving in this direction is minus 2 [SOUND] is equal to
zero. If I combine my terms, I will find that I
have [SOUND] the following equation. Minus 0.5 V sub a plus 1.5 V sub b equal
4. And again, I've got an equation in terms
of Va and Vb. I can now write this as a linear
equation, a matrix equation. Looking at this, [SOUND] I want to solve
for Va and Vb. [SOUND].
I want to set up the equation like this. To write this first equation into the
matrix form, I've got a 0.71 times Va. So that's going to be a 0.71 times Va.
And then, minus 0.5 times Vb. So minus 0.71 time Va minus 0.5 times Vb
is equal to zero. Now the other equation can be written in
the same way. I've got a minus 0.5 times Va, so the
coefficient minus 0.5 times Va and 1.5 times Vb is equal to 4.
So the first row is this first equation and the second row is the second
equation. Many of you might have an equation solver
on your calculator. Or if you don't, there are there are
places that you can go on the internet. In fact, I just did that before I, I
started this. I went on the internet and I just, I just
searched for a linear equation solver and I found one where I can just plug in
these values, these matrix values, and plug in what it's equal to and I was able
to solve for the variable here. And when I did that, I found that Va
[SOUND] and Vb are equal to 2.454, 3.485. And ultimately what I wanted to solve for
was V0, and I just found that right there.
So V0 is equal to Va, which is 2.454. And that solves this problem.
So, just to summarize how we solved it, the first thing we had to do was select a
ground node. And then we identified our other node
voltages. We had two additional nodes.
I wanted to find equations involved that involve only those unknowns, the V sub a
and V sub b as my only unknowns, which are my node voltages.
I use a KCL at each node and then I just solve it.
I came up with two equations and two unknowns.
I chose to solve it using a matrix method.
You don't need to use a matrix method, you can just solve these by substitution.
I should mention that there's a separate tutorial video that I'm going to shoot
[INAUDIBLE] that shows how to solve this by hand.
matrix inversion of two variables by hand.
Otherwise, you don't need to solve it by hand.
If you have a calculator that does it, then by all means use that.
Thank you.