[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.