So, let's take a look at this circuit. 
We want to do a Thevenin equivalent 
method of solving for, in this particular 
case, I'm interested in this current. 
The current through that resistor. 
Now there are different ways of doing 
Thevenin Equivalent. 
There's this source transformation 
method. 
And then there's a method where we're 
finding the R thevenin or V thevenin and 
current directly. 
So that, of the two methods, I cannot use 
the method of solving for V thevenin, or 
R thevenin. 
The reason is I got a dependent source, 
this, this box here, this diamond box 
means that this is a dependent source. 
This current depends on this current 
right here, this i is that current. 
But this is, this is and this is also a 
current source as well, but it's a 
dependent current source. 
So I cannot find R thevenin indirectly. 
When we did R thevenin indirectly, we 
said well let's just 0 out all of our 
independent sources. 
So, I can 0 this one out, but I don't 
know what to do with that. 
So, the only way to do the thevenin. 
The only way to do the Thevenin 
Equivalent method is to use this, this 
formula where we find, we know that V 
thevenin Is equal R thevenin times Isc. 
And V thevenin is here, R thevenin is 
here, this is the equivalent circuit. 
And then Isc was the current going 
through it. 
We'll call that a and b, our terminals. 
So, the only way to find R thevenin, is 
to find i short circuit and V thevenin. 
And since I'm interested in this branch 
right here, I'm going to call this a and 
this b. 
And I should mention that we have the 
dependent source in here. 
And I wanted to expose you to dependent 
sources, because dependent sources often 
times come up in models of non-linear 
circuits. 
Like when non-linear circuits like 
transistor circuits are operating in 
their linear range, we oftentimes have 
dependent sources. 
So, it's a good idea to get familiar with 
how to handle them. 
So, going back to solve this, I have to 
solve for V thevenin and i sub sc. 
So if I look at i sub sc, sc stands for 
shirt, short-circuit. 
I short-circuit across the branch a b. 
Everything else I leave the same. 
This relationship between the dependent 
source and this eye through this branch 
stays the same. 
[SOUND] Voltage sto-, source stays the 
same. 
The only thing is, I short circuit that 
branch right there. 
Now, looking at this, so this is the 
circuit I now need to analyze. 
So, if I look at this current is going up 
in this direction into this node. 
This current is going into that node. 
And they're going to add together. 
So, that's i over 2 plus i is 3i over 2, 
has to leave that node. 
Okay, so that current plus this current 
into this node has to equal to the 
current leaving that node. 
I still don't know what i is though. 
And for doing that I'm going to do 
Kirchhoff's Voltage Law around this loop, 
right here. 
So, starting from this point, going in 
the left direction, the clockwise 
direction. 
I hit the, the plus sign a, this voltage 
source first, now I say, a plus 2. 
And then, I've got the current going 
through here times 5, is that voltage 
drop. 
And then this current is 3i over 2 times 
2 is a voltage drop here and then that 
has to equal 0. 
If I solve for i, I would get i is equal 
to minus 1, 4th and if I is equal to 
minus 1, 4th I sub sc is 3 halves times 
that. 
[SOUND] 
Or minus 3, 8ths. 
So, that's this current right there. 
Let's look at the Thevenin Voltage. 
And I'm going to now redraw this circuit 
where I open circuit, this branch and I 
solve for the voltage drop there. 
Let me just redraw the circuit. 
I need to solve for this, this V 
thevenin. 
Well, if we look at this, looking again 
at these nodes, this node here, and 
looking at the current that goes into 
this, well, I have a current of i going 
into this, a value of i going into this 
node from this direction. 
And from this branch, I have a value of i 
over 2. 
So, I've got 3i over 2 going in this 
direction here. 
That's much the, what we found up here, 
3i over 2. 
But you see, there's a problem. 
The problem is that this is 
open-circuited. 
So, if this is open-circuited, that means 
there's 0 current flowing through here. 
That means that this current i has to 
equal 0. 
Well, if the current i equals 0, that 
means there's no voltage drop here, and 
there's no voltage drop here. 
So, if I look at the voltage around this 
loop right here that means that the V 
thevenin has to balance this voltage 
source right here. 
And since they're both really in the same 
direction, as I go around the voltage, 
the loop in this direction, and did a 
KVL, I would get a plus 2 and then a plus 
V thevenin. 
And that means that they're going to have 
to have negative signs of one another. 
So, V thevinin minus 2 volts. 
So, if I want to redraw the thevanin 
equivalent circuit from up here, I just 
redraw it with my values in there. 
Plus minus, minus 2 volts, R thevenin is 
found from taking V thevenin minus i sub 
sc. 
And I can actually write that here. 
And that is equal to 16 over 3 ohms. 
Two volts, 16 over 3 ohms, that's a, 
that's b. 
So, this is this part of the circuit. 
Let me show it in red right here. 
I've taken this part of the circuit to 
the left and simplified it to this right 
here. 
Then if I want to connect it up with the 
rest of the circuit which is this final 
branch. 
Which is 1 0, and now I'm interested in 
solving for I sub a. 
I sub 0, which is the current through 
that, that branch. 
So, then I just look at the, this is a 
simple circuit now. 
I've got 2 resistors in series and the 
voltage. 
So, I have i0 is equal to the voltage 
which is minus 2, divided by this 
resistance. 
Which is, 19, if I add these together I 
have 19 over 3. 
In other words, it's minus 6 over 19 
amps. 
And that's that current through here. 
So, just to summarize this problem, I 
wanted to do Thevenin Equivalent. 
So, I have to identify what branch I want 
to isolate and keep that branch as is. 
And take the rest of the circuit, and 
simplify it to a Thevenin or Norton 
circuit. 
In this particular case, because I had a 
dependent source in there, I had to use 
this equation, here, to find R thevenin. 
So, I had to find i sub sc, which is 
finding the current if I short circuit 
across those terminals. 
And I had to find V sub thevenin, which 
is the voltage if i open circuit across 
those terminals. 
And then I plug those, and then I find R 
from those two values. 
If I have V and I, I find R, and I just 
redraw my thevenin equivalent circuit. 
And then add in that final branch, and 
then solve for whatever parameter I was 
interested in solving for.