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.