00:01
Okay, we want to find v0 using thevenin's theorem.
00:05
So thevenin's theorem tells us we want to, so for instance here, we're going to find the voltage across this resistor.
00:11
What we do is we take the resistor out of the circuit and then calculate the thevenin equivalent circuit, and we'll put that back in.
00:24
So the first step is to redraw the circuit.
00:27
Circuit.
00:28
So the thevenin equivalent circuit has a voltage source and a battery and a battery resistor connected in series between the points a and b.
00:43
And so what we have to do is we have to calculate what that resistor is and what the voltage source is.
00:49
Okay.
00:49
So first of all, to find the thevenin resistance.
00:54
We redraw this circuit again, but now what we're supposed to do is we're supposed to short out, actually we're supposed to open the current sources, we're supposed to put in shorts wherever there's a voltage source.
01:12
So we redraw the circuit like this, okay? and so our next step is to calculate the resistance distance between the points a and b.
01:58
All right.
02:00
So i want to point something out.
02:02
The 2k resistor here that's horizontal, that is shorted out.
02:07
We look at the wires here, shorts out that resistor.
02:11
So we just ignore it.
02:15
And then we see that these two resistors, the 2k and the 1k, they are in parallel.
02:26
So their equivalent resistance is 2 times 1 over 3.
02:33
It's 2 plus 1.
02:37
That will be in k.
02:43
So that's 2 thirds k.
02:49
So that our thevenin resistance has that 2 thirds k in series with this 1 k.
03:03
So that's 1 .66k ohms for our thevenin resistance.
03:19
Alright, then to find the thevenin voltage, what we have to do is go back to this circuit.
03:29
And we have to solve it and find the voltage between a and b.
03:37
Alright, so here's how i'm going to do that.
03:42
I'm going to put in, i'm going to use the mesh current method, call that current j1, we'll call this current j2, we'll call this current down here j3.
04:14
So we actually have, we get two loops from kirchhoff's loop law, corresponding to j1 and j3.
04:29
You can never use the loop law when you have a current source in the loop.
04:36
However, j2 is just two milliamps because i even drew it in the right direction because the current source here determines what j2 is.
04:55
All right.
04:56
And then now let's look at the loops.
05:00
So loop number one, we follow that loop current and we go around like this so we get minus 2k times j1 we get another 2k and that's j3 my heart j1 minus j3 and then we get minus 1k j1 minus j2 adds up to 0 then then our second loop, which is actually loop three, remember j2 we already know, okay? so we get this for vx.
06:05
All right, so let's rewrite our two equations.
06:09
So we got equation one gives us minus two j1...