00:01
You were supposed to solve and find vx using superposition.
00:07
The problem with using superposition is we basically have to solve three circuits.
00:12
And i'm not sure i have enough space to do that.
00:17
But let's give it a try.
00:20
So in working with superposition, what you want to do is you want to isolate each source independently and basically solve the circuit for each of the three sources.
00:32
And then once you're done with that, you add up your vxs to get your answer.
00:43
So here's the first circuit.
00:47
And so when you don't use a source, if it's a voltage source, you write it as a short.
00:54
And when it's a current source, you write it as open.
00:59
So basically ignore that.
01:15
Okay.
01:17
And then here's our vx1.
01:21
All right.
01:22
So i'm going to do this.
01:34
These two, those two resistors are in parallel and their equivalent resistance is, it's the product over the sum.
02:06
So i get these equations for this little circuit, and then i can solve those.
02:12
So we get these currents and it tells us vx1.
02:23
So now we get this circuit for this one, for this current source.
02:27
All right and then these two they're in parallel so that's 30 times 60 or 30 plus 60 is 30 times 60 is 1800 divided by 90 is 20 and then these two they're also in parallel so that's 20 times 30 times 30 over 20 plus 30, which we saw before is 12.
03:20
So that gives us this circuit.
03:52
Okay.
03:57
And of course, these two, they're in series now.
04:03
So that's 30 -owned.
04:12
So let's go with i -4 and i -5.
04:22
And we see that 6 amps is i -4 plus i -5.
04:45
And then we have a single loop that goes around the outside edge.
04:51
Remember, you can't include a current source and a loop equation.
04:56
So we got 30 i4 minus 12i 0...