The topic of this problem is Thevenin's Analysis, and we're going to work with circuits with independent sources. The problem is to use Thevenin's Equivalent Circuit from Thevenin's Theorems to determine Vout in the circuit shown below. The circuit has a current source and a voltage source, both are independent sources. It also has three resistors, we have a load resistor on the right hand side of the circuit where we're measuring the output voltage V sub 0 across. And so if we know that, we're going to use Thevenin's Equivalent Circuit to determine Vout, we first have to remember what that looks like. Our Thevenin's Equivalent Circuit is this, it has V open circuit, and it has R Thevenin's associated with it, that's our right hand side of the circuit. On the left hand side we would have our load, in our case, our load is a 6k resistor, but it would be R sub L for the resistance across the load. So what we need to do is we need to take our circuit as shown here, everything except for the load, which is 6k, take everything else and reduce it to an equivalent open circuit voltage and an equivalent R Thevenin's. Then we can draw our Thevenin’s Equivalent Circuit and we could then add our 6 kilohm load resistance and we can easily solve for Vout. So this is yet another way that we can use to solve circuits. And so, we're going to first find our Voc. So we're going to take our circuit, which we have initially and we're going to redraw our circuit to determine Voc. And so to do that, we take our load and we replace it with an open circuit. That's what Voc represents. It represents the load voltage with the load taken out and replaced by an open circuit. So if we did that, here's what we would have. We would have something that looks like this. Taking out the 6K resistor, we still have everything else intact. So we have both the 3 volt source at the top, and the 2 milliamp source in the middle, as well as the 1 and 2 k resistors on the left hand side of the circuit. That is our Voc. So we know that Voc is going to be equal to the voltage from top to the bottom of our circuit. So the voltage at this point in our circuit is going to be Voc minus 3 volts. And so if we want to find Voc, we might take this as a node one and we would note that Voc = V sub 1 + 3 volts. That's our V open circuit. So if we can find V sub 1, then we can find Voc. What is V sub 1? The 2 milliamp source in the center of our circuit under the circuit condition is only going to flow back to the left through the 1K and the 2K resistor. So we have plus to minus drop as these current flows through the 2K and plus to minus as it flows through the 1k using the passive sign convention. So our 2 milliamp current source produces a voltage V sub 1, Which is equal to 2 milliamps times (2k+1k). And so that gives us V1, which is equal to 2mA times 3 kilohm or 6 volts. So we end up with a Voc equal to 6V + 3V and that gives us 9V. We also need to find R Thevenin's in our equivalent circuit. So now we have a Voc for our equivalent circuit, it's 9 volts. We need to find an R Thevenin's for our circuit. The way that we do that is by analyzing our circuit once again. We take our original circuit, and to analyze it for R Thevenin's, what we do is we take the load out and we look back into our network and solve for the equivalence resistance of our network. In that operation we open circuit all the current sources and we short circuit all the voltage sources. So let's redraw our circuit with all that in mind. We have our 1 kilohm resistor, our 2 kilohm resistor, we have a short circuit for the voltage source. And remember, we've taken the load out so the load is not part of our circuit. And we open circuit the current source, and so we have an open circuit for the current source. And if we loop back into our network with our 1K resistor and 2K resistor on the left hand side, we are able to find R Thevenin's equivalent resistance. So in this case it's pretty straightforward, R Thevenin's equivalent resistance comes out to 3 kilohms. So we can take that and we can redraw our circuit above for the Thevenin's Equivalent Circuit. And so it's going to be +-6 volts for the source. We have a Thevenin's resistance of 3 kilohms, so it's a 3 kilohm resistor here and we put our load back in and our load is 6 kilohms. And we're looking for the voltage across that 6 kilohm resistor. And using voltage division, this problem becomes very easy. We know that Vout is equal to 6V times (6k divided by 6k plus 3k). So our Vout in the end comes out to be 6 volts times, I'm sorry. Voc is not 6 volts, Voc is 9 volts from below. I looked at the wrong number. So Voc is 9 volts, we put the 9 volts in. This should be 9 volts instead of 6 volts. So we're going to take that and remove that and put in a 9 and then we end up with a Vout equal to 6 volts. So, the circuit that we've drawn is the Thevenin's Equivalent Circuit for our problem. And we found that by first analyzing our circuit for Voc, that's what this step was about. And we analyzed the circuit for R Thevenin's, and that's what this bottom part of our analysis was about. We took those values, put them back into our circuit, recognizing that Voc was 9 volts and then we solved for Vout using voltage division. It was the 9 volts times the division of that voltage across the 6k and the 3k resistors in series with another, and this gives us the final Vout equal to 6 volts.