The topic of this problem is superposition analysis, and we're going to work with circuits with independent sources. The problem is to determine V out, or V sub out, on the circuit shown below. It has a 6 kilo ohm resistor on the right-hand side which is our load resistance. A circuit has two independent sources. It has a two-milliamp current source and a three-volt voltage source. When we analyze this problem using superposition analysis. Our approach will be to look at the contribution from each one of the sources individually, and then to sum them to get the total V sub out. So if I wanted to take this problem and rewrite it in terms of a problem where we're just getting in the contributions of the current source. We could solve for V out in that new circuit. And that would give us part of our total V out from the circuit shown here. And then we could also write another circuit where we're looking at the contribution from the three-volt source. Find its contribution to V out. Sum the one from the two-milliamp source and from the three-volt source together, to get our total V out. And so, that's the concept of super-positioning. In fact, that's the approach we're going to use on this problem. We're going to take the problem, we're going to look at the contribution from the two-milliamp current source first. And then the contribution from the three-volt source. So, let's redraw the circuit looking at the contribution from the two-milliamp source first. We redraw it and when we redraw these circuits, the trick is to take all the other volted sources and treat them as short circuits like we've done here. And, all of the current sources, if we were to have them in the circuit, and treat those as open circuits. And if we do that, then we can find the contribution from an individual source that we have chosen, to find the voltage across. So this is a V out prime which is coming from just the 2 milliamp source. We still have the one kilo ohm resistor and the two kilo ohm resistor. On the left-hand side of the circuit, we still have the six kilo ohm. All we've done is we've cancelled out all of the other sources in this circuit. And we're looking for V out. So, we immediately see that this problem lends itself to current division. We have a two-milliamp source. It's flowing upwards, through the center lag, and then it splits to the left, and to the right. And so, if could find the current, which is going to the right, through the six kilo ohm resistor, then we can find V out prime. because we know, V out prime in this case is going to be 6K times this current, which is flowing through this 6K resistor, which is called I6K. And so this current is I6K. And we know I6K is in current division. I6K is going to be two-milliamps. The part of two milliamps which flows to the right. And so that's going to be the sum of the voltages on the other side, which is three kilo ohm. 1 kilo ohm plus 2 kilo ohm, divided by that same resistance, 3K plus 6K. That is the current flowing through the 6K resistor. So we can plug that into our V out prime equation because it's just a number now and we can solve for V out prime. And if we do that, we get a V out prime equal to four volts. Now we will look at the contribution of the three-volt source, and we're going to eliminate the two-milliamp source. So we're going to redraw our circuit one more time and this time, for the three-volt source and the contribution of that source to our output voltage. So again, we draw our resistors as we had before. Our 1K and 2K resistors, we still have our 6K load resistance like this. And we're measuring the output voltage across V out, double prime in this case. We have a 1K resistor here, a 2K resistor at the top. And our current source is open circuited. So we have our volted source that we're looking for the contribution from. We've taken away our current source, treating it as an open circuit and we're looking for V out. So again, this problem is another one where it's pretty straight forward. We're looking for V out double prime and what's it equal to. It's equal to 6K of the resistance, times current through that resistor. And that current is going to be three volts. Divided by the sum of the resistances as that voltage is dropped across. So it's going to be 9K. That gives us V out double prime. So V out double prime using this approach, gives us two volts. So if we want to find the total V out, V out total from our circuit above is going to be a sum of a contributions to the V out, from each one of the individual sources. So, that is 4 volts plus 2 volts, and so we end up with the V out equal to 6 volts.