0:03

The topic of this problem is superposition analysis, and

Â we're going to work with circuits with independent sources.

Â The problem is to determine V out, or V sub out, on the circuit shown below.

Â It has a 6 kilo ohm resistor on the right-hand side which is our

Â load resistance.

Â 0:28

When we analyze this problem using superposition analysis.

Â Our approach will be to look at the contribution from each one of the sources

Â individually, and then to sum them to get the total V sub out.

Â So if I wanted to take this problem and rewrite it in terms

Â of a problem where we're just getting in the contributions of the current source.

Â We could solve for V out in that new circuit.

Â And that would give us part of our total V out from the circuit shown here.

Â And then we could also write another circuit where we're

Â looking at the contribution from the three-volt source.

Â Find its contribution to V out.

Â Sum the one from the two-milliamp source and

Â from the three-volt source together, to get our total V out.

Â And so, that's the concept of super-positioning.

Â In fact, that's the approach we're going to use on this problem.

Â We're going to take the problem, we're going to look at the contribution from

Â the two-milliamp current source first.

Â 1:24

And then the contribution from the three-volt source.

Â So, let's redraw the circuit looking at the contribution from

Â the two-milliamp source first.

Â We redraw it and when we redraw these circuits, the trick is to take all

Â the other volted sources and treat them as short circuits like we've done here.

Â 1:51

And if we do that, then we can find the contribution from

Â an individual source that we have chosen, to find the voltage across.

Â So this is a V out prime which is coming from just the 2 milliamp source.

Â We still have the one kilo ohm resistor and the two kilo ohm resistor.

Â On the left-hand side of the circuit, we still have the six kilo ohm.

Â All we've done is we've cancelled out all of the other sources in this circuit.

Â And we're looking for V out.

Â 2:27

So, we immediately see that this problem lends itself to current division.

Â We have a two-milliamp source.

Â It's flowing upwards, through the center lag, and then it splits to the left, and

Â to the right.

Â And so, if could find the current, which is going to the right,

Â through the six kilo ohm resistor, then we can find V out prime.

Â because we know, V out prime in this case is going to be 6K times this current,

Â which is flowing through this 6K resistor, which is called I6K.

Â And so this current is I6K.

Â And we know I6K is in current division.

Â I6K is going to be two-milliamps.

Â The part of two milliamps which flows to the right.

Â And so that's going to be the sum of the voltages on the other side,

Â which is three kilo ohm.

Â 1 kilo ohm plus 2 kilo ohm,

Â divided by that same resistance, 3K plus 6K.

Â That is the current flowing through the 6K resistor.

Â So we can plug that into our V out prime equation because

Â it's just a number now and we can solve for V out prime.

Â And if we do that,

Â we get a V out prime equal to four volts.

Â Now we will look at the contribution of the three-volt source, and

Â we're going to eliminate the two-milliamp source.

Â So we're going to redraw our circuit one more time and this time, for

Â the three-volt source and the contribution of that source to our output voltage.

Â So again, we draw our resistors as we had before.

Â Our 1K and 2K resistors, we still have

Â our 6K load resistance like this.

Â And we're measuring the output voltage across V out, double prime in this case.

Â We have a 1K resistor here, a 2K resistor at the top.

Â And our current source is open circuited.

Â So we have our volted source that we're looking for the contribution from.

Â We've taken away our current source, treating it as an open circuit and

Â we're looking for V out.

Â 4:53

So again, this problem is another one where it's pretty straight forward.

Â We're looking for V out double prime and what's it equal to.

Â It's equal to 6K of the resistance, times current through that resistor.

Â And that current is going to be three volts.

Â 5:21

That gives us V out double prime.

Â So V out double prime using this

Â approach, gives us two volts.

Â So if we want to find the total V out, V out total from our

Â circuit above is going to be a sum of a contributions to the V out,

Â from each one of the individual sources.

Â So, that is 4 volts plus 2 volts, and so

Â we end up with the V out equal to 6 volts.

Â