0:03

The topic of this problem is Thevenin's Analysis, and

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

Â The problem is to use Thevenin's Equivalent Circuit

Â from Thevenin's Theorems to determine Vout in the circuit shown below.

Â The circuit has a current source and

Â a voltage source, both are independent sources.

Â It also has three resistors, we have a load resistor on the right hand side of

Â the circuit where we're measuring the output voltage V sub 0 across.

Â And so if we know that, we're going to use Thevenin's Equivalent Circuit to determine

Â Vout, we first have to remember what that looks like.

Â 0:43

Our Thevenin's Equivalent Circuit is this, it has V open circuit, and

Â it has R Thevenin's associated with it, that's our right hand side of the circuit.

Â On the left hand side we would have our load, in our case,

Â our load is a 6k resistor, but it would be R sub L for

Â the resistance across the load.

Â So what we need to do is we need to take our circuit as shown here,

Â everything except for the load, which is 6k, take everything else and

Â reduce it to an equivalent open circuit voltage and an equivalent R Thevenin's.

Â Then we can draw our Theveninâ€™s Equivalent Circuit and

Â we could then add our 6 kilohm load resistance and

Â we can easily solve for Vout.

Â So this is yet another way that we can use to solve circuits.

Â 1:44

So we're going to take our circuit, which we have initially and

Â we're going to redraw our circuit to determine Voc.

Â And so to do that, we take our load and we replace it with an open circuit.

Â That's what Voc represents.

Â It represents the load voltage with the load taken out and

Â replaced by an open circuit.

Â So if we did that, here's what we would have.

Â We would have something that looks like this.

Â Taking out the 6K resistor, we still have everything else intact.

Â 2:36

So we know that Voc is going to be equal to the voltage

Â from top to the bottom of our circuit.

Â So the voltage at this point in our

Â circuit is going to be Voc minus 3 volts.

Â And so if we want to find Voc,

Â we might take this as a node

Â one and we would note that

Â Voc = V sub 1 + 3 volts.

Â That's our V open circuit.

Â So if we can find V sub 1, then we can find Voc.

Â What is V sub 1?

Â The 2 milliamp source in the center of our circuit under the circuit condition

Â is only going to flow back to the left through the 1K and the 2K resistor.

Â So we have plus to minus drop as these current flows through the 2K and

Â plus to minus as it flows through the 1k using the passive sign convention.

Â 4:00

Which is equal to 2 milliamps times (2k+1k).

Â And so that gives us V1, which is equal

Â to 2mA times 3 kilohm or 6 volts.

Â So we end up with a Voc equal to

Â 6V + 3V and that gives us 9V.

Â 4:32

We also need to find R Thevenin's in our equivalent circuit.

Â So now we have a Voc for our equivalent circuit, it's 9 volts.

Â We need to find an R Thevenin's for our circuit.

Â The way that we do that is by analyzing our circuit once again.

Â We take our original circuit, and to analyze it for R Thevenin's,

Â what we do is we take the load out and we look back into our network and

Â solve for the equivalence resistance of our network.

Â In that operation we open circuit all the current sources and

Â we short circuit all the voltage sources.

Â So let's redraw our circuit with all that in mind.

Â We have our 1 kilohm resistor, our 2 kilohm resistor,

Â we have a short circuit for the voltage source.

Â And remember, we've taken the load out so the load is not part of our circuit.

Â And we open circuit the current source, and so we have an open circuit for

Â the current source.

Â And if we loop back into our network with our 1K resistor and

Â 2K resistor on the left hand side,

Â we are able to find R Thevenin's equivalent resistance.

Â So in this case it's pretty straightforward,

Â R Thevenin's equivalent resistance comes out to 3 kilohms.

Â 6:07

We have a Thevenin's resistance of 3 kilohms,

Â so it's a 3 kilohm resistor here and

Â we put our load back in and our load is 6 kilohms.

Â And we're looking for the voltage across that 6 kilohm resistor.

Â And using voltage division, this problem becomes very easy.

Â We know that Vout is equal to

Â 6V times (6k divided by 6k plus 3k).

Â So our Vout in the end comes out to be 6 volts

Â times, I'm sorry.

Â Voc is not 6 volts, Voc is 9 volts from below.

Â I looked at the wrong number.

Â So Voc is 9 volts, we put the 9 volts in.

Â This should be 9 volts instead of 6 volts.

Â So we're going to take that and

Â remove that and put in a 9 and

Â then we end up with a Vout equal to 6 volts.

Â So, the circuit that we've drawn is the Thevenin's Equivalent Circuit for

Â our problem.

Â And we found that by first analyzing our circuit for Voc,

Â that's what this step was about.

Â And we analyzed the circuit for R Thevenin's, and

Â that's what this bottom part of our analysis was about.

Â We took those values, put them back into our circuit, recognizing that Voc

Â was 9 volts and then we solved for Vout using voltage division.

Â It was the 9 volts times the division of that voltage across the 6k and

Â the 3k resistors in series with another, and

Â this gives us the final Vout equal to 6 volts.

Â