0:03

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

Â we're working with circuits with independent sources.

Â The problem is to find the Thevenin's equivalent circuit for

Â the circuit shown below and use it to determine V sub 0.

Â V sub 0 is the voltage that dropped across the six kilo ohm load resistor

Â on the right hand side of the circuit.

Â We have two independent sources in this problem.

Â We have a 2 miliamp source and a 6 volts source.

Â So we know when we're doing Thevenin's analysis,

Â what we first remember is that we need to find the Thevenin's equivalent circuit.

Â The general representation for the Thevenin's equivalent circuit is

Â shown here where we have Voc as part of the Thevenin's equivalent circuit,

Â and R Thevenin's, as part of Thevenin's equivalent circuit.

Â 0:55

Remember when we did the derivation of Thevenin's equivalent circuit conditions.

Â We separated the circuit into a circuit A,

Â which is the circuit which drives the load resistor which is circuit B.

Â So we have our load resistor RL, in our case is six kilo ohm

Â resistor which is our circuit B, and the rest of our circuit looking back into it.

Â We replace it by an equivalent circuit which contains the open circuit and

Â R Thevenins's.

Â So we need to determine the open circuit in R Thevenin's.

Â That's the bulk of the transformation that we need to make from our initial circuit

Â to a simpler circuit that then would be very easy to find things like V out.

Â So RV out is across the load resistor and so we can see just upon inspection,

Â that we can use voltage division to find V out,

Â if we know Voc and we know our Thevenin's and our sub L.

Â 1:56

So let's start with finding Voc, that's the first thing we want to find.

Â We could start with R Thevenin's if we wanted to,

Â but we'll start with Voc in this problem.

Â Voc is the open circuit voltage with the load removed.

Â So take our six kilo ohm resistor out.

Â We look back into our circuit from the leads on

Â either side of the six kilo ohm resistor and determine what the voltage is.

Â That's an open circuit voltage for the load.

Â We've taken the load out, and we've replaced it with an open circuit.

Â So let's redraw the circuit above for the Voc condition.

Â So we still have the six-volt source on the left-hand side of the circuit.

Â We have the two-milliamp source on the left-hand side of the circuit.

Â We still have our two and four kilo ohm resistors in the center of the circuit.

Â 3:06

So, if we can solve for Voc, then we can solve for

Â the first variable in our R Thevenin's equivalent circuit.

Â So how do we find Voc?

Â We know the Voc is going to be the sum of voltage drop

Â across the four kilo ohm resistor and

Â across the two kilo ohm resistor at the bottom center of the circuit.

Â 3:28

So how do we find those?

Â We can do that using various techniques, but for this problem,

Â let's use loop analysis.

Â So if we use loop analysis, the first thing we do for loop or

Â mesh analysis is we assign meshes and we assign mesh currents.

Â So that's our first mesh clockwise around the upper left-hand loop, and

Â here's our second mesh.

Â It's the lower left hand loop and going to call it loop two, and

Â have current I sub 2.

Â So we know that Voc in this case is going to be equal to what?

Â It's going to be 4 kilo ohms times I1,

Â because that's the current that's flowing through the four kilo ohm resistor.

Â And then we have the two kilo ohms at the bottom of the circuit, and

Â the voltage drop across there is going to be 2K times I2.

Â So if we can find I1 and I2, then we can find Voc.

Â So let's write our first loop equation, so this is the equation for loop one.

Â The upper left hand loop.

Â Starting at the lower left hand corner,

Â that loop been traveling around in a clockwise fashion.

Â We first encountered the six-volt source and

Â the negative polarity of that six-volt source.

Â So it's minus 6 volts.

Â 4:39

For that voltage drop across the source.

Â We then come around and encounter the four kilo ohm resistor with just I1 flowing

Â through it so the voltage drop there as we just had talked about is

Â four kilo ohms times I sub 1.

Â And then we have one more resistor in this loop, it's the 2 kilo ohm resistor, and

Â we know the voltage drop across it.

Â If we're adding up the voltages in the clock-wise fashion,

Â it will be plus to minus 4 positive current flowing into the two kilo ohm

Â resistor using the pass of sign convention.

Â So, in this case,

Â this voltage drop across two kilo ohm resistor is going to be 2K I1,

Â which is flowing the same direction as we're summing, minus I2.

Â So it's going to be 2K, I sub 1 minus I sub 2,

Â and that's equal to zero.

Â That's our first equation for loop one or mesh one.

Â Now, if we go down the mesh two.

Â And we look at mesh two, we see that if we start at the lower left-hand corner,

Â we're going around this loop.

Â We first encounter this two-milliamp source and we are adding voltage drops.

Â And we don't know what the voltage drop is across this source.

Â And in fact, we can write it in our equation as maybe a V2 milliamp, but

Â that would add another unknown to our equations.

Â And we only have two loops, we'll come up with two independent equations for

Â those loops.

Â So if we added this third unknown,

Â then we wouldn't have enough equations to solve for our unknowns.

Â So instead of doing that,

Â we recognize that this loop has a current source which is just I2.

Â 6:18

So we instantly see from inspection of our loop that I2 is equal to two milliamps.

Â So we can use equation one and

Â equation two from our two loops to find I1 if we wanted to.

Â And if we do that, and we find that I1 is equal to five-thirds milliamps.

Â 6:42

All right, so we can find I1 with our two equations.

Â Once we have those, we know that we can find Voc.

Â And so Voc plugging I1 and

Â I2 into our equation that we have for

Â Voc, Voc comes out to be 32 over 3 volts.

Â That's the voltage Voc.

Â So now up in our equivalent circuit model, we can replace Voc by a number,

Â which is 32-thirds of volts.

Â The other part of what we need for

Â R Thevenin's equivalent circuit is R Thevenin's.

Â So again, we go back to our initial circuit, which is on the lower or

Â the upper left-hand side of our screen.

Â And we, analyze that circuit for R Thevenin's,

Â similar to the way that we did for Voc.

Â So let's do that.

Â We'll do it over here on the right-hand side of our screen.

Â 7:40

So we take our circuit and we redraw it for the R Thevenin's condition.

Â And when we do that, when we're finding R Thevenin's,

Â we're looking back into our circuit.

Â We take the load out and we look back into our circuit and we replace all the voltage

Â sources by short circuits and we replace all the current sources by open circuits.

Â So let's redraw this condition for this problem.

Â 8:14

So, first, voltage source is a short circuit.

Â Current source is a open circuit and we're going to take out our six kilo ohm load.

Â Because we're trying to find the Thevenin's equivalent resistance for

Â circuit A, which is the circuit that we have initially minus the load.

Â We take the load out and we're looking for

Â R Thevenin's looking back into this circuit that I'm redrawing.

Â So, it's going to be R Thevenin's and we have a four kilo ohm resistor at the top,

Â a two kilo ohm resistor at the bottom, and

Â another two kilo ohm resistor on the left-hand side.

Â 9:09

So what is R Thevenin's in this case?

Â We have a two kilo ohm and a four kilo ohm,

Â which are in parallel with one another.

Â So these 2K and 4K resistors here are on parallel with one another, tied to

Â the same point at one end, and they're tied to the same point at the other end.

Â Okay, so there is, in parallel with one another.

Â So R Thevenin's is 4k and parallel with 2k.

Â That parallel combination of resistors is in series

Â with the 2k at the bottom of our circuit.

Â So we would add that to find R Thevenin's.

Â So R Thevenin's for this problem comes out to be

Â ten-thirds kilo ohm.

Â So that's R Thevenin's.

Â So we'd go back up to R Thevenin's equivalent circuit, and

Â now we have something we can use to solve the problem

Â 10:37

And again, we're looking for V out.

Â So this is our equivalent V out to what we would get from analysis of

Â our circuit that we initially had for the problem.

Â And so, if we solve for this, we can simply use voltage division to find V out.

Â V out is going to be our source,

Â 32-thirds volt which is divided between

Â our R Thevenin's of ten-thirds K, and

Â our load resistance of 6K.

Â So we use voltage division to do that calculation.

Â So it's going to be 6 kilo ohms divided by ten-thirds kilo ohms, plus 6 kilo ohms.

Â And so, if we do that calculation,

Â we end up with 48 divided by 7 volts.

Â 48-sevenths volts for V out.

Â