0:03

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

we're working with circuits with independent sources.

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

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

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

on the right hand side of the circuit.

We have two independent sources in this problem.

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

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

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

The general representation for the Thevenin's equivalent circuit is

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

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

0:55

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

We separated the circuit into a circuit A,

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

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

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

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

R Thevenins's.

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

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

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

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

that we can use voltage division to find V out,

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

1:56

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

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

but we'll start with Voc in this problem.

Voc is the open circuit voltage with the load removed.

So take our six kilo ohm resistor out.

We look back into our circuit from the leads on

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

That's an open circuit voltage for the load.

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

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

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

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

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

3:06

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

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

So how do we find Voc?

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

across the four kilo ohm resistor and

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

3:28

So how do we find those?

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

let's use loop analysis.

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

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

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

here's our second mesh.

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

have current I sub 2.

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

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

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

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

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

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

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

The upper left hand loop.

Starting at the lower left hand corner,

that loop been traveling around in a clockwise fashion.

We first encountered the six-volt source and

the negative polarity of that six-volt source.

So it's minus 6 volts.

4:39

For that voltage drop across the source.

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

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

four kilo ohms times I sub 1.

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

we know the voltage drop across it.

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

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

resistor using the pass of sign convention.

So, in this case,

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

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

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

and that's equal to zero.

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

Now, if we go down the mesh two.

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

we're going around this loop.

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

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

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

that would add another unknown to our equations.

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

those loops.

So if we added this third unknown,

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

So instead of doing that,

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

6:18

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

So we can use equation one and

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

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

6:42

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

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

And so Voc plugging I1 and

I2 into our equation that we have for

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

That's the voltage Voc.

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

which is 32-thirds of volts.

The other part of what we need for

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

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

the upper left-hand side of our screen.

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

similar to the way that we did for Voc.

So let's do that.

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

7:40

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

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

we're looking back into our circuit.

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

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

So let's redraw this condition for this problem.

8:14

So, first, voltage source is a short circuit.

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

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

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

We take the load out and we're looking for

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

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

a two kilo ohm resistor at the bottom, and

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

9:09

So what is R Thevenin's in this case?

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

which are in parallel with one another.

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

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

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

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

That parallel combination of resistors is in series

with the 2k at the bottom of our circuit.

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

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

ten-thirds kilo ohm.

So that's R Thevenin's.

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

now we have something we can use to solve the problem

10:37

And again, we're looking for V out.

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

our circuit that we initially had for the problem.

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

V out is going to be our source,

32-thirds volt which is divided between

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

our load resistance of 6K.

So we use voltage division to do that calculation.

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

And so, if we do that calculation,

we end up with 48 divided by 7 volts.

48-sevenths volts for V out.