Thevenin-equivalence

Hi Rikard, It's many, many years since I studied this stuff, so I could be making an error too, but I end up with an Rth of 92.19 ohms, (92.18749896 ohms), and an Ith of 43.389831mA. The error of 0.07 ohms might be due to rounding during my calculations, or I might be totally wrong here. Does this correlate with any of your (wrong) answers? (I won't explain how I came to this answer, since it's wrong.) I do, however, get the correct Vth of -4v, with A=12V and B=16V. I've inserted my reasoning in your text above.

... johnny

Something's fishy with your equations. Your diagram shows I1 flowing through R2 and R3 and I2 flowing through R4 and R5, but you then choose to make a mesh by including Vth at AB as a circuit element. Then you write your equations with I2 flowing through R2 and R3 (which it doesn't according to the diagram, and if you make the meshes as you say, there should be different mesh currents in R2 and R3.

If you want to write consistent equations you should use a consistent model.

Here's how I would go about this problem (there are several ways to arrive at the solution, but everyone has their own favorite approaches).

First I would note that there's an open circuit at AB, so we have a couple of voltage dividers formed by the pairs R2&R3 and R4&R4. So it would be nice to know the voltage at the top of these dividers. For this purpose we redraw the circuit as follows (combining R2+R3 and R4+R5):

Now, 100||300 is 75 Ohms, so we have a simple divider:

V1 = 32V*75/(25+75) = 24V

Fine. We have 24V at the top of the two voltage dividers. We can now fine the voltages at A and B quite easily since they're just trivial voltage dividers also:

Va = 24V*50/(50+50) = 12V

Vb = 24V*200/(200+100) = 16V

So we have:

Vth = Va - Vb = 12V - 16V = -4V

That takes care of the open circuit (thevenin) voltage.

To find the short circuit current, short out AB and redraw showing the three resulting meshes:

The equations:

-50i1 + 150i2 - 0i3 = 0V (2)

-50i1 - 0i2 + 250i3 - 0V (3)

Solve for i2 and i3:

i2 = (32/295)A = 108.475mA

i3 = (96/1475)A = 65.085mA

Then Iab = i3 - i2 = -43.390mA

So the Thevenin resistance is Vth/Iab = 4V/43.29mA = 92.2 Ohms

Do you notice that R1 is in parallel with the other two branches? You can combine the series resistors and will have three parallel branches.

Solve the parallel combination.

Use this resistance to find the current with the source voltage.

Draw the current flow. Current times R1 will allow you to determine what voltage you have across the other two branches.

Find the voltage across R4 and R5. The difference is VAB

Tom

I would have seen it, if it wasn't for A and B. Since they're between the R2/R4 and R3/R5 respectively, they can't be in parallel.

Or can they?

Please correct me if I'm wrong. What I mean is, if I replace R2/R4 with a serial one, then the point A will be lost since it's supposed to be in between.

After considering your explaination of finding Vth, I find it perfectly clear of what I did wrong. The problem is, the lecturer usually includes the open circuit voltage in his mesh traversions and it seems to work for him.

But by reading your post I understand that since no current flows through A and B, I cannot use Kirchoff's laws of current to find the voltage and that I have to use voltage dividing instead.

Reading through this example a couple of times, I'm getting a grip of what you're doing. I think the big hurdle for me is when I've got a lot of linear equations like this, and so eliminate all variables except one, and then going backwards.

Thank you Greg for taking time replying. Atleast you got me halfway back on track. I think I will have to practice more on the linear equations kind of math.

-- Rikard.

him.

You _can_ include the Vth element in a mesh style of solution, but it adds one extra unknown (Vth) so one extra equation is required. In other words, the set of linear equation to solve is slightly more complex. This may not be a problem if you've access to a computer to take care of the grunt work, but I find that simple, small steps are best when working by hand.

So, let's take a look at what the mesh equations would look like if you include the Vth unknown. You still have three meshes as I detailed previously, only this time we include the Vth as a circuit element between points A and B.

-50i1 + 150i2 = Vth

-50i1 + 250i3 = Vth

Since we have four unknowns, we need one more equation. Since AB is actually an open circuit, we must have

i2 + i3 = 0

You can solve the above four equations to find Vth.

You'd then need to set Vth = 0 in the first three equations (to "force" a short circuit between A and B) and solve for the currents i2 and i3 again, since then i2 will not equal i3, and the short circuit current from A to B will be given by i3 - i2.

As I said above, personally I like to solve these sorts of problems in short, simple steps so it's less likely to make a mistake when under pressure.

Another approach to finding the Thevenin resistance in this circuit would be to perform a Delta-Y conversion of R1,R2, and R3 (after replacing the voltage source with a short circuit) to yield a circuit amenable directly to reduction to a single resistance:

Ra = R1*R2/(R1 + R2 + R3) = 10 Ohms Rb = R1*R3/(R1 + R2 + R3) = 10 Ohms Rc = R2*R3/(R1 + R2 + R3) = 20 Ohms

Giving the following equivalent circuit:

Which leads to:

and

so that Rth = 92.2 Ohms as before.

Yes, Point AB will be lost, until you find the main current. Then you will back up and reconstruct the original circuit and distribute the current. The is an infinite resistance between AB. It is not there as far as this step is concerned. Do It.

Tom

On Sat, 07 Oct 2006 12:00:53 GMT, in message , Rikard Bosnjakovic scribed:

Referencing "Electric Circuit Analysis," Taber/Silgalis:

"Step 1. Remove the element or elements in the circuit that were designated as the load."

In your case, the load is between A and B, correct?

"Step 2. Determine the open-circuit voltage across terminals A-B, call it Voc."

This is what you are attempting to do by your re-write. However, you over complicate the process by introducing mesh equations and then mixing them with current node equations. Simplify your thought process by using the Voltage Divider Law.

----------------------- C | | | > 25 > 50 > 100 > R1 > R2 > R4 | | | | + A + B | | | _ 32 > 50 > 200 - E > R3 > R5 | | | ----------------------- D

First, find VC.

Rp = R2+R3 || R4+R5 = 75

VC = E * Rp / (R1 + Rp) = 24

Now find the two voltages, A and B.

VA = VC * R3 / (R2+R3) = 12

VB = VC * R5 / (R4+R5) = 16

Vth = VA - VB = -4

"Step 3. Replace the voltage supply E by its internal resistance... Find the series resistance...."

In this case, as you found out, the resistance network is difficult to calculate. But, knowing Vth, and redrawing the circuit, current values can be set up and Ith found.

--> Ith A -------------------------- | --> | > 50 I2 > 50 Vth > R2 > R3 | R1 | c +---------^^^---------+ D | 25 |

-4 > 100 > 200 > R4 I3 > R5 | E - R1*I - R2*I1 - R3*(I1-Ith) = 0 (3) M2: R3*(I1-Ith) - R5*(Ith+I2) = 0 => R3*(I1-Ith) = R5*(Ik+Ith)

Why go to all that complication? The solution is obvious and trivial:

Firstly, R2, R3, R4, and R5 are a bridge circuit, so we can treat it as two potential dividers, R2/R3, and R4/R5.

We need to know the voltage at point R1/R2/R4, relative to the negative terminal of the battery.

R2/R3, and R4/R5 simplify to 300 ohms in parallel with 100 ohms, or 75 ohms.

So, in effect we have 75 ohms in series with 25 ohms, and the voltage at R1/R2/R4 is 75/(75+25) = 75/100 = 3/4 times 32 volts, which is 24 volts.

The voltage at point A will be 50/100, ie. half this, which is 12 volts.

The voltage at point B will be 200/300. or 2/3 times 24 volts, which is 16 volts.

Hence the voltage at A with respect to B is 12 - 16, or -4 Volts.

No need for complicated mesh or nodal analysis.

Maybe they don't teach circuit recognition anymore :-(

Now tell me what the voltage between A and B would be if R4=200 ;-(

Now for the Thevenin equivalent resistance:

This is defined as the Thevenin equivalent voltage divided by the current which will flow if the output is shorted, so:

Short points A and B

The circuit now reduces to a simple potential divider, 25 ohms, 33.33 ohms, and 40 ohms, from which we can calculate the voltages at R1/R2/R4, and (R2/R3, shorted to R4/R5)

The voltage across R3 will be 13.017V The voltage across R2 will be (23.864-13.017) = 10.847V

The current in R2 will be 10.847/50 = 216.94mA The current in R3 will be 13.017/50 = 260.34mA

Hence the current in the short circuit is 260.34 - 217.94 = 43.4mA

The Thevenin output resistance is therefore 4 / 43.4e-3 = 92.166 ohms.

Which is as near the published result as makes no difference, allowing for rounding errors.

Agreed. I think I used mesh equation because the book (or lecturer, I don't remember) told us to. The way you and others showed it, by using voltage divider instead, resulted in a much simpler circuit.

I do however expect that I lack knowledge for mesh equations and I need to study them more. I fear that some of the exam questions will be something like "Do this and that with mesh equations", and if I solve it in other ways I will most likely fail that question.

This description and your path of thought made perfect sense to me, and I finally managed to solve this part on my own. I even tried with different values and got the answer correct.

Thank you.

I did however have some problems in trying to find out how you got these three equations. When I drew the same circuit as above, I got no less than

At top-left point; Ith entering, I1 going right, I2 going down. At point C, I2 entering from above, I3 going right, I4 going down. At point D, I1 entering from above, I3 entering from right, I5 going down. At bottom-left point, Ith leaving left, I4 entering from above, I5 entering from right.

I.e.:

Some of them are not linearly dependant, so I can reduce them:

Ith = I1 + I3 + I4 (1+2) Ith = I1 + (I5 - I1) + I4 (1+2+3) Ith = I4 + I5

So I take it you reduced the number of unknowns to three, and put into three equations. I think this makes sense.

However, can you please elaborate how you got the constants?

The first row for example. -4 on the left, I take it you start the mesh walk at point A. I'm not at all sure about the right side of the equation.

They're not part of that mesh path, are they?

Quick 'n dirty guide to writing mesh equations:

1. Identify and label (number) the meshes in the circuit.
2. Identify the mesh currents by sketching in a clockwise pointing current arrow in each mesh.
3. Write the mesh equations by inspection as follows. a. Create a 'blank' matrix equation that you will proceed to fill in. It will have as many rows as there are meshes. So for example, for a three mesh circuit you would write a 3x3 equation:

| | |i1| | | | |*|i2| = | | | | |23| | |

b. The entries in the matrix position ij are: i=j : The sum of the resistances around mesh i i/=j: Negative sum of the resistances shared by meshes i and j

c. The entries in the column vector on the right are: Row i is the sum of the voltage rises due to active voltage sources as you proceed clockwise around mesh i.

Example:

There are four meshes which are numbered. I didn't sketch in the currents because ascii's a bitch.

The diagonal entries will be: The Voltage vector will be: (1,1): 50+100+30 = 180 Ohms (1): +12V (2,2): 200+60+30 = 290 Ohms (2): -16V (3,3): 300+40 = 340 Ohms (3): +16V + 8V = 24V (4,4): 60+40+300 = 400 Ohms (4): -7V

The off-diagonal entries:

(1,2): -30 Ohms The complete equation: (1,3): 0 Ohms (1,4): 0 Ohms |180 -30 0 0 ||i1| | 12V| (2,1): -30 Ohms |-30 290 0 -60 ||i2| |-16V| (2,3): 0 Ohms | 0 0 340 -40 ||i3| = | 24V| (2,4): -60 Ohms | 0 -60 -40 400 ||i4| | -7V| (3,1): 0 Ohms (3,2): 0 Ohms Note that there is a symmetry about (3,4): -40 Ohms the diagonal that simplifies filling (4,1): 0 Ohms out the matrix. (4,2): -60 Ohms (4,3): -40 Ohms

With a bit of practice you should be able to write down the matrix equations by inspection.

On Sun, 08 Oct 2006 14:27:05 GMT, in message , Rikard Bosnjakovic scribed:

OK, the problem is that my drawing is crude. I meant the three currents to mean three mesh currents, and I had a comment to that effect. I mean Ith to be the leftmost mesh current, so the voltage drops around that mesh are:

-4 (the applied voltage) = 150Ith (100 + 50 ohms multiplied by Ith)

- 50I2 (subtract 50 ohms times I2 because the I2 mesh current is in the opposite direction)

- 100I3 (subtract 100 ohms times I3 for the same reason)

The other two equations take the positive value as the total of the respective mesh current times the total resistance in the loop, subtracting the opposite direction currents through the shared resistances.

Does that make sense?

On Sun, 08 Oct 2006 14:27:05 GMT, in message , Rikard Bosnjakovic scribed:

Redrawn:

A -------------------------- >---| | |------->------| | | > 50 ^ I2 V > 50 Vth | > R2 |------- R3 | | R1 |

-4 V +---------^^^---------+ D | | 25 | | > 100 |---->--------| > 200 Ith | > R4 ^ I3 V > R5

It's been a long time, but as I recall whenever a teacher or text book wanted to teach Kirchhoff's laws or mesh circuits, etc, they would provide a problem with at least 2 power supplies in it. Your problem only has one power supply, so I doubt if it was meant to be solved using mesh circuits. This is a simple voltage divider problem. It can be solved using the basic formulas at:

The voltage drop behind R1 and over the two parallel branches is:

V(branches) = 32 (75/100) = 24V

The voltage drop over R3 is:

V(R3) = 24 (50/100) = 12V

The voltage drop over R5 is: V(R5) = 24 (200/300) = 16V

The potential difference is +/- 4V depending on your reference.