Proof of Thevenin's Theorem

[brian]

Do any of you have a proof for Thevenin's Theorem, or alternatively, do you know where I can find one? I'll admit that I haven't been to the library yet, but I thought it might be quicker to go this route.

I would like to see it because on a recent EE homework, I got 2 1/2 points deducted for not removing a circuit element and then solving for the Thevenin equivalent. The method I used gives the correct answer, but it is harder to see intuitively. So, I would like to see the proof so that I can understand why the method I used is correct and then present that as justification to my teacher.

I realize that the last paragraph is pretty vague, so if you're curious about the problem and the details, let me know and I can post a scan of the problem on the web, or email it to you for analysis.

Thanks for your time and/or responses!

Brian

[Mike Burgundy]

try the library on theory books, or maybe here:
http://users.telenet.be/educypedia/electronics/electricitycircuits.htm
http://horizon.ece.utexas.edu/~jacome/ee411/lectures/SL5.pdf
http://engr.calvin.edu/courses/engr204/2000/examples/superposition/theory.htm
http://slv.math.adams.edu/nehring/phys221/labs/thevenin.pdf
http://www.mit.edu:8001/people/dimitrib/thevenin.pdf
No idea though if these are of any value.
If your method is sound, you should be able to defend it to your teacher. Be aware of having the right answer, but having got there the wrong way (ie with a different set of data it could all go pear-shaped)

[kroushl]

To find the thevenin equivalent of a circuit, find the open circuit voltage across the terminals you're finding the equivalent for and then the closed circuit current through the terminals.  I prefer the mesh current method for circuit analysis, but use whatever you want to get the right values.  The thevenin source is the open circuit voltage you find and the thevenin resistance (or impedance, if you're working with complex numbers) is Vopen/Iclosed.

Also, the voltage source must be independant.  If you have dependant sources in the problem, you must find a way to solve the circuit to produce an independent source.

Hope it helps...

If it doesn't work out, would you mind posting your problem?

Later,
Brad

[Transmogrifox]

You can probably prove it to yourself easily.  Try messing with these variables.

Rs == resistor in series with a voltage source.
Rp == resistor in parallel with current source

Rp==Rs is the first premise you will make.

Zl = any arbitrary impedance series connected to the voltage source in series with the resistor, or parallel connected to the current source in parallel with the resistor.

V = IR

Write the expression for the voltage across Zl in both cases and if you use the thevenin equivalent, both will be equal...

or, solve for V source and I source in each case and you will find that Thevenin's theorem is the relationship between the two.

I hope that helps.

[brian]

Next post is a scan of the problem. The specific problem is to find the voltage across the 6 Amp source. The series resistor on the top of the circuit is 3 ohms. The parallel elements (l to r) are a 6A source, a 3A source, and a 6 ohm resistance. The method the book and the teacher wanted me to use was to remove the 6A source from the circuit and solve the remaining part using Thevenin's. Then, I would replace the 6A source and solve for the voltage across it.

As you can see, what I did was place the Thevenin terminals 'a' & 'b' in parallel with the current sources and the 6 ohm resistance. Then, the Thevenin equivalent voltage, which is the open-circuit voltage across the terminals would be equivalent to the voltage across the 6A source. I think this is valid and so did the teacher.

So, my point was that Thevenin's Theorem does not require a circuit element to be removed, so how could I lose points for that reason? Basically, we are in agreement with that also. Her problem was that she has never solved circuits using Thevenin's without removing an element. She kept telling me that the Thevenin resistance was superfluous in my solution. Her assertion was that I solved the circuit using Nodal Analysis and not Thevenin's. I agreed that I used Nodal, but in the process of finding the Thevenin voltage, you HAVE to use either Nodal or Mesh Analysis. I maintained that this solution was more general, because you could analyze the effect of any load between terminals 'a' and 'b'.

Finally, the problem she was having was seeing where the 6A source had 30 Volts across it. Because we only learned the application and not the proof, I couldn't really justify my answer other than by saying that originally, the 6A source is in parallel with terminals 'a' and 'b' and since the Thevenin voltage across 'a' and 'b' is 30 Volts, the voltage across the 6A source is 30 Volts.

Thanks for your help.

Brian

[brian]

[travissk]

You can prove Thevenin by using superposition: the following is more or less from a textbook I have, Analysis and Design of Linear Circuits.

Consider a  circuit made up of two subcircuits, source (S) and load (L)

    v+
(S)==(L)
    v-

Where the = are two wires carrying current from S to L. Disconnect the load circuit and apply a test current i1. Then, to find the corresponding v1 (voltage across those two wires with L removed), use superposition to turn off i1 and leave all sources inside S on. This leaves an open circuit, so V1 = V(oc).
Now turn i1 back on and turn all sources inside S off. The resulting circuit is linear, so v2 = R(eq)*(-i1) by ohm's loaw.

Actually a picture would really help here; I can try to make one up in Paint Shop later tonight.

If you followed that (sorry that it's about as clear as mud), then you can add the two voltages through superposition to get the total voltage:
Vt = v1+v2 = v(oc)-R(eq)*i1

Which gets you the same thing Thevenin's does.

I'm not sure how much of a "proof" this is; it's more of a derivation but with some pictures it's actually pretty simple, just isolating two things and summing their respective results to get the total voltage.

[Boofhead]

Presumably the idea is to find the Thevenin equivalent of the this whole circuit.   It can be done very easily if you understand the idea behind the Thevenin equivalent circuit - that is any linear circuit  (involving only R's and sources) can be represented by a voltage source and series resistance.

Also to demonstrate you understand the idea you should not be using nework reduction, node voltage solutions or super position for that matter.

Here's the main ideas:

- The Thevenin resistance is the open circuit voltage divided by the short circuit current.  Once you find any two of these variables the other follows.

- The Thevenin resistance is the the resistance looking into the output terminals when the current sources are open circuited and the voltage sources are short circuited.


For that cirucit the short circuirt current is trivial to find:

Isc = 18/3 + 6 + 3 = 15A

The resistance looking into terminals A and B is 3ohm // 6ohm = 2ohm, so Rth = 2 ohm.

Vth (= Voc)  = Isc * Rth = 30V

The choice of using Isc or Voc is up to you both are valid, it's just some circuits are easier one way.  In this case the current method is much simpler and is a more direct application of Theven's idea.  To get the open circuit voltage you need to use superposition or a circuit solution - IMHO it's not the intention to do this.

In practice all methods lead to the same answer but usual one is much easier - the skill is know what tools a best suited to the problem.

Yet another easy solution is to apply the Norton <-> Thevenin transformation on the 18V source and 3ohm resistor. That gives you a Norton source of 18/3 = 6A in parallel witha 3 ohm resistor.  From that you   add the currents and parallel the resistances to get a total Norton cirucit then convert that back to Thevenin.  This method is quick but it's not a good as the above method which lets you do the whole problem in your head if you think about it.

[eliktronik]

You actually made a small mistake when you calculated vth. You said vth = (node voltage eq. for v1) = 0... Which might have made the professor think you didn't quite understand what was going on, even though you found v1 and got the right answer, since v1 = vth.

[brian]

actually, i think the equation is correct. i applied nodal analysis and wrote a KCL equation for which the sum of the currents out of the node = 0. i didn't, however, explicitly note that so i understand what you are saying about Vth = 0.

we didn't cover the short circuit current method that Boofhead mentioned, but it makes sense and is a quick way to solution. we also didn't cover source transformation, although i did read that section and i applied it to a few of the homework problems.

i have already discussed this with my teacher and she's giving me 2 points back so i guess what i'm really interested in now is whether or not i'm right about the voltage across the 6A source = Vth. I know it gives the correct answer in this case, but i'd like to know if it is just a lucky coincidence in this case, or if this method is valid in general.

I know the voltage across the 6A source is 30 volts. Do you agree with my assertion that Vth = Vx ?

THanks,
brian

[Boofhead]

Do you agree with my assertion that Vth = Vx ?

I couldn't read the symbols on the scan properly, but if it's the voltage across the 6A source then yes it has to be for the reasons you quoted: it's across the A/B terminals.  This is true provide the 6A source is left in situ and nothing is else connected to A/B.

Edit:

Forgot to mention, IMHO, the short circuit idea and the source transformation idea are the key tools in solving problems with Thevenin.  I honestly can't see how the course could be tought like without these - I'm sure it's all bundled in the same place in the text and for good reason too.

[brian]

Boofhead,

You're right, it is bundled in there. They present a sort of roundabout way of doing it before they introduce the Rth = Voc / Isc, but it is in there. When they introduce Norton's, they call it a dual of Thevenin's and then go on to explain that "There is a systematic duality possible in circuit analysis which replaces the current sources with voltage sources, voltage with current, mesh equations with nodal equations, etc."

The text stresses the mesh and nodal methods and then treats Thevenin's and Norton's as more or less special cases of mesh and nodal. You remove a circuit element and replace it with terminals a and b. Then you apply mesh or nodal to solve for voltages and currents, etc.

My teacher was upset about my solution because she says she has never used Thevenin's to solve a circuit without removing a circuit element and because of that, she couldn't be sure that my solution was correct. She told me that for the purposes of the class, I should solve Thevenin's problems by removing an element and using the procedure outlined in the text. Unfortunately, I've been running into problems blindly following textbook procedures and I would rather understand the means to solution than just get the right answer. Such is life, I suppose.

Thanks again for all of your help!

Brian

[Boofhead]

duality possible

The duality stuff is normal, transforming the circuit (or parts of it) into it's dual sometimes lets you see things clearer, or lets you solve the problem easier.

Thevenin's and Norton's as more or less special cases of mesh and nodal.

There's two aspects to Thevenin/Norton.  The first is a tool to bend circuits around to make the solution very simple - like your problem for example.  When I look at a circuit I try to find "opportunities" to bend parts of it around to make life easier, sometimes it does't help but when it does it's great.  The second aspect is bending circuits around so you can apply some *mechanistic methods* of writing mesh and node equations, this is just mechanical process in itself-it doesn't use Thevenin/Norton as a tool for solving circuits.

without removing a circuit element

That's kind of restrictive an unnecessary, you don't have to have another circuit element/source connected to the Thevenin outputs to use Thevenin-  it's perfectly  calid to transform the whole circuit and have an open set of terminals.

Believe it or not I used all those circuit techniques to solve real problems, it's not just crud in the textbook, it's very fundamental stuff.  I'll admit I use PSPICE a lot these days on specific circuits.

Good luck with it all.

[Brian Marshall]

the whole point of a theorem is that it cant be proven.

it goes in this order.....

hypothesis
theory
theorem
law

That's what i remember anyways.

[Boofhead]

the whole point of a theorem is that it cant be proven.

I don't think that's true, theorems are *always* proven - actually mathematicians spend their life proving theorems.

Thevenin's theorem follows from superposition.  Superposition follows from the definition of a linear circuit.

The things you can't prove are axioms/postulates.  The laws of physics can be written as postulates - "they just are".

[Brian Marshall]

1. An idea that has been demonstrated as true or is assumed to be so demonstrable.
2. Mathematics A proposition that has been or is to be proved on the basis of explicit assumptions.

sounds like probably maybe to me.  Baically as i understand the defininition it is believed to be true, and no evidence exists to say it isnt.