Is there any algebraic proof for Thevenin's theorem?
Yes. Have you searched for it? https://en.wikipedia.org/wiki/Thévenin's_theorem
Wikipedia did not provide any algebraic proof, they only provided a statement and an examples, which is not clear
Yes, but the references in the Wikipedia article cite this
Johnson, D.H. (2003a). "Origins of the equivalent circuit concept: the voltage-source equivalent"(PDF).Proceedings of the IEEE91 (4): 636–640
I have checked that paper but there i can only see a statement of the theorem and a little introduction to what thevenin did then, though i they made reference to a circuit that proof the theorem, there is no algebraic or a more vigorous mathematical proof that shows that.
http://www.ohio.edu/people/starzykj/network/Class/ee616/homework/equivalent circuits.pdf has a proof, but does not prove the superposition theorem.
http://nptel.ac.in/courses/108105053/pdf/L-08(GDR)(ET) ((EE)NPTEL).pdf section L8.5
Maybe think of it as a derivation rather than a proof.
I think thats a verification rather than a derivative, in a derivative you assume, you dont have to quote any part of the theorem until you completely derive it. However that was insightful, when i did derive it myself, i did not realize other people have verified it. and its only when you derive it that you will understand the shortcomings of the theorem, it will be very difficult to realize it when you only try verifying it. In verifying it i found out that only the voltage across the resistive circuit other than the equivalent circuit remains consistent with the original circuit, but the voltages across the resistors in the equivalent circuit, become inconsistent with the original circuit, hence for very complicated purely resistive circuits, it will be wrong to apply Thevenin, especially with circuits with more than 4 rows and colums of nodes
Are you saying you can devise a linear circuit to which Thévenin's Theorem does not apply?
Yes, i will upload it here as soon as i device it
this should be interesting
the point of the thevenin circuit is that the load will see the same voltage and current. The voltage across the equivalent resistor means nothing
Interesting indeed. Can nonlinear behavior arise from linear components? I suspect not. But how to prove that is not obvious.
It is impossible to create a 2 port network from fixed resistors, fixed voltage sources and fixed current sources that will not behave identically at the two ports as the thevenin equivalent for that network. Period. End of Issue.
Of course, because the circuit topology is different.
Comparing the behavior internally is irrelevant/meaningless. I'm not sure what you are even getting at with that. It only matters what happens at the two ports. Why do you care anything about the internal nodes?
I am not comparing it internally for no reason, you need to understand that by finding the equivalence circuit which is infact not equivalent to the actual circuit--it only tells you the equivalence of the circuit for a particular load--your new circuit would be misrepresenting the history of the circuit save one. It is best to explain a problem without including another problem. Imust say that at this point i must reveal the real purpose of this thread, i have been able to derive Thevenin's equation myself, by assuming i know nothing of it. I derived it using algebra and i published. The theorem might be useful but it actually assumes a false state to solve a problem. The other methods icluding Kirchoff solve circuit problems without assuming a false state. So many undergraduate studets find it hard to understand Thevenin, this is because if you present the equivalent circuit to me after applying Thevenin, i would find every other thing wrong except the load. The problem is find the voltage across a load. What you do is find another voltage inorder to get the voltage across the load, what you have done is create another problem, there was initially a voltage, hence it is almost irrelevant to find another one, Find the equivalent circuit for the problem below with R7 as the load below , i am sure you would opt for something else.
I think you misunderstand the meaning of equivalent. The equivalent behaves the same as the original only from the point of view of the two external terminals.
In power systems analysis, I can make you an equivalent of Canada as seen from two terminals in Niagara. There is no claim that it models everything inside Canada. (Joke about Canada omitted :-)
I very much understand the meaning of the equivalent, it only means an equivalent circuit for the same voltage across the load, like i said earlier on i have an algebraic derivation for it. If you also say it behaves the same way as the original i would say that statement is not clear as thevenin;s voltage only assumes the same propery of the original voltage across that particular load, everywhere else its just inconsistent. The above question is just a simple example that i think you should solve using thevenin, i am now trying to derive aa more complex circuit with two or more voltages which would prove inapplicable with thevenin. I think i should have given this thread a different title "Shortcomings of Thevenin;s theorem". I hope you understand me
There is no "everywhere else". It is meaningful only at the terminals.
All Thevenin's and Norton's theorems amount to is that, for a linear circuit, the relationship between the voltage across any two points in the circuit and the current through those same two points (those two points are the "terminals") has a linear relationship that can be written as:
Vx = Vth - Ix*Req
This is the "load line" for the circuit between those two terminals.
But this is exactly the same relationship that you get if you replace the circuit with a voltage source with an output of Vth in series with a resistance of value Req -- so these two circuits are equivalent as far as what happens between the two terminals.
Tnk God you now understand me, i am infact not saying Thevenin is wrong, he was rigt on point with his specifics, but it not that thevenin helps you to find voltage across the load directly, it rather finds another voltage.
thats not correct. It does find the voltage across the load.
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