Talk:Thévenin's theorem

Example requires further explanation
Why (in the current example picture) is R1 ignored when calculating the Vth? —Preceding unsigned comment added by 24.205.234.129 (talk • contribs)


 * It isn't. Why do you think it is?  Looking at the circuit from the output, you see R1 + ...  Let me add  the equation to the article. — Omegatron 20:32, 22 May 2006 (UTC)
 * I added the equation with Rs in it. Clearer now? — Omegatron 20:38, 22 May 2006 (UTC)

I can see where it looks like R1 was ignored. You are probably trying to assume that the voltage at A is 0 V. However, you are trying to find the voltage at A, so you can't assume that it is 0 V.  Therefore R1 is in an open circuit by itself, so the voltage must be the same on both sides of R1 because of Kirchhoff's Voltage Law. /n0:04 CST 06 Oct 2006

Example is (was) wrong
A reader's note: I believe the example is wrong -- according to my calculations, Vth should be 13.8V, not 172.5V. Please check.
 * I too, have done this calculation and got 13.8V. (When you think about it, it can't possibly be MORE than 15 volts). I think the diagram is too confusing, as a simple example, to begin with. (I would use 1k, 1k, 2k). I'd like to replace it with another diagram, but I don't know how to go about doing that. Can you wiki pictures? Or will I have to change the tags? I'm going to post this in "Pages needing attention." - Sim
 * Changed the diagram. This new version is a .jpg file. I have not deleted the old .png file, as I do not know the standard for such things. Sim 15:04, Feb 13, 2004 (UTC)
 * Diagrams should be in PNG, for the record. — Omegatron 20:40, 22 May 2006 (UTC)

Accent mark?
should this be thévenin's theorem with an accent mark? - Omegatron 15:29, Dec 23, 2004 (UTC)

Example not clear
V_{AB} example is not clear, i.e. it isn't obvious which of Rs correspond to which resistors on the diagram. - imbaczek 17 Apr 2005


 * oops. looks like i left out the numbers.  fixing... - Omegatron 22:34, Apr 17, 2005 (UTC)


 * I added more labels. did i miss anything? - Omegatron 23:08, Apr 17, 2005 (UTC)

note about non-resistive parts
For generalized impedances, the non-resistive parts are apparently folded into the voltage source, so the voltage is a function of frequency. - Omegatron 03:34, Jun 25, 2005 (UTC)

Conversion to norton
source not that its hard to calculate on your own. Comments on my pretty large changes? Fresheneesz 05:12, 14 April 2006 (UTC)

Impendances - not just resistances
I think it would be very very very helpful if this were generalized to impedances. I assume this is the right place to do that. I also assume that this would work with impedances (given that the voltage source would be sinusoidal in some way like omegatron said). Can this stuff be extrapolated to the frequency domain? Fresheneesz 05:15, 14 April 2006 (UTC)


 * Yes it will work with impedances but the equivilent circuit will contain a component whose impedance is a non standard function of frequency. The source in the equivilent circuit will ofc only have outputs at the frequencies the original sources did. Plugwash 10:45, 14 April 2006 (UTC)


 * I thought the impedance was standard and the source varies with frequency.  — Omegatron 12:58, 14 April 2006 (UTC)


 * No both would need to vary with frequency as the impedance of the parts in the original circuit will vary with frequency. The equivilent series impedance is determined by removing the sources from the circuit so if the original impedances are functions of frequency (which they will be unless they are purely resistances) then the resulting impedance will also be a function of frequency. As for the sources they will only exist at spot frequencies being zero everywhere else)Plugwash 23:17, 14 April 2006 (UTC)


 * I would have assumed that impedances would be standard complex impedances, and that the voltage source would be time varying. Complex impedances take into account the frequncy of the source, but of course I'm not an expert so varying complex impedances may well be neccessary in that case.
 * Is thevenin's theorem used in frequency domain, or does it make the circuit more cumbersome? How would the steps have to change to find the equivelant in the frequency domain? Fresheneesz 02:50, 15 April 2006 (UTC)


 * I meant a single complex impedance for all frequencies, but now that I think about it, it makes sense that they would both vary with frequency. Someone told me that the impedance part doesn't change, but I have no real reason to believe that. — Omegatron 05:00, 15 April 2006 (UTC)
 * I knew you meant complex : ). But I didn't know if he took it that way. Fresheneesz 05:39, 15 April 2006 (UTC)
 * Yes this theorem holds even if to be replaced it with AC parameters, such as reactance of L and C. It is called Ho-Thevenin's theorem (鳳-Thevenin's theorem) which is an extension of Thevenin's one, and is well known in Japan. It is described in the Japanese version of Wikipedia. Discharger1016 (talk) 15:08, 5 December 2020 (UTC)

Doonesbury
It's not notable. Albert Einstein is notable, Garfield is notable, but a mention of Einstein in a Garfield strip would certainly not be notable, and would not be in the Einstein article. — Omegatron 15:12, 21 May 2006 (UTC)
 * Normally, I would be the first to agree with you, there are too many articles in Wikipedia whose "in popular culture," section is both useless and distracting. A prime example is the article on the poem "O Captain! My Captain!," whose "popular culture," section is longer than any other, save the poem itslef.  However, in this case, I would like to beg your indulgence to disagree with you.  In your analagy, "Einstein," is a bit too notable---references to him in popular culture are a dime a dozen.  Unfortunately, Thévenin does not enjoy quite the same reputation.  Myself, even though I have had enough physics classes to qualify for a minor in the subject, the Doonsebury strip is the first time I have heard of this therom.  I looked it up because to me "which is the current source, and which is the voltage," did not make any sence.  Not that the article helped clarify my confusion, but I did learn something new.  As will, I think, a future peruser of the article from this very small "popular culture," section---it is a very rare mention of this subject in the broader world.  --VonWoland 19:08, 21 May 2006 (UTC)

Distinguishing Thévenin and Norton circuits
Re: a recent comment added to the article and removed by me:

Viewed from outside "the black box", there's no "electrical" way to distinguish a Thévenin circuit from a Norton circuit. In the real world, of course, you can tell the difference, at least eventually: With no load applied, the Thévenin black box consumes no power so it doesn't get hot and can keep outputing its voltage (into no load) forever. By comparison, the Norton circuit always dissipates power so it gets warm and, short of magic or perpetual motion, eventually runs down.

But viewed in the short run with voltmeters, ammeters, voltage sources, current sources, and/or resistors, there's no way to distinguish what's inside "the black box".

Atlant 20:47, 18 January 2007 (UTC)


 * Just like the Doonesbury cartoon, it's basically nonsense, since there's no such thing as an ideal current source or an ideal voltage source; and if there was an ideal voltage source, who is to say what its power dissipation would be? Certainly not zero, even for a battery.  But it's interesting nonetheless.  On the other hand, it only belongs in the article if there's a source we can quote for the analysis.  Do we have one? Dicklyon 22:36, 18 January 2007 (UTC)


 * Sure we have a reference: Horowitz and Hill, The Art of Electronics, page 11; this is basic EE stuff and any EE text will confirm the Thévenin/Norton equivalence. And, of course, if it's an ideal voltage source, it has no resistance so no power dissipation.


 * Atlant 00:52, 19 January 2007 (UTC)


 * What do H&H say about distinguishing a box by its temperature? And where is it defined that an ideal voltage source has no power dissipation?  Does the standard idealization have anything to say about the internal structure or power source of it?  I suppose that's a reasonable idealization, which could also be applied to an ideal current source, but do they do so in that book?  If so, then distinguishing the boxes via instruments would be pretty easy, as long as one of the instruments could measure temperature or thermal flux in some way. Dicklyon 01:09, 19 January 2007 (UTC)


 * H&H define an ideal voltage sources and ideal current sources on page 9.


 * Atlant 01:20, 19 January 2007 (UTC)


 * OK, I checked it out on Amazon. Sure enough, nothing there is relevant to the question of whether one can distinguish the black boxes by their power dissipation or temperature.  They do not suggest that an ideal voltage source has zero power dissipation: "A perfect voltage source is a two-terminal black box that maintains a fixed voltage drop across its terminals,..." which allows arbitrary and lossy machinery inside the black box, which is why I said the thing was "basically nonsense"; it could be made meaningful, but I don't know any sources that do so, so let's drop it. Dicklyon 04:10, 19 January 2007 (UTC)


 * Friends, i am an electronics and communication engineer. I have to say that the question of an existant ideal voltage source doesn't arise. This is because, Thevenin has never told that there exists one. He has clearly explained that, "Once the internal resistance of a voltage source is accounted for by considering an equivalent resistance in series with the voltage source we consider, we can idealize the voltage source". It is as simple as that. Any queries can be mailed to my e-mail id- kalyan.themusicdude@gmail.com — Preceding unsigned comment added by KALYAN T.V. (talk • contribs) 04:39, 18 December 2010 (UTC)

Conversion to a Thévenin equivalent - wrong?
The following comment was added on my talk page Fresheneesz 17:49, 14 May 2007 (UTC) :


 * The direction of the current source in "Conversion to a Thévenin equivalent" is incorrect. It would generate a negative voltage across the output terminals. The Thevenin equivalent circuit generates a positive voltage. —The preceding unsigned comment was added by THeeren (talk • contribs) 17:20, 14 May 2007 (UTC).


 * I don't think he's correct, but I'm putting his comment here just to make sure. Fresheneesz 17:49, 14 May 2007 (UTC)


 * I don't even remember what I was talking about. I can't find it anymore, so I assume you are correct. THeeren (talk) 18:30, 16 January 2008 (UTC)

Other "equivalents"
I was wondering if theres other equivalent circuits (other than thevenin's and norton's) - ones perhaps with capacitances and maybe inductances. A model with capacitances would be very useful in a lot of cases, so I would think they exist. Fresheneesz 17:52, 14 May 2007 (UTC)
 * Thevinins and nortons do generalise somewhat, you can have a source with a periodic waveform represented by a mix of sinewave components in series with a frequency dependent impedance. Plugwash 17:59, 14 May 2007 (UTC)


 * Thevenin/Norton's theorem applies to general complex impedances, not only resistors. So you can use them to find equivalent circuits for ones with inductors and caps. Roger 18:01, 14 May 2007 (UTC)


 * (edit conflict) Equivalent circuits can only "usefully" exist at a single frequency, but given that constraint, any circuit can be reduced to a single vector impedance and a single voltage or current source (which may be an ac voltage or current). When you think about it, of course, the classical Thevénin and Norton circuits are exactly what I just described with the operating frequency = 0 (and the simplifying fact that, at dc, all of the vector impedances look like pure resistances or open circuits).


 * The reason you can't do this for multiple, fully-general frequencies is that you need to take into account all the various resonances of the Ls and Cs.


 * Atlant 18:02, 14 May 2007 (UTC)

"For single frequency AC systems the theorem can also be applied to general impedances" ?
Why say only for a single frequency? You can find the Thevenin equivalent for any network at any frequency, except in the general case you'll end up with a $$Z_{th}$$ that's a function of frequency. I think that condition should be changed in the article as its a bit misleading. Roger 18:40, 14 May 2007 (UTC)
 * I wanted to get some clarification about frequency responding systems. Lets start with a single frequency voltage source at the input side of a black box. First open circuit the output side and measure the output voltage. This voltage will have a magnitude and phase shift from the input, and is the Thevenin voltage source of the circuit. Second, short circuit the output and observe the short circuit current which will have a magnitude and phase. Then divide the complex Thevenin voltage by the complex short circuit current to get the complex Thevenin equivalent impedance. Alternatively one could short all voltage sources, open all current sources, and directly find the complex Thevenin equivalent impedance. I believe that this is what was previously explained in this talk section, I just needed some clarity.


 * Now lets get complex and say that we find the complex Thevenin equivalent voltages and impedances for a wide range of frequencies. Next I have a non-sinusoidal or aperiodic input signal and take a Fourier Transform to find the frequencies that compose this signal. Using the Thevenin models that correspond to each frequency appearing in the non-sinusoid/aperiodic signal I would like to be able to model the output for any impedance load. This would work by putting a specific fixed impedance on the Thevenin models for each frequency found in my input signal, then by using superposition of the output waveforms from each included frequency's Thevenin model find the actual output waveform. Have I gone off the deep end, or will this actually work? Thanks. Dude202 06:16, 27 June 2007 (UTC)
 * Lets see. If your signal is a periodic voltage and has a Fourier series representation, then you can expand it as a sum of sinusoids. Using superposition, the current that flows when you connect this signal to a linear network will have the same form, except each component sinusoid will have a amplitude and phase shift. In theory you replace the network by an infinite number of ideal narrow band filters, each with the same specific amplitude and phase response at the respective component frequency. Keep in mind that the input signal must be periodic, the network must be linear and you'll need alot of those narrow band filters (depending of course on what the input signal looks like). Your method of finding the Thevenin equivalent method sounds also sounds correct. Roger 15:56, 27 June 2007 (UTC)

Cramer's rule
The article has recently had added a proof by Cramer's rule. I have doubts that what is stated is actually a proof. I agree that proving that the load current is in the form;


 * $$I_L = \frac {E}{R_L + R} $$

is equivalent to proving Thevenin's theorem. However, it is merely stated without proof that this is the result from using Cramer's rule. Doubtless Cramer's rule can be used in a proof of Thevenin and the article can say this (ideally with a suitable citation) but it should not be claiming to present a proof when no proof is given.  Sp in ni ng  Spark  10:29, 17 July 2010 (UTC)


 * I think so too. There's no need calculating to prove this theorem. The linearity of the network is the sole basis of it. There are lots of proof wich are better and simpler than this. http://fourier.eng.hmc.edu/e84/lectures/ch2/node3.html http://netlecturer.com/NTOnLine/T08_THEOREMS/p03ThevA.htm http://www.eng.fsu.edu/~depriest/EE3003Lab/circuit_theorems.ppt —Preceding unsigned comment added by 87.91.111.183 (talk) 09:32, 3 September 2010 (UTC)

what is the specific advantage in using thevenin theorem in solution of dc circuits. —Preceding unsigned comment added by 196.46.245.44 (talk) 17:42, 16 December 2010 (UTC)

On determining Rth
In the part

"Step 2 could also be thought of as:        2a. Replace voltage sources with short circuits, and current sources with open circuits.         2b. Calculate the resistance between terminals A and B. This is RTh. "

It should be noted that this method only works for circuits with independent voltage/current sources. (in case the circuit has dependent sources, some other method must be used) — Preceding unsigned comment added by 187.107.63.44 (talk) 10:47, 11 July 2011 (UTC)

Proof
The alleged proof seems to open with a statement that itself amounts to Thévenin's theorem. It is unsourced. All the proofs I have seen involve examining the behaviour with added external sources, either voltage, or current. I vote to delete it until something better can be written.  Spinning Spark  08:45, 2 November 2012 (UTC)


 * When I was reading this page, I found a proof in the first external link,


 * http://tcts.fpms.ac.be/cours/1005-01/equiv.pdf


 * which I did not understand. Then I realized one can use superposition theorem to construct a solution of the problem and then use the uniqueness of the solution of classical electrodynamics to show it is the only solution. I just figure this out when I was studying again some concepts and theorems in electric circuit for work. I actually don't know any textbook which contains a proof. I will try to write down the arguments in more details. Gamebm (talk) 17:17, 20 November 2012 (UTC)
 * (Edit) I added few sentences and read through it again, it seems clear to me, and maybe not so to other. I would suggest you to propose which part of the proof contains ambiguity or mistake, therefore should be improved or corrected, rather than saying it is not in the same style as those textbooks you have read and therefore it is bad and should be removed. Honestly, to me, I would prefer such things (a correct but not perfectly written proof, rather than nothing) in other places when I look for topics that I am not familiar with in wiki. My experience with wiki in the field of quantum field theory (such as QCD or string theory) told me how those things I work with are so terribly done, yet I do appreciate so much the knowledge I am able to access to other branches I am not familiar with (and logically I believe they are poorly done in the eyes of an expert.) Gamebm (talk) 17:39, 20 November 2012 (UTC)


 * I took a look at the textbooks you have cited (This is what I should have done before came to wiki), and would like to give few comments. I cannot access the page with proof of the firs one [1]. The proof given in the second text book is somewhat incomplete or limited. The proof is to explain why any black box behave exactly like a voltage source with some internal resistance when it is connected to anything (of course, linear circuit), the book examine the case when it is connected to an impedance (N2), but why not an impedance connected to some voltage source, or several impedance forming a star circuit with some source embedded, so it can not be simplified trivially that itself may need some help from Thevenin's theorem. The author did not try to discuss such a situation (in fact it is not necessary, but unfortunately what the author did just eliminate the possibility to discuss in a general context, his proof is heavily based on this one external impedance assumption), therefore it is incomplete. In this proof, he did not explicitly discuss how and why the black box should behave exactly like something else, though the case is analysed in terms of superposition of two configurations, and after think it over, I figure the (limited) proof can be done using those plots. But I really missed the points/arguments why one thing is equivalent to another where the author really should emphasized (for instance, why inverting the inserted source will generate exact the same effect of the black box, for any external impedance in question, while the superposition analysis only deals with a given impedance. Well, this is because the voltage of this inserted voltage source is determined by the open circuit voltage, therefore this very voltage works for any value of impedance, in order words, it is intrinsic characteristic/paramter of the black box, but the author really should have pointed this out explicitly.). It seems to me more of some statement/instruction than a proof of the theorem itself. I like the third textbook more. The author does exactly what I have in my mind, one should discussion the voltage and current relation of the black box (since this contains all the information one can exact from it, and in fact nothing more!), following this line of thought, once obtained, one can show the voltage-current relation is exactly the same as a simple voltage source with some series resistance (they are equivalent as a consequence). Then the arguments in the textbook are quite brief, (even more than what I did :) ), and to me the arguments are kinda close to what I did in the page, i.e., the resultant V-I curve is linear superposition of several V-I curves of simpler problems. Two things one need to think over, (1) what are those problems, confirm they are simpler enough to be solved (2) why the superposition of all of them indeed gives the V-I curve in question. (I did not explicitly discuss them, the author didn't either, I assume who want to think this problem over should know this in the first place.) Meanwhile, one should take into consideration that the black box can be complicated. In the textbook, the author considered that there can be voltage sources as well as current sources in the black box. (In the proof on wiki page, I only considered voltage source, but it seems that the generalization is quite intuitive). Then the difference from his proof is that the author divided the problem in a different way than I did: he did not try to divide it into the summation of some open circuit problems with one close circuit problem (as in the wiki page), however the argument works out quite the same way, except in this case, it is not straightforward to obtain the equivalent voltage of the black box (which is the summation of all the open circuit voltage of each individual source). Gamebm (talk) 18:39, 20 November 2012 (UTC)
 * First of all, I don't think the introduction of uniqueness theorem is helpful here. This applies widely to all of linear circuit theory and inserting it here is an unnecessary complication and just adds confusion.  If it is needed to say it at all, it should be said at the end of the proof, it is certainly does not constitute "stage 2" of the proof.  By the way, the link goes to a dab page making it doubly confusing and unhelpful to the reader.
 * My principle objection is that the proof starts with
 * "one can always write down its voltage as a linear function of the corresponding current as follows
 * $$V = V_\mathrm{Eq}-Z_\mathrm{Eq}I \!$$
 * This is, in fact, a statement of Thevenin's theorem so it is hardly surprising that it results in a "proof".  Spinning Spark  19:46, 20 November 2012 (UTC)
 * If uniqueness is implied, it can be (and should be) removed, but I kinda realized it is not so trivial. It is not the uniqueness theorem of electrostatics (no current in electrostatics), neither the uniqueness of solution of Maxwell's equations (current and charge distributions are usually considered as given boundary conditions in this case). In fact, I don't see a wiki page for this.
 * You are correct, if one obtains a linear relation between the current and voltage of the black box, he is only one small step away from Thevenin's theorem itself. However, to obain this relation is not trivial (a linear system does not immediately imply that a relation between any two quantities of the system must be linear) and the proof mainly focuses on how to establish such a relation using superposition of simpler problems. If there is a better way to present this, please improve it. And again, I think a correct but maybe not perfectly written proof is better than no proof. Gamebm (talk) 14:35, 21 November 2012 (UTC)

Thévenin's theorem lead changes
What follows is copied from recent User:Spinningspark Talk, at his suggestion.

Hello, I am proposing edit to Thévenin's theorem lead along the following lines:

In circuit theory, Thévenin's theorem for linear electrical networks may be stated in contemporary generalized form as follows (see image at right): The theorem was independently derived by German scientist Hermann von Helmholtz in 1853 and by French telegraph engineer Léon Charles Thévenin (1857–1926) in 1883. This theorem in widely used as a circuit analysis simplification technique to convert any circuit's voltage sources and impedances to a Thévenin equivalent. Thévenin's theorem is commonly used to study circuits' initial-condition and steady-state response; use of the theorem may in some cases be more convenient than use of Kirchhoff's laws..
 * 1) Any active terminal pair a-b composed of combinations of impedances and ideal voltage and current sources can be represented by electrically equivalent series connection of an ideal voltage source V with an impedance Z between terminals a-b, combinations involving either all DC sources or all resistive impedances, or both, being special cases.
 * 2) The voltage source V is the voltage found at terminals a-b with no external elements connected to the terminals and the impedance Z is the impedance measured at the terminals when all the ideal sources within the terminal pair are set to zero.



Single-frequency AC systems
The term 'single-frequency AC systems' may be too restrictive. See for example http://www.wolframalpha.com/input/?i=harmonic+addition+theorem.Cblambert (talk) 20:00, 3 February 2013 (UTC)

Léon Thévenin's 1883 proof of his theorem
"To show the theorem, we suppose that one introduces in the conductor ABA’ an electromotive force - E, equal and opposite to the potential difference V – V’. Clearly, no other current flows though the conductor ABA'. Thus, the system of electromotive forces - E, E1, E2, . . ., En give instead a new distribution of currents, among which is one where the current through ABA" is null.

"We suppose now that, in the same conductor, one introduces, at the time with the first, a second electromotive force + E, equal to the potential difference V – V’ and in the same sense. By virtue of the principle of the independence of simultaneous electromotive forces, the force + E gives birth to a new current distribution, that simply superimposes in the preceding one. Among these new currents, the one flowing through ABA" is precisely the sought i, because the effect of the forces + E and - E, equal and opposite, cancel each other. The resulting current i is only due to the force + E = V - V', whose consequence is in the branch r, one can by calling R a certain resistance, write, according to Ohm's Law, i = (V – V’)/(r + R). Moreover, the significance of the quantity R immediately appears; it is the resistance of a wire that can replace the primitive network of conductors between the points A and A’, without the undisturbed flow due to a constant electrical source that would exist in the branch r before it was modified. The quantity R has a precise physical significance, and one can call it the resistance of the primitive network measured between the points A and A' considered like electrodes. The statement of the theorem results immediately from this definition."Cblambert (talk) 06:37, 6 February 2013 (UTC)
 * This proof is of course as translated in Johnson 2003a.
 * This proof is, without however making mention of Thévenin proof, exactly the same approach reflected in Brenner & Javid, pp. 270-273, including in terms of full page of step by step diagrams.Cblambert (talk) 20:36, 6 February 2013 (UTC)

Notation is not clear to a general reader
The notation after the phrase "Calculating equivalent resistance:" R1 + [ (R2 + R2) || R4 ]  seems obscure and confusing. For example, I do not find this notation in H&H, Delaney, or any other textbook on my shelf. Can you please express this in a way that a non-initiate will understand without a lot of additional and unnecessary searching? — Preceding unsigned comment added by DrKN1 (talk • contribs) 13:06, 16 October 2017 (UTC)
 * I added some text that I hope clarifies that R1||R2 means the effective resistance of R1 in parallel with R2. The notation is common. Johnuniq (talk) 23:19, 16 October 2017 (UTC)

Not really opposite
"The replacements of voltage and current sources do the opposite of what the sources themselves are meant to do. A voltage source creates a difference of electric potential between its terminals; its replacement, a short circuit, equalizes potential. Likewise, a current source generates a certain amount of current, whereas an open circuit stops electric flow altogether."

Here I feel that the opposite of a + 5 Voltage source would be -5 Voltage source not 0 Volt. Similarly with the current source. The opposite of matter, is antimatter while it's absence would be vacuum.

Aditya 09:48, 16 November 2017 (UTC) — Preceding unsigned comment added by Aditya8795 (talk • contribs)


 * I agree. Opposite is not the correct word.Constant314 (talk) 17:07, 16 November 2017 (UTC)

I agree too. Alej27 (talk) 22:23, 1 January 2021 (UTC)

Proof added
The section titled "Proof" did not actually contain a proof, and the text that was there, I found completely opaque and unenlightening. The actual proof (as given by Thevenin) is actually simpler to understand than this section! So, I replaced the section with the actual proof. It then made sense to move this section nearer the beginning, which I did. LyleHoward (talk) 20:24, 3 February 2024 (UTC)

Helmholtz's proof
Since Helmholtz was the first to prove this theorem (by 30 years!) I felt it was important to add a section detailing his proof. LyleHoward (talk) 15:50, 7 June 2024 (UTC)