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)

I think Req is wrong. It should be R1+((R2+R3)||R4). It is showing R4+((R2+R3)||R1).

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. #note about non-resistive parts — 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)

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. SpinningSpark 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)