# Limitation of the Thevenin's equivalent circuit

simulate this circuit – Schematic created using CircuitLab

Thevenin's theorem independently derived in 1853 by the German scientist Hermann von Helmholtz and in 1883 by Léon Charles Thévenin (1857–1926) allows any two-terminal linear network to be reduced to an equivalent circuit consisting of a single voltage source in series with a single resistor. This is known as the Thevenin equivalent circuit.

But the Thevenin equivalent circuit doesn't make sense in some situations. For example, if we connect a load with low enough resistance Thevenin Equivalent may not behave as the original circuit especially if the original circuit had resistors parallel with the load. I couldn't find any online material that especially mentions this limitation. So I want to know if that is a valid observation (probably implicit?).

Edit: It is a wrong observation and I understood it only after I drew the diagram. Stupid AI (yes, multiple) reassured me, every time I asked this, my observation is correct. When asked to cite sources all minus one told me they couldn't find any and derived it from basic circuit logic. The one AI that did cite the source fabricated an imagined statement (hallucination probably) from an existing source. That is how good AI is.

• Please show a schematic drawing that illustrates the case where you see Thevenin does not apply. Words are words. But a picture says so much more. Commented Jul 19, 2023 at 5:26
• You got the Thevenin voltage correct, but not the Thevenin resistance. The new resistance should be 5k, not 10k. And everything remains valid. Where is the invalidation? Commented Jul 19, 2023 at 8:57
• I still see the same 10k in the right side schematic. But it's just a detail. Thevenin works just fine, so far. Regarding the "stupid AI" I can only tell you that the ChatGPT one is totally stupid. It has no clues, at all. Think of it more as a stew of all kinds of factoids that can be given various associations to each other, but none of it with any intelligence applied, at all. And it is also a very confident AI, which makes it still worse. You can't believe anything it says. Like a broken clock, it is right twice a day. But only by accident of fortune. It does write as if it knows. Commented Jul 19, 2023 at 9:26
• It will fabricate references, too. And it will get part of author's names right. Back when they allowed it, I would ask it for the DOI for something. And it would try, and then try, and then try. And then after I told it that it failed over and over again, it would finally just say that it doesn't know of one. But before? It acted like it was absolutely certain about the DOI, the name of the paper, the author, etc. It lied. Simple as that. Do not try to learn anything from it, except words you might later look up on a real site and then read about. It does get some useful terms right. Commented Jul 19, 2023 at 9:29
• Understood. Just as an added note to the wise: You'll get many opinions. But you may want to learn to chart your own course or else be forever blown around by the storms of opinion here. (And that's yet another opinion, of course. ;) Commented Jul 19, 2023 at 11:35

simulate this circuit – Schematic created using CircuitLab

In your second circuit, the Thevenin resistor should be 5 kΩ, the parallel combination of the two resistors in your first circuit.

That can easily be seen by looking at the open circuit voltage and short circuit current from the source.

On my left circuit, the open circuit voltage is 50 V, given by the 2:1 voltage division of R1 and R2.

On my right, the short circuit current is controlled only by R4. RL1 and R3 are shorted, so are not relevant. The short circuit current is 10 mA.

Now if we want to form the Thevenin equivalent of this source between A and B terminals, we will need a 50 V source. Now obviously its series resistance needs to be 5kΩ to have the same short circuit current flowing.

Once you are experienced in doing Thevenin conversions, you do the following to work out the equivalent series resistance - replace all voltage sources with short circuits, replace all current sources with open circuits, calculate the remaining series/parallel combinations of all the resistors. As you can see, shorting V1 puts R1 and R2 in parallel.

Some people don't like the idea of 'shorting voltage sources', but it's just a mis-naming of what we are really doing. Voltages sources, in ideal circuit theory like this, already have zero output resistance. You can draw whatever current you like from a voltage source, and the voltage will stay the same - that's the definition of zero output resistance. What we are doing then is adjusting the voltage of the source down to zero, so that we can compute the resistance of the rest of the network, without the voltage getting in the way.

Here's a very good exercise to work through, both to get some intuition, and for you to understand what's happening should you try to debug some circuit by driving it from this simple controllable source.

simulate this circuit

What is the Thevenin circuit for this simple variable voltage source?

You know V is variable, but R_Thevenin is variable too. It's at its maximum of 2.5 kΩ at mid voltage out. It's always lower than that elsewhere, and it's zero at full voltage and zero voltage.

• @DukeWilliam The thevenin resistance should be 5k. Do you understand why it's 5 and not 10k? It's perhaps more useful to put two terminals on the circuit, one at each end of the load resistor, and then refer to those. Commented Jul 19, 2023 at 8:44
• @DukeWilliam I was referring to the thevenin equivalent of all that is not RL. Do the simple bits first. Replace RL by short circuit, what current flows? Replace RL by open circuit, what voltage do we get? So what should be the value of R_thev_equiv be when used with the 50 V source to provide the same behaviour as R1 and R2 with V1? That's how you calculate what the thevenin reistsance should be. Commented Jul 19, 2023 at 9:00
• @DukeWilliam The thevenin equivalent resistance is 5k. Your circuit still shows 10k (as at 2023-07-19 09:07:34 UTC). Your circuit is wrong. Try calculating with the short circuit / open circuit method I gave, that's as clear an intuitive ground truth as I can think of. Thevenin is quite happy with any loading on its equivalent source, whether low resistance, short, open, non-linear (non linear loading is OK- non linear elements in the Thevenin source are not allowed as jp314 pointed out). You took account of R1 to get the correct voltage, why are you ignoring it when deriving the resistance? Commented Jul 19, 2023 at 9:09
• @DukeWilliam No, it's correct, but it's not the same as your example. It's more complicated, and you've simplified it incorrectly. R2+R3 = 2k. R4 // that 2k = 1k, the thevenin equivalent of R2,3,4 with 7.5 V. Now R1 goes in series with that 1k, giving a total of 2k for the final diagram. Once again, the acid test is 1) open circuit it, what's the voltage, 2) short circuit it, what's the current? Note how they've put those terminals there that I suggested you should use on your diagram! Commented Jul 19, 2023 at 9:16
• I would love to put terminals there but there are no terminals in the circuit lab drawing tool or at least I could not find it. If I short-circuit RL in the original circuit then the current will be 10mA corresponding to the original 10K resistor. To be 5k, I should be short-circuiting the Thevenin equivalent voltage source. Is that what you suggest probably? Commented Jul 19, 2023 at 9:30

The Thevenin (and Norton) equivalent only applies to linear circuits. It also does not mean that the power dissipation in the equivalent circuit is the same as in the original.