Limitation of the Thevenin's equivalent circuit

Question

^{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.

Answer 1

^{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.

Answer 2

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.