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In a lecture by Prof. Anant Agarwal at 36:00, he intuitively proves Thévenin's Theorem using the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

If we consider the voltage across points a & b, we will be able to guess the form of the answer using the superposition theorem as the sum of voltage sources and current sources multiplied by a corresponding factor, so:

$$e= \alpha _1 V_1 + \alpha _2 V_2 + ... + \beta_1 I_1 + \beta_2 I_2 +... + \mathbf i\, \mathbf R_{th} \\ e= \sum \alpha_n V_n + \sum \beta_n I_n + \mathbf i\,\mathbf R_{th}$$

The primary terms \$\sum \alpha_n V_n + \sum \beta_n I_n\$ ultimately form a voltage, so we can write them as \$V_{th}\$ and so the total circuit can be reduced to a voltage source (\$V_{th}\$) and a resistor (\$R_{th}\$) in series with a current source.

schematic

simulate this circuit

If the current source in the circuit was replaced by a resistor:

schematic

simulate this circuit

then there would be no \$\mathbf i \mathbf R_{th}\$ term in the equation for \$e\$. So we will just be able to reduce the circuit into a voltage source \$V_{th}\$ in series with the current source.

Absurd Thévenin's equivalent \$\downarrow\$

schematic

simulate this circuit

So how will you keep up the argument when there is no current source?

I know I'm obviously wrong in two places:

  1. I'm wrong when I say that the source \$V_{th}\$ and \$R_{th}\$ are in series with the current source.
  2. That the new Thévenin's equivalent is absurd.

But I don't know why. In short, prove Thévenin's circuit intuitively with the superposition theorem.

In almost every proof I visited, they introduce a current source between the nodes (under study). But in the actual circuit there may not be a current source, but we can use Thévenin's theorem to find the current across any resistor or generally two nodes. So, why do they introduce a test current source to prove the theorem?

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). \$\endgroup\$
    – Dave Tweed
    Commented Mar 17, 2019 at 21:31
  • \$\begingroup\$ @jonk I finally found the answer, the theorem holds even when the circuit has no current source because, we can replace any circuit element(linear or non linear) with a equivalent current source. Substitution theorem youtu.be/8ZVZ5D7JUNA \$\endgroup\$ Commented Mar 22, 2019 at 2:31

2 Answers 2

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The Thévenin equivalent is not a closed circuit; it will have an output node. So the final Thévenin circuit will be Vth in series with R3, whose other terminal will connect to the rest of the circuit.

It does not matter what components the original circuit used, what matters is what the resistance seen at the node will be, and what the voltage without any load will be. Knowing these two, the voltage and current at the output can be calculated.

The Thévenin equivalent is not the same circuit, it just looks equivalent from the outside word, i.e. looking into the terminals of the circuit. This simplifies so many calculations! It is a very important concept used every day in analog circuit design.

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  • \$\begingroup\$ Yeah okay, but then, my question is how to prove Thevenin's theorem using superposition for the second circuit (the second with no current sources), please take a look at 36:00 of the lecture \$\endgroup\$ Commented Mar 18, 2019 at 15:50
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What matters is that the circuit is linear throughout, so that the voltage and current are a linear combination of each other. This means that every voltage and current source makes a contribution that linearly adds up with the others. Resistances and dependent sources that are linear force a certain linear relationship between voltage and current at every point in the circuit.

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