# Intuition on Thévenin's Theorem

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

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.

simulate this circuit

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

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\$$

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?

• 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). Commented Mar 17, 2019 at 21:31
• @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 Commented Mar 22, 2019 at 2:31