# Intuition on Thevenin's Theorem

In the following lecture by Prof. Anant Agarwal at 36:00, he intuitively proves Thevenin'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 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 $$\\sum \alpha_n V_n + \sum \beta_n I_n\$$ ultimately form an voltage, so we can write them as $$\V_{th}\$$ and so the total circuit can be reduced to an Voltage source( $$\V_{th}\$$) and a Resistor($$\R_{th}\$$) in series with current source. simulate this circuit

My Question

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 an voltage source $$\V_{th}\$$ in series with the current source.

Absurd Thevenin'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 at 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 new Thevenin's equivalent is absurd .

But I don't know why? In short, prove Thevenin's circuit intuitively with 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 Thevenin'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?

PS: How to decrease the size of the circuit?

• 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). – Dave Tweed Mar 17 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 – Aravindh Vasu Mar 22 at 2:31