[Physics] Thevenin theorem and its proof

Question

In the proof of this theorem a test current source is attached to the terminals of a network called $N$. We want to know the equivalent of network $N$. Then we calculate the potential at this terminal which is:

$$\Delta V = V_\text{th} + R_\text{th} I_\text{external} \, .$$

$V_\text{th}$ is the potential due to the network and $R_\text{th} I_\text{external}$ is the potential due to the external current source and equivalent resistance of the network.

The key point in the proof is that in order to calculate the potential of the network we do not consider the external current source and open circuit it. This is understandable because we are using superposition principle.

However, what if we connect the network $N$ to another external network called $E$. Then in order to calculate the potential at the terminal we cannot open circuit the terminal because when we ignore the sources in the external network $E$ it does not mean that the terminal is open now. Maybe still current goes to the external network $E$. How can we prove the theorem in this case?

Another issue is that when instead of current source we use another voltage source. Then the potential at terminal should be the same as that of the voltage source. How can we get the equivalent circuit when we have a voltage source at the terminals.

Edit

Here is the Thevenin theorem form Wikipedia:

1. Any linear electrical network with voltage and current sources and only resistances can be replaced at terminals A-B by an equivalent voltage source $V_\text{th}$ in series connection with an equivalent resistance $R_\text{th}$.

2. The equivalent voltage $V_\text{th}$ is the voltage obtained at terminals A-B of the network with terminals A-B open circuited.

3. The equivalent resistance $R_\text{th}$ is the resistance that the circuit between terminals A and B would have if all ideal voltage sources in the circuit were replaced by a short circuit and all ideal current sources were replaced by an open circuit.

4. If terminals A and B are connected to one another, the current flowing from A to B will be $V_\text{th}$/$R_\text{th}$. This means that $R_\text{th}$ could alternatively be calculated as $V_\text{th}$ divided by the short-circuit current between A and B when they are connected together.

Answer 1

I wrote up a proof of Thevenin's Theorem going through each step in detail, inspired by another question. One of the answer's referenced a text by Kendall Su "Fundamentals of Circuits, Electronics, and Signal Analysis", Appendix A.1, page 568).

For the general form of Thévenin's theorem we allow unlimited resistors and unlimited independent voltage or current sources. (Dependent sources add a small complication not covered here. They basically get grouped in with the resistors.)

Select any two nodes inside the circuit and bring them out to a port. The goal is to prove the voltage and current can be related like this: $v = \text V_\text T + \bf Ri$, where $\bf R$ stands for some expression composed of resistance values.

Assume a linear network composed of,

• any number of resistors,
• $N$ voltage sources: $\text V1 \ldots \text V_N$,
• $M$ current sources: $\text I1 \ldots \text I_{M-1}$ internal current sources plus an external current source $\text I_M$ connected to the port.

Now apply the principle of superposition. The original circuit can be decomposed into $N+M$ sub-circuits, one for each source. Each sub-circuit contains a single source and some arbitrary network of resistors. (All the other sources have been suppressed.)

For the $N$ sub-circuits with a voltage source, the voltage at the port is always a scaled version of the internal voltage source. The scale factor is a dimensionless ratio of resistances, $\bf A$,

$v_n = \bf A_n \text V_n, \quad$ where $n$ ranges from $1 \ldots \text N$

For $M-1$ sub-circuits with an internal current source the voltage at the port is always the Ohm's Law product of the current source $\text I_m$ times some resistance expression $\bf R$.

$v_m = \text I_m \bf R_m, \quad$ where $m$ ranges from $1 \ldots \text M-1$

The last current source is the external $\text I_M$, which produces this voltage,

$v_M = \text I_M \bf R_M$

where $\bf R_M$ is the equivalent resistance looking into the port when all the internal sources are suppressed.

We've found all the separate contributions to the port voltage. Now superimpose (add them all up),

$\displaystyle v = \sum_{n=1}^N v_n + \sum_{m = 1}^{M-1} v_m + v_M$

$\displaystyle v = \sum_{n=1}^N A_n \text V_n + \sum_{m=1}^{M-1} \text I_m \bf R_m + \text I_M \bf R_M$

The first two summation terms produce voltage values based on the internals of the circuit, with nothing connected to the port. For every one of these sub-circuits the external current source was suppressed. The combination of these two summation terms gets a special name, $v_{oc}$, which stands for open-circuit voltage,

$v_{oc} = \displaystyle \sum_{n=1}^N A_n \text V_n + \sum_{m=1}^{M-1} \text I_m \bf R_m$

$v_{oc}$ is the voltage that appears at the port when the port is left open.

With this new variable name we rewrite the superposition equation,

$v = v_{oc} + \text I_M \bf R_M$

The whole arbitrary circuit boils down to this equation. It's exactly what we want. This voltage-current relationship allows us to construct a Thévenin equivalent circuit,

$\text R_\text T = \bf R_M$

$\text V_\text T = v_{oc}$

$v = \text V_\text T + i\,\text R_\text T$

Both the original complex circuit and the little Thévenin equivalent obey the same $i$-$v$ equation. Done! We proved any circuit composed of resistors, voltage sources, and current sources can be reduced to a single voltage source and a single resistor.