# Conceptual Question About Hemholtz-Thévenin Theorem

I've been taught that for the Hemholtz-Thévenin's method I need to add an independent voltage source with a current flowing and at the same time "turn off" independent sources of voltage and current to calculate the equivalent resistance.

Then I should just calculate the open circuit voltage on the terminals with the independent sources as they were on the first place.

First, I find difficult to understand this concept, and my second question is: If I have a resistor in series with the independent current source I "turned off" on the first step, does it mean the current in that resistor is also 0 or is it another current as if it was a cable?

I'm basing my question on the following excercise and it's solution: Find equivalent resistance ($$\R_{eq}\$$) seen from point A: (Don't mind the spanish).

Solution:

Thanks!

## 1 Answer

For the first part of your question:

The basic idea is to inject a 'test' current into the terminals and see what voltage this produces. As the circuit is linear, the effect of this 'test' source can be separated from that of any other independent sources by switching the others off. This means replacing all independent voltage sources with short circuits, and all independent current sources with open circuits.

We can then find the value of the equivalent resistance that produces this voltage. The easiest choice of test current is just 1A. So the equivalent resistance can be determined by picturing what paths the current can take through the circuit, and combining resistors appropriately in series or parallel.

This resistance tells us the circuit's response to external circuitry, but we still need to find the open circuit voltage to have a circuit that is completely equivalent.

For the second part of your question:

If we turn off a current source, it is replaced with an open circuit. So in the circuit in your question, the resistor that is in series with the current source is now only connected at one end so zero current will flow.