Thévenin's theorem exercise, resistances

I've some problem identifying the $$\R_{th}\$$

simulate this circuit – Schematic created using CircuitLab

Removing the load and short circuiting the Voltage source we get :

simulate this circuit

I've drawn the equivalent to better state my thought process :
First of all, in both cases $$\R2\$$ doesn't matter since it is short circuited.
For the first picture, if we consider a current in the circuit it wouldn't flow towards $$\A\$$ right? (Since it is an open circuit) And hence we'll consider $$\R1\$$ and $$\R3\$$ in series?
In the equivalent though, it somehow feels like a current will go from $$\A\$$ to the circuit. But I guess it's the same thing, it's an open circuit so current only inside and thus $$\R1\$$ and $$\R3\$$ not in parallel right?
Or do we imagine a wire between $$\A\$$ and $$\B\$$ as is the case with Norton?

In the process of finding $$\E_{th}\$$ I separated the circuit in two :

simulate this circuit

$$\R_{th}\$$ is $$\R1\$$ but I couldn't figure out $$\E_{th}\$$ $$E_{th}+R_1i_1-E=0 \Leftrightarrow i_1=\frac{E-E_{th}}{R_1}$$ $$E-R_2i_2=0 \Leftrightarrow i_2=\frac{E}{R_2}$$ And that's it, can't go further, The $$\C\$$ -> $$\R_1\$$ -> $$\R_2\$$ -> $$\D\$$ -> $$\C\$$ KVL is not useful. (I've 0 current values btw)

Between point $$\A\$$ and point $$\B\$$ the current can take two independent paths.

As I try to show here:

All this means that the $$\ R_1\$$ and $$\R_3\$$ are connected in parallel.

As for the $$\E_{TH}\$$ you do not need to separate anything.

simulate this circuit – Schematic created using CircuitLab

All you need to do is to find the voltage across the $$\R_3\$$ resistor.

And you do not have to worry about the $$\R_2\$$ resistor. As he will do not have any influence and $$\E_{TH}\$$ voltage because $$\R_2\$$ is connected across the ideal voltage source $$\E_1\$$. And the source fixes the voltage difference across the $$\R_2\$$ resistor.

So you are left with this circuit:

simulate this circuit

• Yes, bless you! – user201599 Oct 21 '18 at 9:31
• My problem for the first part was not knowing whether we would consider a probable circuit between $A$ and $B$ or not, and from your answer I see that we would, an imaginary wire perhaps just like with Norton's theorem. – user201599 Oct 21 '18 at 9:33

I think I got this : (the voltage source has a value $$\E\$$)

$$\R_2\$$ will vanish because apparently that's what happens when you have a resistor in parallel with a voltage source hence we can transform the voltage source which is is now in series with a resistor into a current source $$\I_o=\frac{E}{R_1}\$$, right after that we find the equivalent resistance $$\R=R_1//R_3\$$ and now we can transform the whole thing into another voltage source $$\E'=\frac{E.(R_1//R_3)}{R_1}\$$ in series with $$\R\$$.

$$\E'=E_{th}\$$ and $$\R=R_{th}\$$