Thévenin's theorem exercise, resistances

Question

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)

Thank you for your time!

Answer 1

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}

Answer 2

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}\$