Q:

Calculate current in branch using Thévenin's theorem

E1 = 20 V
E2 = 5 V
R1 = 10 Ω 
R2 = 20 Ω 
R3 = 10 Ω 
R4 = 10 Ω 
R5 = 5 Ω 
Calculate Ix.

It's obvious that the Thévenin's theorem should have been used, but the implementation is a bit confusing for me in this case. So far I even don't know how to find an substitute impedance outside of the branch with Ix.

A:

This is how I would split up the circuit using Thévenin. Next step is calculating \$U_{Th1}\$, \$U_{Th2}\$, \$R_{Th1}\$, \$R_{Th2}\$ and drawing the resulting circuit. Then solving \$I_x\$ should be easy.

simulate this circuit – Schematic created using CircuitLab
As a response to your comment: Shuffle the symbols around a bit ... Nothing changed! You can change positions of components in series:

simulate this circuit

