The application of Thévenin theorem should give a equivalent circuit in terms of current and voltage of any linear circuit consisting of independent sources and resistances.
Given this sample circuit:
The voltage drop between A and B is \$50 V\$. We calculate the Thévenin equivalent between those ports:
\$V_{th}\$ is \$50 V\$, the same voltage drop that we had between A and B in the initial circuit.
Given that in the second circuit we also have a resistance with more than 0 ohms the voltage drop between A and B will not be \$50 V\$, being not equivalent.
What I am doing wrong? Which concept is escaping to my mind?