Thevenin Equivalent does not seem equivalent for me

Question

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?

Answer 1

In the equivalent circuit, with no load there is no voltage drop across the 25 ohm resistor (since there is no current through it), so the output is 50 volts.

Now let's say you add a 50 ohm load. The top circuit has a lower branch equivalent to 25 ohms (two 50's in parallel, and will give an output of $$ V = 100 (\frac{25}{25 + 50}) = 33.3 \text{ volts} $$ The equivalent has 50 ohms between A and B, so $$ V = 50(\frac{50}{25 + 50}) = 33.3\text{ volts}$$ So they really are equivalent.

Answer 2

Denote the current \$I_L\$ as the current leaving the A terminal and returning to the B terminal. By KVL, the voltage \$V_{AB}\$ for the equivalent circuit is given by

$$V_{AB} = 50V - I_L\cdot R_{th}$$

But, for the open-circuit case, \$I_L = 0\$ by definition thus

$$V_{AB(OC)} = 50V - 0\cdot R_{th} = 50V$$

just as is the case for the actual circuit.