# Thevenin Equivalent does not seem equivalent for me

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?

• Think of it like terminal equivalent. If it produces the same current and voltage across the terminals, it is equivalent. The black box approach. Apr 22, 2016 at 0:03
• The open circuit voltage of the second circuit is 50V provided you take NO CURRENT from the output. The voltage drop across the 25 Ohms will be zero (V = 0x25) (Anything times zero is sero). You have assumed (wrongly) that this non zero resistance will somehow reduce the output voltage with open circuit output. Apr 22, 2016 at 11:34
• @ Felix Fernandez You wrote: "...Thévenin theorem should give an equivalent circuit in terms of current and voltage of any linear circuit consisting of independent sources...". The word "linear" is, no doubt, well-founded but the word "independent"? In terms of the Thévenin's theorem, it is irrelevant whether the sources within the linear circuit in question are independent or controlled. Oct 25, 2017 at 12:44

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.

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.