Question: We are given the following circuit and told to calculate the equivalent voltage of everything to the left of the a and b terminals: 

[![enter image description here](https://i.stack.imgur.com/ECjj0.png)](https://i.stack.imgur.com/ECjj0.png)

It omits the load resistor and lays the circuit out as such for mesh analysis: 

[![enter image description here](https://i.stack.imgur.com/SQkc2.png)](https://i.stack.imgur.com/SQkc2.png)

It claims that voltage across terminals a and b, the voltage we are looking for, is the same as the potential in between the 4 ohm and 1 ohm resistor. Shouldn't that 1 ohm resistor cause the potential to change a little?

Answer: 
You're trying to find the Thevenin equivalent voltage of the network. 

This is the voltage that the network produces when the output is open. 
When the output is open, no current flows out or in to the `a` terminal.
Therefore no current flows through the 1 ohm resistor.

Since no current is flowing through the 1 ohm resistor, the voltage across it is 0, by Ohm's law.
I think its an optical illusion and that's why it looks odd. the 4 ohm and 12 ohm resistors are whats important here. You connect the RL at that junction, so all that matters is that you get the 4 + 12 = 16, and 32/16 = 2A, confirmed by the 2A symbol there.

then its whatever voltage ratio you get. if E = IR then voltage is 2 amps x 12 ohms = 24v at your RL point there

the 1 ohm will cause the current to decrease but i think the voltage will be 24v. in real equipment they will only be able to supply so much current, so in real circuits you would need to watch the 1 ohm there, but in theory with no load defined, its assumed the other parts of this equation are perfect (right?)