Greetings,

Simple Thevenin problem.
1) Working towards a Thevenin equivalent resistance. The small blue line is the removed voltage source. Why are the 2 resistors removed?


2) We found Rthev so now we bring back the voltage source. Is it possible to just move the resistance like that to the other side of the source?

 

1) An ideal voltage source has zero internal resistance. This 0Ω will override any resistance placed in parallel with the voltage source.

2) In the analysis of any circuit we are interested in what goes on between two nodes. Every dot in your circuit diagram represents a node. At the +ve side of the 90V source you can see three nodes that are connected to each other. Since these three are connected they are essentially the same node. You have shown the components wired as a ladder network. You could have drawn the same circuit using a star connection.



The same thing happens on the -ve side of the 90V source.

Once you reconfigure the nodes using a star connection perhaps you can visualize that the behaviour of the circuits does not change regardless of where you place the two resistors in parallel with the voltage source.

 

Greetings,

Simple Thevenin problem.
1) Working towards a Thevenin equivalent resistance. The small blue line is the removed voltage source. Why are the 2 resistors removed?
View attachment 247604

2) We found Rthev so now we bring back the voltage source. Is it possible to just move the resistance like that to the other side of the source?
View attachment 247605

I like MrChips explanation that outlined this well. Here is another interesting take on this idea.

If you do a full symbolic nodal analysis of the circuit including all resistors, those two resistors will 'drop' out of the final solution so they will not appear anywhere once the solution is simplified algebraically. This is pretty interesting i think because it proves that those two resistors do not change any voltage any place in the circuit, at least in theory. The only thing you might want to follow up on is the power consumed in those two resistors because they are certainly dissipating energy.

Now be aware i used the phrase above, "at least in theory". That is because in pure theory (that we use much in circuit analysis) the two resistors do not affect any voltage, and that is because of what MrChips said about the voltage source having zero internal resistance. If it did have internal resistance, then those two resistors would change all the voltages in the circuit because they would be partially loading down the voltage source because the internal resistance would appear to be another resistor in series with the voltage source and so one terminal would be at the bottom and the other at the top of the resistor that is now in series.
If you want to see how this works out, simply place a small value resistor in series with the voltage source and do the analysis again, then compare to when you did not have that extra resistor in there.

 

It is oftentimes useful to designate a ground node when analyzing a circuit, and it is typical to treat the negative power terminal as the circuit ground. Also, it is sometimes useful to label individual power points rather than show power bus connections. So a circuit that looks like this:

... can be analyzed as a circuit that looks like this:

 

Greetings,

Simple Thevenin problem.
1) Working towards a Thevenin equivalent resistance. The small blue line is the removed voltage source. Why are the 2 resistors removed?
View attachment 247604
The power supply that was removed and replaced with a short, has removed them

2) We found Rthev so now we bring back the voltage source. Is it possible to just move the resistance like that to the other side of the source?

Sure. Notice there is one net connecting all three items on top and another net connecting the other side.
View attachment 247605