What's the total amount of resistance in this kind of circuit?

[Reneee]

TL;DR Summary

Homework Statement::
(for another Problem) I just need to know what the total amound of resistance would be for the following circuit?

I am not sure how to treat the extra connections K12 and K34.
My first guess would be (R1 and R2) are parallel and so are (R3 and R4) and (R5 and R6) ,so
R_{total}= {R1*R2} /{R1+R2} +...

Or maybe the behaviour is like this:
(R1, R3, R5) are in series , and (R2 R4 R6) are, thus
R_{total}= (R1+R3+R5)*(R2+R4+R6) / (R1+R3+R5)+(R2+R4+R6)?

 

[gneill]

Your first method is correct. (R1, R2, R3) cannot be in series because where they connect to each other there's a third connection that could potentially add or subtract current from the connection point.

 

[Reneee]

alright, thank you!

 

[gneill]

You're very welcome!

 

[sophiecentaur]

Summary:: Homework Statement::
(for another Problem) I just need to know what the total amound of resistance would be for the following circuit?

I am not sure how to treat the extra connections K12 and K34.

Wired together, as in the diagram, the pairs are at the same potential so you have three pairs of parallel connected resistors. Easy when you can simplify a circuit in this way.
The problems would arise if K1 and K2 were connected by resistors with non zero value.