Been working with this all day and can't quite suss it out. Reduced it down as much as I could and was given the answer is Rt=10Ω. But... R2 is giving me headaches. Can't quite see what I'm missing...

 

Lots of 3D confusion.

R2 and R5 are shorted out, R1, R3, R4 and R6 are connected across the inputs,

\[ R_{T} = \frac{1}{1/R_1+1/R_3+1/R_4+1/R_6} = 10 \]

 

I see that R2 is shorted but not R5. Ahhh... Now I see it, K, thanks!

 

One thing that can really help with a problem like this is to color the nodes different colors. Any resistor that is connected to the same color node on both sides is shorted.



This makes it very evident that R2 and R5 are shorted out and that the remaining four are in parallel.

 

Been working with this all day and can't quite suss it out. Reduced it down as much as I could and was given the answer is Rt=10Ω. But... R2 is giving me headaches. Can't quite see what I'm missing...
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Hi there,

As others have already pointed out, when you consider the 6 edges that are not resistors but just shorts, the circuit looks like it becomes greatly simplified.
If that wasn't the case though and you had a complete cube where all edges were actually resistors of different values, a simpler way to solve it is to start by solving for the voltages at the two nodes that R1 is connected to. You can then go on to figure out the value. If you did not have an input to the circuit and just needed to calculate the equivalent resistance, you could use an assumed current like 1 amp or possibly an assumed voltage like 1v, then go from there. That's a very general method although it does take some work.

 

This makes it very evident that R2 and R5 are shorted

Neat trick! That I hope I can remember.