Calculating Equivalent Resistance in a Hexagonal Prism Circuit

In summary: I find that by connecting a 1 A current source between P and Q, solving the circuit using a systematic method such as nodal analysis, and then remembering Thévenin's theorem (that the equivalent resistance the current source "sees" can be derived from V = IR, V being the voltage drop in the current source and I being its current, 1 A), you can solve for the equivalent resistance R.Can you explain why there is no balanced Wheatstone bridge?I have to say ... its a different question ... i was bored of those cubes ! LOL This is my attempt:just extended branches a little :tongue:I just simply used symmetry !And
  • #36
cupid.callin said:
there won't be any current through them right? so we can just remove them. Is that wrong ?:confused:

I don't think that you can assume that there will be no current through them.
 
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  • #37
gneill said:
Sure. I've only labeled some of the resistances, the rest should be obvious by color code and symmetry.

Did you get 17R/16?
 
  • #38
Abdul Quadeer said:
Did you get 17R/16?

Nope. A bit too small.
 
  • #39
Here's the next step in the simplification. The diagram is still symmetrical, so the Delta-Y transformations at either end will be the same...
 

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  • #40
Yup. That's what I had & did.
 
  • #41
I got it - 23R/20
Thanks!
 
  • #42
Abdul Quadeer said:
I got it - 23R/20
Thanks!

Yes, that's it. Attached is the summary of the simplifications used.
 

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  • HEX1.jpg
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