Total resistance through cubic circuit

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SUMMARY

The discussion centers on calculating the equivalent resistance of a cube formed by 12 resistors, with each edge of the cube containing one resistor. Participants explore using Kirchhoff's laws, specifically the principles of current and voltage at nodes and loops, to derive the solution. The suggestion to visualize the cube in a flattened format is proposed as a potential method to simplify the analysis. Ultimately, the focus is on applying Kirchhoff's laws effectively to determine the equivalent resistance between two diagonally opposite corners of the cube.

PREREQUISITES
  • Understanding of Kirchhoff's laws (current and voltage)
  • Familiarity with Ohm's Law (V=IR)
  • Basic knowledge of electrical circuits and resistors
  • Ability to visualize three-dimensional circuit configurations
NEXT STEPS
  • Study the application of Kirchhoff's laws in complex circuits
  • Learn about equivalent resistance in series and parallel resistor configurations
  • Explore techniques for visualizing and simplifying three-dimensional circuits
  • Investigate the use of symmetry in circuit analysis to simplify calculations
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone interested in advanced resistor network analysis.

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Homework Statement



Suppose a resistor R lies along each edge of a cube (12 resistors in all) with connections at the corners. Find the equivalent resistance between two diagonally opposite corners of the cube


Homework Equations



\SigmaI = 0 at a node (a junction)

\SigmaV = 0 through a loop

V=IR

The Attempt at a Solution




Not quite sure how to do this one algebraically. Would I just add up all the V/Is? how?
 
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Not quite sure where to start...can anyone help me?
 
How about opening the cube out so it's flat and looking at it that way.
 
Interesting...is that necessary though, or could I just apply kirchhoffs laws to it as it is?
 
I count from anyone node:
3 paths to 3 nodes which then connect by:
6 paths to 3 nodes which then connect by:
3 paths to 1 node.

connecting the 3 nodes together with wires should change nothing.
 
are you saying just ignore those 6 connections?
 
I think the post I was responding to was deleted.
 

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