Equivalent resistance of a cube

The three answers

1. Across a body diagonal: R = 5r/6. The 3 corners next to a share one potential, and so do the 3 next to b. Three r in parallel, six in parallel, three in parallel: r/3 + r/6 + r/3.
2. Across a face diagonal: R = 3r/4. The two corners on the far side of the face, off the line from a to b, share one potential by symmetry. Merge them and reduce in series and parallel.
3. Across an edge: R = 7r/12. The edge ab is in parallel with the rest of the cube, which by the same merging works out to 7r/5. Then r and 7r/5 in parallel give 7r/12.

Why it works

1. Feed a current I in at corner a and take it out at the opposite corner b.
2. By symmetry it splits equally into the 3 edges at a, I/3 each, and the 3 corners they reach share one potential.
3. From those corners it spreads into 6 edges, I/6 each, then gathers into the 3 edges at b, I/3 each.
4. Along any path from a to b the drop is (I/3)r + (I/6)r + (I/3)r = (5/6)Ir, so R = 5r/6.
5. The face and edge cases merge corners of equal potential in the same way. Kirchhoff's laws confirm 3r/4 and 7r/12.

More on circuits

• Resistance between A and B
• Series and parallel
• Kirchhoff's laws solved example
• Wheatstone bridge
• Resistance in parallel
• Kirchhoff's current law