If you had a cube of capacitors (each of the 12 sides was a capacitor of capacitance C) how would you find the equivalent capacitance across two outer diagonal corners?

Observe that the three vertices adjacent to one of connection points of your grid always have the same potential, due to symmetry. Since they do, you can fuse them all together into one. The same goes for the other three vertices adjacent to the other connection point.

Now you have reduced your problem to a simple configuration: (C || C || C) + (C || C || C || C || C || C) + (C || C || C) = 3C + 6C + 3C = 6/5 C

Here + means serial connection and || means parallel connection.

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TopNewestThis webpage solves the problem with 12 resistors instead of 12 capacitors.

However, the techniques to solve this problem should be very similar.

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The link was very useful thank you....

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Run some AC through your grid.

Observe that the three vertices adjacent to one of connection points of your grid always have the same potential, due to symmetry. Since they do, you can fuse them all together into one. The same goes for the other three vertices adjacent to the other connection point.

Now you have reduced your problem to a simple configuration: (C || C || C) + (C || C || C || C || C || C) + (C || C || C) = 3C + 6C + 3C = 6/5 C

Here + means serial connection and || means parallel connection.This picture may help you understand: Cube graph

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Can you explain it more precisely

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