Capacitor Cube

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?

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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

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

Jimmy Kariznov - 7 years, 10 months ago

The link was very useful thank you....

Rajath Krishna R - 7 years, 10 months ago

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

Ivan Stošić - 7 years, 10 months ago

This man is just superb. Solved with ease

jatin bhatoya - 3 years, 1 month ago

Can you explain it more precisely

Divyansh sen - 3 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$