Twelve capacitors, each having a capacitance C, are connected to form a cube . Find the equivalent capacitance between the diagonally opposite corners such as A and B.

Explanation

Consider the figure given below, let us assume a battery of potential difference V is connected at points A and G. The charge distribution is shown in the figure. We have to find equivalent capacitance between points A and G. All three capacitors connected to the point A are same so charge is equally divided among them. We have assumed total charge 6Q is supplied by the battery. It is first divided equally among three capacitors, hence 2Q each, and then from these branches this charge is equally divided between two capacitors hence Q each. Equivalent capacitance will
`C_(eq)=6Q//V " " ...(i)`

Let us apply Kirchhoff.s law for loop IJABFGI:
`V-2Q//C-Q//C-2Q//C=0`
`rArr V=5Q//C`
`rArr Q=CV//5 " " ...(ii)`
Substituting Q equation (ii) in equation (i) we get the following:
`C_(eq)=6C//5`
Alternate solution:
We can see that, due to same capacitance and charge in the capacitors of branches AD, AE and AB, electric potentials of points B, E and D are the same. Using the same logic electric potential of points C, H and F are also the same. We can put points of same potential at the same location and redraw the network as follows:

Hence, using above redrawn network we can find the answer as before by the series-parallel method.