Find the capacitance of the circuit shown in Fig, between points `A` and `B`.

Explanation

Taking the advantage of symmetry of the problem charge distribution may be made, as shown in the figure.
In the loop, `12561, -Delta varphi = 0`
or `(q_(2))/(C_(2)) + (q_(2) - q_(1))/(C_(3)) - (q_(1))/(C_(1)) = 0`
or `(q_(1))/(q_(2)) = (C_(1) (C_(3) + C_(2)))/(C_(2) (C_(1) + C_(3)))` ....(1)
Now, capacitance of the network,
`C_(0) = (q_(1) + q_(2))/(varphi_(A) - varphi_(B)) = (q_(1) + q_(2))/(q_(2)//C_(2) + q_(1)//C_(1))`
`= ((1 + q_(1)//q_(2)))/(((1)/(C_(2)) + (q_(1))/(q_(2) C_(1)))` ....(2)
From Eqs (1) and (2)
`C_(0) = (2 C_(1) C_(2) + C_(3) (C_(1) + C_(2)))/(C_(1) + C_(2) + 2 C_(3))`