Find the potential difference between points `A` and `B` of the system shown in Fig. if the `emf` is equal to `E = 100V` and the capacitance ratio `C_(2)//C_(1) = eta = 2.0`.

Explanation

Let us distribute the chagres, as shown in the figure.
Now, we know that in a closed circuit, `-Delta varphi = 0`
So, in the loop, `DCFED`,
`(q_(1))/(C_(1)) - (q_(1))/(C_(1)) - (q_(2))/(C_(2)) = xi`, or, `q_(1) = C_(1) [xi + q_(2) ((1)/(C_(1)) + (1)/(C_(2)))]` ...(1)
Again in loop `DGHED`,
`(q_(1))/(C_(1)) + (q_(1) + q_(2))/(C_(2)) = xi` ...(2)

Using Eqs (1) and (2), we get
`q_(2)[(1)/(C_(1)) + (3)/(C_(2)) + (C_(1))/(C_(2))] = (xi C_(1))/(C_(2))`
Now, `varphi_(A) - varphi_(B) = (-q_(2))/(C_(2)) = (xi)/(C_(2)^(2)//C_(1)) (1)/([(1)/(C_(1)) + (3)/(C_(2)) + (C_(1))/(C_(2)^(2))])`
`varphi_(A) - varphi_(B) = (xi)/([(C_(2)^(2))/(C_(1)^(2)) + (3 C_(2))/(C_(1)) + 1]) = (xi)/(eta^(2) + 3 eta + 1) = 10 V`