Find the potential difference between points `M` and `N` of the system shown in figure, if the emf is equal to `E = 110 V` and the capacitance ratio `C_1/C_2 is 2`.

Answer

`550 volt`

Explanation

Let `C_(1)=C` and `C_(2)=2C` and charges on different cacitors have been shown. Net charge on isolated system should be zero.
.
Hence, `q_(1)-q_(2)-q_(3)=0` (i)
For left loop,
`E-q_(3)/(2c)-q_(1)/C=0` (ii)
or `E-q_(3)/(2C)-((q_(2)+q_(3))/C)=0` (using `q_(1)=q_(2)+q_(3))`
or `E-q_(3)/(2C)-q_(2)/C-q_(3)/C=0`
or `E-q_(2)/C-3/2q_(3)/C=0` (iii)
For right loop,
`E-q_(2)/(2C)-q_(3)/(2C)=0`
or `2E-q_(2)/C+q_(3)/C=0` (iv)
Solving Eqs. (iii) and (iv), we get
`q_(3)=(-2CE)/5`
So `V_(MN)=(-q_(3))/(2C)=(2CE)/(5xx2C)=E/5=(110)/5=22 V`.