A network of four capacitors of capacity equal to `C_(1) = C, C_(2) = 2C, C_(3) = 3C` and `C_(4) = 4C` are connected to a battery as shown in the figure. The ratio o fthe charges on `C_(2)` an `C_(4)` is

Options

A. `(22)/(3)`
B. `(3)/(22)`
C. `(7)/(4)`
D. `(4)/(7)`

Answer

B

Explanation

The charge in capacitor `C_(4),q_(4) = C_(4) xx V = 4 CV`
The capacitors `C_(1), C_(2)` and `C_(3)` are connected in series.
Let the resultant of these ca[acotors is `C'`.
`(1)/(C') = (1)/(C ) + (1)/(2C) + (1)/(3C)`
`= (6 +3+2)/(6C) = (11)/(6C)`
`rArr C' = (6C)/(11)`
Now C' and `C_(4)` from parallel combination giving
`C'' = C' + C_(4)`
`= (6C)/(11)+4C = (50C)/(11)`
Net charge `q = C'' V`
`= (50)/(11)CV`
Total charge flowing through `C_(1), C_(2)` and `C_(3)` will be
`q' = q' - q_(4)`
`= (50)/(11)CV - 4CV = (6CV)/(11)`
Since, `C_(1), C_(2)` and `C_(3)` are in series combination hence, charge flowing through these will be the same. Hence,
`q_(2) = q_(1) = q_(3) = q' = (6CV)/(11)`
Thus, `(q_(2))/(q_(4)) = (6CV//11)/(4CV) = (3)/(22)`