A capacitor `C_(1) =1.0 muF` is charged up to a voltage V=60 V by connecting it to battery B through switch (1) Now `C_(1)` is disconnected from battery and connected to a circuit consisting of two uncharged capaictors `C_(2)=3.0 mu F and C_(3)=6.0 mu F` through switch (2) as shown in the figure The sum of final charges on `C_(2) and c_(3)` is :

Options

A. `20muC`
B. `36muC`
C. `40muC`
D. `54muC`

Answer

C

Explanation

`C_(23)=(C_(2)C_(3))/(C_(2)+C_(3))=(3xx6)/(3+6)=2muF,` common potential `V=(C_(1)V_(1)+C_(23)V_(23))/(C_(1)+C_(23))=(1xx60+2xx0)/(1+2)`
`V=20`Volt `therefore q=C_(eq)V=2xx20muC, q=40muC`.