A condenser of capacitance `10 muF` has been charged to `100V`. It is now connected to another uncharged condenser in parallel. The common potential becomes `40 V`. The capacitance of another condenser is

Options

A. `2.5 mu F`
B. `5 mu F`
C. `10 muF`
D. `15 mu F`

Answer

D

Explanation

`C_(1)=10 muF, C_(2)=?`
`V=100 V, V'=40 V`
`V'=(C_(1)V)/(C_(1)+C_(2)) , 40 =(40xx100)/(10 +C_(2))`
`C_(2)=15 mu F`