A capacitor of capacitance `C_(o)` is charged to a potential `V` and then isolated. A small capacitor `C` is then charged from `C_(o)`, discharged and charged again, the process being repeated `n` times. Due to this, potential of the large capacitor is decreased to `V`. Find the capacitance of the small capacitor :

Options

A. `C_(0)((V_(0))/V)^(1//n)`
B. `C_(0)[((V_(0))/V)^(1//n)-1]`
C. `C_(0)[(V/(V_(0)))-1)]^(n)`
D. `C_(0)[(V/(V_(0)))^(n)+1]`

Answer

B

Explanation

Potential of larger capacitor after the first charging is
`V_(1)=(C_(0)V_(0))/((C+C_(0)))`
After second chageing potential is
`V_(2)=(C_(0)V_(1))/((C+C_(0)))=((C_(0))/(C+C_(0)))^(2)V_(0)`
After nth charging, potential is
`V_(n)=((C_(0))/(C+C_(0)))^(n)V_(0)`
But `V_(n)=V` So `C=C_(0)[((V_(0))/V)^(1//n)-1]`