Question

A capacitor of capacitance `C_(0)` is charged to a patential `V_(0)` and then isolated. A small capacitor C is then charged from `C_(0)`, discharged and charge again, the process being is decreased to V. Find the value of C.
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 1

Correct Answer - B
Potential of larger capacitor after the first chargin g is
`V_(1)=(C_(0)V_(0))/(C+C_(0))`
After second charging, potential is
`V_(2)(C_(0)V_(1))/((C+C_(0)))=(C_(0)/(C+C_(0)))^(2)V_(0)`
Aftre `n^(th)` 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]`.