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    				| 1. | 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» 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]`. | |