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    				| 1. | A parallel plate capacitor of capacity `C_(0)` is charged to a potential `V_(0), E_(1)` is the energy stored in the capacitor when the battery is disconnected and the plate separation is doubled, and `E_(2)` is the energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is dounled. find the ratio `E_(1)//E_(2)`. | 
| Answer» Correct Answer - 4 (4) Initial energy : `E=1/2 C_(0)V_(0)^(2)` (i) `E_(1)=2E`, Because charge remains the same and the capacitance becomes half. (ii) `E_(2)=E/2`, because potential trmains the same and the capacitance becomes half. So` `E_(1)/E_(2)=4`. | |