Saved Bookmarks
| 1. |
A capacitor of `2 muF` is charged to a potential of 4V using a battery, and then the battery is disconnected and the changed capacior is connected to an uncjharged caspacitor of `4 muF` capacitance. When the equilibrium is established the total energy stored in the capacitors isA. `16 muJ`B. `(16)/(3)muJ`C. `(32)/(3)muJ`D. `(32)/(9)muJ` |
|
Answer» Correct Answer - B `V=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))` `U_(F)=1/2(C_(1)+C_(2))V^(2)` |
|