Saved Bookmarks
| 1. |
The capacitance of a variable capacitor joined with a battery of 100 V is changed from 2muF to 10 mu F. What is the change in the energy stored in it ? |
|
Answer» `2xx10^(2) J` `U_(1)= (1)/(2) xx2xx10^(-6)xx(100)^(2) =10^(-2)` J The energy stored in capacitor of `10 mu F ` `U_(2) = (1)/(2)xx10xx10^(-6)xx(100)^(2)` `U_(2) = 5xx10^(-2)` J `:.` The change in stored energy `U = U_(2)-U_(1)= 5xx10^(-2) - 1 xx10^(-2)` `:. U = 4xx10^(-2) `J |
|