A coil resistance `20 W` and inductance `5 H` is connected with a `100 V` battery. Energy stored in the coil will be

Options

A. `41.5J`
B. `62.50 J`
C. `125J`
D. `250J`

Answer

B

Explanation

`U=(1)/(2)Li^(2)=(1)/(2)L((E)/(R ))^(2)=(1)/(2)xx5xx((100)/(20))^(2)=62.50J`