A coil of inductance `L=50muH` and resistance = `0.5Omega` is connected to a battery of emf = 5V.
A resistance of `10Omega` is connected parallel to the coil. Now, at the same instant the connection of the battery is switched off. Then, the amount of heat generated in the coil after switching off the battery is

Options

A. 1.25 mJ
B. 2.5 mJ
C. 0.65 mJ
D. 0.12 mJ

Answer

D

Explanation

Energy, `U=(1)/(2)Li_(0)^(2)=(1)/(2)L((V)/(R))^(2)`
`=(1)/(2)xx50xx10^(-6)((5)/(0.5))^(2)J=2.5mJ`
Now, after switching off the battery this energy is dissipated in coil (resistance = `0.5Omega`) and resistnace `(10Omega)` in the ratio of their resistances (`H=i^(2)Rt` or `HpropR`). Therefore, heat generated in the coil.
`((5)/(0.5))^(2)((0.5xx10)/(0.5+10))2.5mJ=0.12mJ`