A simple `LR` circuit is connected to a battery at time `t = 0`. The energy stored in the inductor reaches half its maximum value at time

Options

A. `(R )/(L)` In `[(sqrt(2))/(sqrt(2) - 1)]`
B. `(L )/(R)` In `[(sqrt(2)-1)/(sqrt(2))]`
C. `(L )/(R)` In `[(sqrt(2))/(sqrt(2) - 1)]`
D. `(R)/(L)` In `[(sqrt(2)-1)/(sqrt(2))]`

Answer

C

Explanation

`U_(max) = (1)/(2) LI_(0)^(2)`, `U = (U_(max))/(2)`
`rArr (1)/(2) LI^(2) = (1)/(2) [(1)/(2) LI_(0)^(2)] rArr I_(0)^(2)[1 - e^(-t//tau)]^(2) = (I_(0)^(2))/(2)`
`rArr e^(-t//tau) = 1 - (1)/(sqrt(2)) = (sqrt(2) - 1)/(sqrt(2))`
`rArr -(t)/(tau) = In ((sqrt(2) - 1)/(sqrt(2))) rArr t = tau In ((sqrt(2))/(sqrt(2) - 1))`