An inductance `L` and a resistance `R` are first connected to a battery. After some time the battery is disconnected but `L` and `R` remain connected in a closed circuit. Then the current reduces to `37%` of its initial value in

Options

A. `R//L`
B. `R//L`
C. `L//R`
D. `1//LR`

Answer

C

Explanation

`i_(0)=E//R`
`i=i_(1)e^(-(Rt)/(L))`
`0.37i_(0)=i_(0)e^(-(Rt)/(L))`
`e^(-(Rt)/(L))=(1)/(0.37)=2.7`
`(Rt)/(L)=1n(2.7) implies (Rt)/(L)=1 implies t=(L)/(R)`