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. RL second
B. `(R)/(L)`second
C. `(L)/(R)`second
D. `(1)/(LR)`second

Answer

C

Explanation

Time constant, `tau=(L)/(R)`
It is the time interval during which the current after opening an inductive circuit falls to `37%` of its maximum value.