An inductor `(L = 100 mH)`, a resistor `(R = 100 Omega)`, and a battery `(W = 100 V)` are initially connected in series as shown by the figure. After a long time the battery is disconnected after short circuiting the points `A` and `B`. The currents in the circuit `1 mm` after the short circuit is

Options

A. `0.1 A`
B. `1 A`
C. `(1)/(e) A`
D. `e A`

Answer

C

Explanation

`I = I_(0)e^(-Rt//L)`
`= (E)/(R )e^(-Rt//L)`
`= (100)/(100) e^(-) (100)/(100 xx 10^(-3)) xx 10^(-3)`
`= (1)/(e)A`