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

Options

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

Answer

D

Explanation

Here, `L = 100 mH = 100 xx 10^(-3) H = 0.1 H`
`R = 100 ohm, E = 100 V`
`I_(0) = (E)/(R ) = (100)/(100) = 1 A`
When short circuiting is done, battery is disconnected. Current in RL circuit decays. At any time t.
`I = I_(0) e^((-R)/(L)) = 1 xx e^((-100)/(0.1) xx10^(-3)) = e^(-1) = (1)/(e) ampere`