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. `eA`
B. `0.1A`
C. `1A`
D. `1//eA`

Answer

D

Explanation

`i_(0)=E//R=(100)/(100)=1A` , `tau=(L)/(R)=(100xx10^(-3))/(100)=10^(-3)sec`
`i=i_(0)e^(-t//tau)`
`i=i_(0)e^((10^(-3))/(10^(-3)))=i_(0)e^(-1)=1//eA`