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

Answer

A

Explanation

Initially, when steady state is achieved, i = (E)/(R)
Let E is short circuited at t = 0. Then
At t = 0, `i_(0) = (E)/(R)`
Let during decay of current at any time the current flowing is `- L (dl)/(dt) - iR = 0`
`Rightarrow (di)/(i) = -(R)/(L) dt Rightarrow int_(i_(0))^(i) (dl)/(i) = int_(0)^(t) - (R)/(L) dt`
`log_(e)(i)/(i_(0)) = -(R)/(L) t rightarrow i = i_(0) e^(-(R)/(L)t)`
`rightarrow i = (E)/(R)e^(-(R)/(L)t) = (100)/(100) e^((-100 xx 10^(-3))/(100 xx 10^(-3))) = (1)/(e)`