An inductor (L = 100 mH), a resistor `(R = 100Omega)` 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 points 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

During decay of current,
`i=i_(0)e^(-(Rt)/(L))=(E)/(R)e^(-(Rt)/(L))=(100)/(100)e^((100xx10^(-3))/(100xx10^(-3)))=(1)/(e)A`