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. 0.1A
D. `1A`

Answer

A

Explanation

(a) 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 at time the current
Flowing is `-L(di)/(dt)-iR=0`
`implies(di)/(i)=-(R)/(L)dt implies int_(0)^(i) (di)/(i)=int_(0)^(t)-(R)/(L)dt`
`implies log_(e)(i)/(i_0)=-(R)/(L)t implies i=(i_0)e^(-(R/L)t)`
`implies i=(E)/(R)e^(-(R/L)t)=100/100e^(-100xx10^(-3))/(100xx10^(-3))=(1)/(e)`.