a
Initially, when steady state is achieved,

\(i=\frac{E}{R}\)

Let \(E\) is short circuited at \(t=0 .\) Then

At \(t=0, i_{0}=\frac{E}{R}\)

Let during decay of current at any time the current

flowing is \(-L \frac{d i}{d t}-i R=0\)

\(\Rightarrow \frac{d i}{i}=-\frac{R}{L} d t \Rightarrow \int_{i_{0}}^{i} \frac{d i}{i}=\int_{0}^{t}-\frac{R}{L} d t\)

\(\Rightarrow \log _{e} \frac{i}{i_{0}}=-\frac{R}{L} t \Rightarrow i=i_{0} e^{-\frac{R}{L} t}\)

\(\Rightarrow i=\frac{E}{R} e^{-\frac{R}{L} t}=\frac{100}{100} e^{\frac{-100 \times 10^{-3}}{100 \times 10^{-3}}}=\frac{1}{e}\)