SSC JE Electrical Engineering 2014 Set 2 Paper

At time t = 0, when switch ‘s’ is closed, current i(t) flows through circuit.
Now according to kirchoff’s Law
$\mathit{R}\mathit{i}\left(\mathit{t}\right)+\mathit{L}\frac{\mathit{d}\mathit{i}\left(\mathit{t}\right)}{\mathit{d}\mathit{t}}=\mathit{E}$
Taking Laplace on both side we get
$\left(\mathit{R}+\mathit{S}\mathit{L}\right)\mathit{I}\left(\mathit{S}\right)=\frac{\mathit{E}}{\mathit{S}}$
∴ $\mathit{I}\left(\mathit{S}\right)=\frac{\mathit{E}}{\mathit{S}\left(\mathit{S}\mathit{L}+\mathit{R}\right)}$
$=\frac{10}{\mathit{S}\left(0.01\mathit{s}+1\right)}$
$=\frac{1000}{\mathit{s}\left(\mathit{s}+100\right)}$
Now applying partial fraction:
$\mathit{I}\left(\mathit{s}\right)=\frac{\mathit{A}}{\mathit{s}}+\frac{\mathit{B}}{\mathit{s}+100}$
Where $\mathit{A}={\frac{1000}{\mathit{s}+100}|}_{\mathit{s}=0}=\frac{1000}{100}=10$
and $\mathit{B}={\frac{1000}{\mathit{s}}|}_{\mathit{s}=-100}$ $=-\frac{1000}{100}=-10$
$\therefore \mathit{I}\left(\mathit{s}\right)=\frac{10}{\mathit{s}}-\frac{10}{\mathit{s}+100}$
Taking Inverse Laplace Transform:
$\mathit{i}\left(\mathit{t}\right)=10-10{\mathit{e}}^{-100\mathit{t}}$
∴ Option (a) is correct