SSC JE Electrical Engineering 2014 Set 2 Paper

From Kirchoff’s voltage Law
$\mathit{R}\mathit{i}\left(\mathit{t}\right)+\mathit{L}\frac{\mathit{d}\mathit{i}\left(\mathit{t}\right)}{\mathit{d}\mathit{t}}+\frac{1}{\mathit{c}}\int \mathit{i}\left(\mathit{t}\right)\mathit{d}\mathit{t}=\mathit{E}$
where i(t) is the transient current At time t = 0,
Taking Laplace transform on both side:
$\mathit{R}\mathit{I}\left(\mathit{s}\right)+\mathit{S}\mathit{L}\mathit{I}\left(\mathit{s}\right)+\frac{1}{\mathit{S}\mathit{c}}\mathit{I}\left(\mathit{s}\right)=\frac{\mathit{E}}{\mathit{s}}$
∴ $\mathit{I}\left(\mathit{s}\right)=\frac{\mathit{E}}{\mathit{S}\left(\mathit{R}+\mathit{S}\mathit{L}+\frac{1}{\mathit{S}\mathit{C}}\right)}$ $=\frac{\mathit{E}\mathit{C}}{\left({\mathit{S}}^2\mathit{L}\mathit{C}+\mathit{R}\mathit{C}\mathit{S}+1\right)}$
∴ $\mathit{I}\left(\mathit{s}\right)=\frac{50\times 0.08}{0.5\times 0.08\times {\mathit{s}}^2+5\times 0.08\times \mathit{s}+1}$
= $\frac{100}{{\mathit{s}}^2+10\mathit{s}+25}=\frac{100}{{\left(\mathit{s}+5\right)}^2}$
Now taking Inverse Laplace Transform
$\mathit{i}\left(\mathit{t}\right)=100\mathit{t}{\mathit{e}}^{\mathit{–}5\mathit{t}}$
∴ Option (c) is correct.