The potential difference across a 2H inductor as a function of time is shown in figure. At t = 0, current is zero. Choose the correct statement.

$\mathrm{V}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}}\Rightarrow {\int }_0^{\mathrm{I}} \mathrm{dI}=\frac{1}{\mathrm{L}}{\int }_0^{\mathrm{t}} \mathrm{Vdt}$

$\Rightarrow \mathrm{I}=\frac{1}{\mathrm{L}}\text{ [Area under }\mathrm{v}-\mathrm{t}\text{ graph from }\mathrm{t}=0\text{ to }\mathrm{t}=\mathrm{t}\text{ ] }$

$\text{ At }\mathrm{t}=2\mathrm{s},\mathrm{I}=\frac{1}{2}\left[\frac{1}{2}\times 2\times 10\right]=5\mathrm{A}$

$\text{ For }\mathrm{t}=0\text{ to }2\mathrm{s},\mathrm{V}=5\mathrm{t}$

     $\mathrm{I}=\frac{1}{\mathrm{L}}{\int }_0^{\mathrm{t}} 5\mathrm{tdt}=\frac{1}{2}{\left[\frac{5{\mathrm{t}}^2}{2}\right]}_0^{\mathrm{t}}=\frac{5{\mathrm{t}}^2}{4}$

$\text{ For }\mathrm{t}=2\text{ to }4\mathrm{s},\mathrm{V}=-5\mathrm{t}+20$

   $\mathrm{I}=\frac{1}{2}{\int }_0^{\mathrm{t}} (-5\mathrm{t}+20)\mathrm{dt}\Rightarrow \mathrm{I}=\frac{-5{\mathrm{t}}^2}{4}+10\mathrm{t}$

Hence the correct graph is (c).