In a circuit shown in the figure, the capacitor C is initially uncharged, and the key K is open. In this condition, a current of $1\mathrm{A}$ flows through the $1\mathrm{\Omega }$ resistor. The key is closed at time $\mathrm{t}={\mathrm{t}}_0$. Which of the following statement(s) is(are) correct?  
[Given: ${\mathrm{e}}^{-1}=0.36$] 

By applying Kirchhoff’s law
$15-\left({\mathrm{i}}_1+1\right)\mathrm{R}=5+1\left(1\right)={\mathrm{i}}_1\left(3\right)\Rightarrow {\mathrm{i}}_1=2\Rightarrow \mathrm{R}=3$
Potential difference across capacitor before closing the switch 
$$\begin{gathered} \begin{array}{l}\mathrm{Voltage} \mathrm{drop} \mathrm{across} 3\mathrm{\Omega } \mathrm{is}=2\left(3\right)=6\mathrm{V}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\therefore {\mathrm{q}}_{max}=\mathrm{CV}=2\left(6\right)\mathrm{\mu C}\\ {\mathrm{q}}_{max}=12\mathrm{\mu C}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \\ {\mathrm{R}}_{\mathrm{eq}}=\frac{1}{\frac{1}{3}+\frac{1}{3}+1}+3 {\mathrm{R}}_{\mathrm{eq}}=\frac{3}{5}+3=\frac{18}{5}\mathrm{\tau }={\mathrm{R}}_{\mathrm{eq}}\mathrm{C}\\ \mathrm{\tau }=\frac{18}{5}\times 2=7.2\mathrm{\mu s}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{q}=12\mathrm{\mu c}\left(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}}\right)\end{array} \end{gathered}$$
$\mathrm{i}=\frac{12\mathrm{\mu c}}{\mathrm{\tau }}{\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{\tau }}}{\mathrm{i}}_{max}=\frac{12}{7.2}\times 0.36=0.6$