Figure shows a part of complete circuit at $t = 0$ Charge on capacitor as a function of time is given by $q = 3$ $\left(1-e^{-t}\right)$ coulombs. At $t=0,$ mark the correct statement(s}.  

 

Current through branch containing capacitor is

$\frac{dq}{dt}=i_1=3e^{-t}Amp$  

At $t=0,i_1=3A$ in downward direction i.e. towards $P$ Let $V$ is the potential of point $P$ then,   

$20-V=\frac{q}{C}+3\times 1=\frac{3}{1}+3=6\Rightarrow V=14V$

Current through branch containing battery is. 

$I_2=\frac{18-14}{1}=4\mathrm{A}$ again towards $P$. 

Applying $KCL$ at $P$. Current through $R$ is $7 A$. 

Potential difference across $R$ is $V=14V=7R$

$\Rightarrow R=2\mathrm{\Omega }$.