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, $\frac{dq}{dt}=I=3e^{-t}amp$

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

$$\begin{gathered} 20-V=\frac{q}{C}+3\times 1=\frac{0}{1}+3=3 \\ \Rightarrow V=17volt \end{gathered}$$

Current through branch containing 1$\Omega$ is,

$I_2=\frac{18-17}{1}=1A$ towards P.

Applying KCL at P, current through R is 4A and potential difference across R is 17 V.

$4=\frac{17}{R}\Rightarrow R=\frac{17}{4}\Omega$