In the circuit shown below voltmeter and ammeter are ideal, the key is pressed at time t=0. Which of the following statement(s) is (are) true?
 

(A) At t=0, capacitor acts as short-circuit

At $\mathrm{t}\to \infty$, capacitor acts as open circuit & no current flows through voltmeter
 

$$\begin{gathered} {\mathrm{q}}_{\mathrm{x}}=2\mathrm{CV}\left(1-{\mathrm{e}}^{-\mathrm{t}/2\mathrm{CR}}\right) \mathrm{X}=\frac{\mathrm{V}}{\mathrm{R}}{\mathrm{e}}^{-\mathrm{t}/2\mathrm{CR}} \\ {\mathrm{q}}_{\mathrm{y}}=\mathrm{CV}(1-{\mathrm{e}}^{-\mathrm{t}/2\mathrm{CR}}) \mathrm{Y}=\frac{\mathrm{V}}{2\mathrm{R}}{\mathrm{e}}^{-\mathrm{t}/2\mathrm{CR}} \\ \mathrm{\Delta V}=-\mathrm{y}2\mathrm{R}+\frac{{\mathrm{q}}_{\mathrm{x}}}{2\mathrm{C}} \\ =\mathrm{V}\left[1-2{\mathrm{e}}^{-\mathrm{t}/2\mathrm{CR}}\right]=0 \\ \left(\mathrm{C}\right)\mathrm{\tau }=1\sec \end{gathered}$$
So by $\mathrm{i}={\mathrm{i}}_0{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}$ current at t = 1 sec is$={\mathrm{i}}_0/\mathrm{e}$
D) After long time no current flows since both capacitor & voltmeter do not allow it.