In a $CR$ circuit, the growth of charge on the capacitor is

$A$ combination of two identical capacitors, a resistor $R$, and a $DC$ voltage source of voltage $6\; V$ is used in an experiment on a $C-R$ circuit. It is found that for a parallel combination of the capacitors, the time in which the voltage of the fully charged combination reduces to half its original voltage is $10\; s$. For a series combination, the time needed for reducing the voltage of the fully charged series combination by half is: (in $; s$)

Two identical capacitors $A$ and $B$,charged to the same potential $V$,are connected in two different circuits as shown below at time $t=0$. If the charge on capacitors $A$ and $B$ at time $t=CR$ is $Q_{A}$ and $Q_{B}$ respectively,then (Here $e$ is the base of natural logarithm):

After the switch shown in Figure $A$ is closed,there is a current $i$ through the resistance $R$. Figure $B$ indicates current variation curves $a, b, c,$ and $d$ for four sets of values of $R$ and capacitance $C$:
$(i)$ $R_0$ and $C_0$
$(ii)$ $2R_0$ and $C_0$
$(iii)$ $R_0$ and $2C_0$
$(iv)$ $2R_0$ and $2C_0$
Which set goes with which curve?

Find the charge in steady state of the capacitor. (in $nC$)

Under steady state condition, the potential difference across the capacitor in the circuit is . . . . . . $V$.