$A$ capacitor of $4\, \mu F$ is connected to a $15\, V$ supply through a $1\, M\Omega$ resistance. The time taken by the capacitor to charge up to $63.2\%$ of its final charge will be......$s$.

$A$ capacitor of capacitance $C$ is discharging through a resistor $R$. Let $t_1$ be the time taken for the energy stored in the capacitor to reduce to half of its initial value,and $t_2$ be the time taken for the charge to reduce to one-fourth of its initial value. Then the ratio $t_1 / t_2$ 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 $; s$)

In the circuit shown,the switch is shifted from position $1 \rightarrow 2$ at $t = 0$. The switch was initially in position $1$ for a long time. The graph between the charge on capacitor $C$ and time $t$ is:

$A$ capacitor is connected to a cell of $emf$ $E$ having some internal resistance $r$. The potential difference across the

$A$ charged capacitor is allowed to discharge through a resistor by closing the key at the instant $t = 0$. At the instant $t = (\ln 4) \, \mu s$,the reading of the ammeter falls to half the initial value. The resistance of the ammeter is equal to