The figure shows an experimental plot for discharging of a capacitor in an $R-C$ circuit. The time constant $\tau$ of this circuit lies between

In the circuit shown in the figure,what happens in the steady state?

As shown in the figure,if a capacitor of capacitance $C$ is charged by connecting it with a resistance $R$ and a battery of $EMF$ $V$,then the total energy supplied by the battery is:

In the circuit shown,the cells are ideal and of equal emfs $E$. The capacitance of the capacitor is $C$ and the resistance of the resistor is $R$. $X$ is first joined to $Y$ and then to $Z$. After a long time,the total heat produced in the resistor will be:

$A$ 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 $; s$)

The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)