Let $C$ be the capacitance of a capacitor discharging through a resistor $R$. Suppose $t_{1}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_{2}$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_{1} / t_{2}$ will be

The circuit shown has been connected for a long time. The voltage across the capacitor is ............. $V$.

In the circuit diagram shown,$E = 5 \, V, r = 1 \, \Omega, R_2 = 4 \, \Omega, R_1 = R_3 = 1 \, \Omega$ and $C = 3 \, \mu F$. The magnitude of the charge on each capacitor plate is......$\mu C$.

When a capacitor is connected to a battery,

The circuit shown in the figure consists of a battery of $emf$ $\varepsilon = 10 \,V$,a capacitor of capacitance $C = 1.0 \, \mu F$,and three resistors of values $R_1 = 2 \, \Omega$,$R_2 = 2 \, \Omega$,and $R_3 = 1 \, \Omega$. Initially,the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t = 0$.

$A$ filter circuit used in a rectifier has a load resistance of $200 \Omega$ and a capacitance of $15 \mu\text{F}$. The value of the time constant is . . . . . . .