$A$ capacitor of capacitance $C$ is being discharged 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. Find the ratio $t_1/t_2$.

In the given figure,initially the capacitor is uncharged. The ratio of current at $t = 0$ and $t = \infty$ will be:

$A$ capacitor of capacitance $C = 1 \, \mu F$ is suddenly connected to a battery of $100 \, V$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to $50 \, V$ is $.... \times 10^{-4} \, s$. (Take $\ln 2 = 0.69$)

In the circuit shown in the diagram,the capacitances of the capacitors $C_1$,$C_2$,and $C_3$ are $4\ \mu F$,$6\ \mu F$,and $12\ \mu F$ respectively. The switch $S$ remains closed for a long time. When the switch $S$ is opened,which of the following statements will be correct about the current flowing through the battery $B$?

The potential difference between the plates of a parallel plate capacitor is changing at the rate of $10^6\, V/s$. If the capacitance is $2\,\mu F$,the displacement current in the dielectric of the capacitor will be......$A$

Given,
${R_1} = 1\,\Omega, R_2 = 2\,\Omega$
${C_1} = 2\,\mu F, C_2 = 4\,\mu F$
The time constants (in $\mu s$) for the circuits $I, II, III$ are respectively: