Charging and Discharging of a Capacitor Through a Resistance

The space between the plates of a parallel plate capacitor is completely filled with a material of resistivity $2\times {10}^{11} \mathrm{\Omega } \mathrm{m}$ and dielectric constant $6$. The capacity of the capacitor with the given dielectric medium between the plates is $20 \mathrm{nF}$. Find the leakage current if a potential difference $2500 \mathrm{V}$ is applied across the capacitor.

A parallel plate capacitor of capacitance $10 \mu \mathrm{F}$ is connected to a cell of emf $10 \mathrm{V}$ and is fully charged. Now a dielectric slab $\left(k=3\right)$ of thickness equal to the gap between the plates is completely filled in the gap, keeping the cell connected. During the filling process,

At time $t=0,$ a battery of $10 \mathrm{V}$ is connected across points $A$ and $B$ in the given circuit. If the capacitors have no charge initially, at what time (in $\mathrm{seconds}$) does the voltage across them becomes $4 \mathrm{V}$? [Take $\ln 5=1.6, \ln 3=1.1$]

In an $LCR$ circuit, as shown below, both switches are open initially. Now the switch $S_1$ is closed and $S_2$ kept open. ($q$ is the charge on the capacitor and $\mathrm{\tau }=RC$ is the capacitive time constant). Which of the following statement is correct?

The figure shows an experimental plot discharging of a capacitor in an $RC$ circuit. The time constant $\mathrm{\tau }$ of this circuit lies between:

A resistor $R$ and $2 \mathrm{\mu F}$ capacitor in series are connected through a switch to $200 \mathrm{V}$ direct supply. Across the capacitor is a neon bulb that lights up at $120 \mathrm{V}$. Calculate the value of $R$ to make the bulb light up $5 \mathrm{s}$ after the switch has been closed. $\left({\log }_{10}2.5=0.4\right)$

Let $C$ be the capacitance of a capacitor discharging through a resistor $R$. Suppose $t_1$ be the time taken for the energy stored in the capacitor to reduce to half 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 $\frac{t_1}{t_2}$ will be,

In the circuit shown, the capacitor $C_1$ is initially charged with charge $q_0$. The switch $S$ is closed at time $t=0$. The charge on $C_2$ after time $t$ is

The charge on the capacitor in steady-state in the circuit shown (figure) is

The switch $S$ in the circuit diagram (figure) is closed at $t=0$. The charge on capacitors at any time $t$ is,

Initially, switch $S$ is connected to position $1$ for a long time (figure). The net amount of heat generated in the circuit after it is shifted to position $2$ is

Each resistor in the following circuit (figure) has a resistance of $2 \mathrm{M}\mathrm{\Omega }$ and the capacitors have a capacitance of $1 \mathrm{\mu }\mathrm{F}$. The battery voltage is $3 \mathrm{V}$. The voltage across the resistor $A$ in the following circuit in steady state is,

In the circuit shown (figure), the switch $S_2$ is closed first and is kept closed for a long time. Now, $S_1$ is closed, Just after that instant, the current thought $S_1$ is,

In the figure, the charge that flows from $P$ to $Q$ when the switch $S$ is closed is,