$A$ $20 \mu F$ capacitor is connected to a $45 \text{ V}$ battery through a circuit whose resistance is $2000 \Omega$. What is the final charge on the capacitor?

$A$ 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 $t_1/t_2$.

For the $RC$ circuit shown,the resistance is $R = 10.0\ \Omega$,the capacitance is $C = 5.0\ F$ and the battery has voltage $\xi = 12\ V$. The capacitor is initially uncharged when the switch $S$ is closed at time $t = 0$. At some time later,the current in the circuit is $0.50\ A$. What is the magnitude of the charge across the capacitor at that moment (in $C$)?

The circuit shown in the figure consists of a charged capacitor of capacity $3\, \mu F$ and a charge of $30\, \mu C$. At time $t = 0,$ when the key is closed,the value of current flowing through the $5\, M\Omega$ resistor is $'x'\, \mu A$. The value of $'x'$ to the nearest integer is .........

The circuit shown here has two batteries of $8.0 \, V$ and $16.0 \, V$,three resistors of $3 \, \Omega$,$9 \, \Omega$,and $9 \, \Omega$,and a capacitor of $5.0 \, \mu F$. How much is the current $I$ in the circuit in steady state? ................... $A$

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