During the charging of a capacitor,the variation of the potential $V$ of the capacitor with time $t$ is shown as:

$A$ capacitor of capacity $0.1 \mu F$ connected in series to a resistor of $10 M \Omega$ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given, $\log _{10} 2=0.3010$ ) (in $\,s$)

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

The capacitor $C$ is initially uncharged. Switch $S_1$ is closed for a long time while $S_2$ remains open. Now at $t = 0$, $S_2$ is closed while $S_1$ is opened. All the batteries are ideal and connecting wires are resistanceless. Find the $\text{INCORRECT}$ statement.

Find the time constant for the given circuit.

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? ................... $A$