$A$ capacitor of capacitance $C$ is connected in series with a resistance $R$ and a $DC$ source of $emf$ $E$ through a key. The capacitor starts charging when the key is closed. By the time the capacitor has been fully charged, what amount of energy is dissipated in the resistance $R$?

Identify the valid statements relevant to the given circuit at the instant when the key is closed.
$A.$ There will be no current through resistor $R.$
$B.$ There will be maximum current in the connecting wires.
$C.$ Potential difference between the capacitor plates $A$ and $B$ is minimum.
$D.$ Charge on the capacitor plates is minimum.
Choose the correct answer from the options given below $:$

$A$ $500\,\mu F$ capacitor is charged at a steady rate of $100\,\mu C/s$. The potential difference across the capacitor will be $10\,V$ after an interval of.....$s$.

$A$ charged capacitor is allowed to discharge through a resistance of $2 \, \Omega$ by closing the switch $S$ at the instant $t = 0$. At time $t = \ln 2 \, \mu s$,the reading of the ammeter falls to half of its initial value. The resistance of the ammeter is equal to:

The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)

In the circuit shown in the figure,what happens in the steady state?