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

What are the charges stored in the $1 \mu F$ and $2 \mu F$ capacitors in the circuit as shown in the figure once the current $(I)$ becomes steady?

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 $:$

The time constant of the given circuit is:

The plates of a capacitor are charged to a potential difference of $320 \, V$ and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After $1 \, s$ the potential difference between the plates of the capacitor is $240 \, V$,then after $2 \, s$ and $3 \, s$ the potential difference between the plates will be:

Two identical capacitors $A$ and $B$ are charged to the same potential $V$ and are connected in two circuits at $t = 0,$ as shown in the figure. The charge on the capacitors at time $t = CR$ are respectively: