$A$ $4 \mu F$ capacitor and a resistance of $2.5 \, M\Omega$ are in series with a $12 \, V$ battery. Find the time after which the potential difference across the capacitor is $3$ times the potential difference across the resistor. (Given $\ln(2) = 0.693$)

At $t = 0$,switch $S$ is closed. The charge on the capacitor is varying with time $t$ as $Q = Q_0(1 - e^{-\alpha t})$. Find the value of $Q_0$.

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:

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 $:$

$A$ parallel plate capacitor is charged to a potential difference of $50 \, V$. It is discharged through a resistance. After $1 \, s$,the potential difference between the plates becomes $40 \, V$. Then:

In the circuit diagram shown,$E = 5 \, V, r = 1 \, \Omega, R_2 = 4 \, \Omega, R_1 = R_3 = 1 \, \Omega$ and $C = 3 \, \mu F$. The magnitude of the charge on each capacitor plate is......$\mu C$.