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,$\log_{e} 3 = 1.1$)

$A$ capacitor is charged to $320\ V$ and then connected to a resistor. After $1\ s$,the voltage is $240\ V$. What will be the voltage after $2\ s$ and $3\ s$?

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.

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

$A$ resistor $R$ and a $2 \ \mu F$ capacitor in series are connected through a switch to a $200 \ V$ direct supply. Across the capacitor is a neon bulb that lights up at $120 \ V$. Calculate the value of $R$ to make the bulb light up $5 \ s$ after the switch has been closed. (Given: $\log_{10} 2.5 = 0.4$)

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: