$A$ capacitor of $4\, \mu F$ is connected to a $15\, V$ supply through a $1\, M\Omega$ resistance. The time taken by the capacitor to charge up to $63.2\%$ of its final charge will be......$s$.

$A$ capacitor of capacitance $C$ is discharging through a resistor $R$. Let $t_1$ be the time taken for the energy stored in the capacitor to reduce to half of its initial value,and $t_2$ be the time taken for the charge to reduce to one-fourth of its initial value. Then the ratio $t_1 / t_2$ is:

In the circuit shown below,the key is pressed at time $t=0$. Which of the following statement$(s)$ is(are) true?
$(A)$ The voltmeter displays $-5 \ V$ as soon as the key is pressed,and displays $+5 \ V$ after a long time.
$(B)$ The voltmeter will display $0 \ V$ at time $t=\ln 2 \ s$.
$(C)$ The current in the ammeter becomes $1/e$ of the initial value after $1 \ s$.
$(D)$ The current in the ammeter becomes zero after a long time.

An electric bulb, a capacitor, a battery, and a switch are all connected in series in a circuit. How does the intensity of light vary when the switch is turned on?

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:

$A$ charged capacitor is allowed to discharge through a resistor by closing the key at the instant $t = 0$. At the instant $t = (\ln 4) \, \mu s$,the reading of the ammeter falls to half the initial value. The resistance of the ammeter is equal to