In the circuit shown,the switch is shifted from position $1 \rightarrow 2$ at $t = 0$. The switch was initially in position $1$ for a long time. The graph between the charge on capacitor $C$ and time $t$ is:

When the switch $K$ is closed at $t = 0$,what can be said about the current passing through the resistor $AB$ in the given circuit?

$A$ 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

$A$ 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 capacitor $C$ is initially without charge. $X$ is now joined to $Y$ for a long time,during which $H_1$ heat is produced in the resistance $R$. The $X-Y$ connection is removed and $X$ is now joined to $Z$ for a long time,during which heat $H_2$ is produced in $R$.

The circuit shown in the figure consists of a battery of $emf$ $\varepsilon = 10 \,V$,a capacitor of capacitance $C = 1.0 \, \mu F$,and three resistors of values $R_1 = 2 \, \Omega$,$R_2 = 2 \, \Omega$,and $R_3 = 1 \, \Omega$. Initially,the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t = 0$.