In the given circuit diagram,when the current reaches a steady state,the charge on the capacitor of capacitance $C$ will be:

$A$ 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 of capacitance $C$ in the circuit shown is fully charged initially. The resistance is $R$. After the switch $S$ is closed,the time taken to reduce the stored energy in the capacitor to half its initial value is

In an $RC$ circuit while charging,the graph of $\log i$ versus time is as shown by the dotted line in the diagram,where $i$ is the current. When the value of the resistance is doubled,which of the solid curves best represents the variation of $\log i$ versus time?

After closing the key in the given circuit:
$(a)$ Bulb $(2)$ will glow and maintain its brightness.
$(b)$ Brightness of bulb $(1)$ will gradually decrease and at steady state it goes completely dark.

$A$ capacitor is charged to a potential difference of $100\, V$ and then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After $1\, s$,the potential difference across the capacitor plates is $80\, V$. What is the fractional loss of the stored energy?