When a lamp is connected in series with a capacitor,then:

Three identical capacitors are given a charge $Q$ each and they are then allowed to discharge through resistances $R_1, R_2,$ and $R_3$. Their charges,as a function of time,are shown in the graph below. The smallest of the three resistances is

Let $C$ be the capacitance of a capacitor discharging through a resistor $R$. Suppose $t_{1}$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_{2}$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $t_{1} / t_{2}$ will be

$A$ circuit element is placed in a closed box. At time $t=0$,a constant current generator supplying a current of $1\, A$ is connected across the box. The potential difference across the box varies according to the graph shown in the figure. The element in the box is:

$A$ $500\,\mu F$ capacitor is charged at a steady rate of $100\,\mu C/s$. The potential difference across the capacitor will be $10\,V$ after an interval of.....$s$.

As shown in the figure,if a capacitor of capacitance $C$ is charged by connecting it with a resistance $R$ and a battery of $EMF$ $V$,then the total energy supplied by the battery is: