$A$ filter circuit used in a rectifier has a load resistance of $200 \Omega$ and a capacitance of $15 \mu\text{F}$. The value of the time constant is . . . . . . .

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

$A$ combination of two identical capacitors, a resistor $R$, and a $DC$ voltage source of voltage $6\; V$ is used in an experiment on a $C-R$ circuit. It is found that for a parallel combination of the capacitors, the time in which the voltage of the fully charged combination reduces to half its original voltage is $10\; s$. For a series combination, the time needed for reducing the voltage of the fully charged series combination by half is: (in $; s$)

Find the charge on the capacitor $C$ in the following circuit in $\mu C$.

$A$ capacitor of capacity $C$ is charged to a steady potential difference $V$ and connected in series with an open key and a pure resistor $R$. At time $t = 0$,the key is closed. If $I$ is the current at time $t$,a plot of $\log I$ against $t$ is as shown in $(1)$ in the graph. Later,one of the parameters,i.e.,$V, R,$ or $C$,is changed while keeping the other two constant,and the graph $(2)$ is recorded. 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