Two capacitances $C_1$ and $C_2$ are connected in series; assume that $C_1<C_2$. The equivalent capacitance of this arrangement is $C,$ where :

In the following circuit, the resultant capacitance between $\mathrm{A}$ and $\mathrm{B}$ is $1\mu \mathrm{F}$. Then value of $\mathrm{C}$ is

A capacitor of capacitance $5 \mathrm{\mu }F$ is connected, as shown in the figure. The internal resistance of the cell is $0.5 \mathrm{\Omega }$. The amount of charge on the capacitor plate will be

In the given circuit, the potential difference across 3 µF capacitor will be : 

If the potential of $A$ is $10 \mathrm{V},$ then potential of $B$ is:

Capacitance, $C_1= 2C_2 = 2C_3$ and, the potential difference across, $C_1, C_2$ and $C_3$ are $V_1, V_2$ and $V_3$ respectively then:

In the circuit shown in the figure, the battery is an ideal one with emf $V$. The capacitor is initially uncharged. Switch, $S$ is closed at time $t=0$.

The final charge $Q$ on the capacitor is:

In the circuit shown in the figure, the battery is an ideal one with E.M.F. $V$. The capacitor is initially uncharged. Switch, $S$ is closed at time $t=0$.

The final charge, $Q$ on the capacitor, is, what is the current in the steady-state is?

A light bulb, a capacitor, and a battery are connected together as shown here, with the switch $S$ initially open. When the switch $S$ is closed, which one of the following is true?