A $600\,pF$ capacitor is charged by $200\,V$ supply. It is then disconnected from the supply and is connected to another uncharged $600\,pF$ capacitor. Electrostatic energy lost in the process is $.........\,\mu J$.

Consider a simple $RC$ circuit as shown in Figure $1$.

Process $1$: In the circuit the switch $S$ is closed at $t=0$ and the capacitor is fully charged to voltage $V_0$ (i.e. charging continues for time $T \gg R C$ ). In the process some dissipation ( $E_D$ ) occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $EC$.

Process $2$: In a different process the voltage is first set to $\frac{V_0}{3}$ and maintained for a charging time $T \gg R C$. Then the voltage is raised to $\frac{2 \mathrm{~V}_0}{3}$ without discharging the capacitor and again maintained for time $\mathrm{T} \gg \mathrm{RC}$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final

take $\mathrm{V}_0$ as voltage

These two processes are depicted in Figure $2$.

 ($1$) In Process $1$, the energy stored in the capacitor $E_C$ and heat dissipated across resistance $E_D$ are released by:

$[A]$ $E_C=E_D$ $[B]$ $E_C=E_D \ln 2$ $[C]$ $\mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}$ $[D]$ $E_C=2 E_D$

 ($2$) In Process $2$, total energy dissipated across the resistance $E_D$ is:

$[A]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{2} \mathrm{CV}_0^2$     $[B]$ $\mathrm{E}_{\mathrm{D}}=3\left(\frac{1}{2} \mathrm{CV}_0^2\right)$    $[C]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{3}\left(\frac{1}{2} \mathrm{CV}_0^2\right)$   $[D]$ $\mathrm{E}_{\mathrm{D}}=3 \mathrm{CV}_0^2$

Given the answer quetion  ($1$) and  ($2$)

$27$ similar drops of mercury are maintained at $10 \,V$ each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ....... times that of a smaller drop.

In the figure shown, after the switch $‘S’$ is turned from position $‘A’$ to position $‘B’$, the energy dissipated in the circuit in terms of capacitance $‘C’$ and total charge $‘Q’$ is

A parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E _2$. The ratio $E _2 / E _1$ is

A capacitor of capacity $C$ is connected with a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential $V$ again, the energy given by the battery will be