Question

A parallel plate condenser of capacity C is connected to a battery and is charged to potential V. Another condenser of capacity $2\;C$ is connected to another battery and is charged to potential $2\;V$.The charging batteries are removed and now the condensers are connected in such a way that the positive plate of one is connected to negative plate of another. The final energy of this system is:
\[\begin{align}
  & A.0 \\
 & B.\dfrac{25C{{V}^{2}}}{6} \\
 & C.\dfrac{3C{{V}^{2}}}{2} \\
 & D.\dfrac{9C{{V}^{2}}}{2} \\
\end{align}\]

Answer 1

Hint: A capacitor is a two terminal component which stores electrical energy in the form of potential energy later discharges them. This property of capacitors is called the capacitance of the capacitor. Also, capacitors can be connected in series or in parallel circuits with respect to each other.
Formula: $E=\dfrac{1}{2}CV^{2}$

Complete answer:
A capacitor is an electrical device which can store electrical energy, and behaves as a temporary battery. These are used to maintain the power supply. It is one of the main components used in full wave and half wave rectifiers. (symbol: F), named after the English physicist Michael Faraday. Also, 1 farad capacitor, when charged with 1 coulomb of electrical charge, has a potential difference of 1 volt between its plates.
We know that the charge $Q$ produced due to capacitance $C$ and potential difference $V$ is given as $Q=CV$. Also, the energy of the capacitor is $E=\dfrac{1}{2}CV^{2}$.
Given that, the capacitance $C$ is charged to potential $V$, then the charge is given as $Q_{1}=CV$
Also, given, that another capacitance of $2\;C$ is charged to potential $2\;V$, then the charge is given as $Q_{2}=4CV$ is connected to each other then, the net charge, $Q=Q_{2}-Q_{1}=3CV$
Since both the capacitance are connected such that, the positive plate of one is connected to negative plate of another then, both are parallel in connection, then the net capacitance $C=C_{1}+C_{2}=C+2C=3C$
Then the new potential is given as $V=\dfrac{Q}{C}=\dfrac{3CV}{3C}=V$
Then the new energy is given as $E=\dfrac{1}{2}CV^{2}$
$\implies E=\dfrac{1}{2}(3C)V^{2}$

Hence the answer is option \[C.\dfrac{3C{{V}^{2}}}{2}\]

Note:
Also, we know that $C=\dfrac{KA\epsilon_{0}}{d}$ where, $A$ is the area of the capacitance and $d$ is the width of the dielectric $K$and the electric constant $\epsilon_{0}=8.854\times 10^{-12}Fm^{-1}$.Also, the series of capacitor is the sum of reciprocal of its individual capacitors, also remember that capacitor can charge and discharge.