A capacitor of capacitance $C$ is initially charged to a potential difference of V volt. Now it is connected to a battery of $2V$ volt with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be:
$\left( A \right)1.75$
$\left( B \right)2.25$
$\left( C \right)2.5$
$\left( D \right)\dfrac{1}{2}$

Hint: Capacitor is a component that has the capacity to store the energy. It is in the form of electrical charge producing a potential difference across its plates. Calculate the initial and final energy stored in the capacitor and use that to determine the heat generated in the capacitor. Then find the ratio between the two components.

Formula used:
$U = \dfrac{1}{2}C{V^2}$
$U$ is the energy stored, $C$ is the capacitance.

Complete step by step answer:
Capacitor is a component that has the capacity to store the energy. It is in the form of electrical charge producing a potential difference across its plates. Electrons flowing onto the plates are known as charging of capacitors. The potential difference depends on the number of charges present on the plates of the capacitor.
If the DC source is one in a resistance containing circuit, then the current will attain its maximum steady value in zero-time interval. A resistor circuit containing both inductor and capacitor, the current will take some time to attain its maximum peak value. If the battery is removed from the circuit which is having a capacitor or an inductor, then the current takes some time to decay to zero value.
Resonance occurs in a circuit that is when the inductor, capacitor and resistor are connected in series when the supply frequency causes the voltage across the inductor and capacitor to be equal. Q factor will be affected if there is resistive loss. Q factor is a unit less dimensionless quantity.
The initial and final charge is given by
${q_i} = CV$
${q_f} = 2CV$
In this case the battery is connected with opposite polarity hence,
$\vartriangle q = {q_f} - \left( { - {q_i}} \right) = 2CV + CV = 3CV$
Initial and final energy stored in the capacitor
$\Rightarrow {U_i} = \dfrac{1}{2}C{V^2}$
$\Rightarrow {U_f} = \dfrac{1}{2}C{\left( {2V} \right)^2} = 2C{V^2}$
Then the heat generated is $H = \left( {{U_f} - {U_i}} \right) - \vartriangle q\left( {2V} \right)$
$\Rightarrow H = 2C{V^2} - \dfrac{1}{2}C{V^2} - 6C{V^2} = \dfrac{9}{2}C{V^2}$
Thus, the ratio $\dfrac{H}{{{U_f}}} = \dfrac{{9C{V^2}}}{{2\left( {C{V^2}} \right)}} = 2.25$

Hence option B is the correct option.

Note: Electron flowing onto the plates is known as charging of capacitor. The potential difference depends on the number of charges present on the plates of the capacitor. If the battery is removed from the circuit which is having a capacitor or an inductor, then the current takes some time to decay to zero value.