Question

A capacitor of capacitance C has charge Q . It is connected to an identical capacitor through a resistance. The heat produced in the resistance. ( it's answer is Q2/4C).

Answer 1

Dear Student,

Initially, total charge is Q,

Thus energy in the system
$E=\frac{1}{2}\frac{Q^2}{C}\Rightarrow \frac{Q^2}{2C}$

When two similar capacitor's are connected in series, the charge divides into equal share. i.e Q/2
and equivalent capacitance becomes
$C_{eq}=\frac{C\times C}{C+C}\Rightarrow \frac{C}{2}$$C_{eq}=\frac{C\times C}{C+C}\Rightarrow \frac{C}{2}$
Thus,

Energy energy stored in the system

$$\begin{gathered} E\text{'}=\frac{Q{\text{'}}^2}{2C_{eq}}\Rightarrow \frac{{\displaystyle \raisebox{1ex}{$Q^2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{2{\displaystyle \raisebox{1ex}{$C$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\Rightarrow \frac{Q^2}{4C} \\ Thusenergylostfromthesystem(thisistheenergyexperiencedovertheresistor) \\ ∆E=E-E\text{'}\Rightarrow \frac{Q^2}{2C}-\frac{Q^2}{4C}=\frac{{\mathbf{Q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{C}} \end{gathered}$$

Regards.