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 is
$
  {\text{A}}{\text{. }}\dfrac{{{Q^2}}}{{2C}} \\
  {\text{B}}{\text{. }}\dfrac{{{Q^2}}}{{4C}} \\
  {\text{C}}{\text{. }}\dfrac{{{Q^2}}}{{8C}} \\
  {\text{D}}{\text{. Dependent on the value of the resistance}} \\
$

Answer 1

Hint: The heat produced in the resistor will be equal to energy lost when the new capacitor is connected. The charge Q gets equally distributed between the two identical capacitors which decide the total energy of the capacitor in two cases.

Formula used:
The energy stored in a capacitor is given as
$E = \dfrac{{{Q^2}}}{{4C}}{\text{ }}...{\text{(i)}}$
Here E is the energy which is stored in the capacitor.
Q is the charge stored in the capacitor.
C is the capacitance of the given capacitor.

Detailed step by step solution:
Initially, we have one capacitor whose capacitance is C and charge stored in it is Q. The energy stored in the capacitor is given by equation (i) as follows:
${E_1} = \dfrac{{{Q^2}}}{{4C}}{\text{ }}...{\text{(ii)}}$
Now this capacitor is connected to another capacitor of the same capacitance through a resistor. The total charge Q will be distributed between these capacitors since total charge must remain the same or conserved. Therefore, each of the capacitors have a charge equal to $\dfrac{Q}{2}$ as the two capacitors are identical to each other. Now we can calculate the total energy of the two capacitors which is given as
${E_2} = \dfrac{{{{\left( {\dfrac{Q}{2}} \right)}^2}}}{{4C}} + \dfrac{{{{\left( {\dfrac{Q}{2}} \right)}^2}}}{{4C}} = 2 \times \dfrac{{{{\left( {\dfrac{Q}{2}} \right)}^2}}}{{4C}} = \dfrac{{{Q^2}}}{{8C}}{\text{ }}...{\text{(iii)}}$
We notice that when the other capacitor is put in the circuit then the total energy of the capacitors has decreased. The lost energy is converted into heat in the resistor. The amount of heat produced in the resistor is equal to the difference between the initial energy stored in the capacitor and the final energy stored in two capacitors. This can be given as
$
  E = {E_1} - {E_2} \\
   = \dfrac{{{Q^2}}}{{4C}} - \dfrac{{{Q^2}}}{{8C}} \\
   \Rightarrow E = \dfrac{{{Q^2}}}{{8C}} \\
$
Hence, the correct answer is option C.

Note: The total energy of a system is always conserved. The lost energy is converted into some other form of energy through some other component of the circuit. In our case, we have the resistor which acts as a medium to expend the energy lost by the capacitors.