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(A) Initially,the capacitor $C$ connected across $AB$ is charged to $Q$. The initial energy stored is $U_i = \frac{Q^2}{2C}$.When the switch is shifte

Vedclass · Accepted Answer

$\frac{Q^2}{3C}$

Vedclass · Answer

(A) Initially,the capacitor $C$ connected across $AB$ is charged to $Q$. The initial energy stored is $U_i = \frac{Q^2}{2C}$.When the switch is shifted to position $2$,the capacitor $C$ is connected in parallel with two other capacitors,each of capacitance $C$. The total equivalent capacitance of the circuit becomes $C_{eq} = C + C + C = 3C$.The charge $Q$ is now redistributed among these three capacitors. Since they are in parallel,the final potential difference across each is the same. The final energy stored in the system is $U_f = \frac{Q^2}{2C_{eq}} = \frac{Q^2}{2(3C)} = \frac{Q^2}{6C}$.The heat released is the difference between the initial and final energy: $	ext{Heat} = U_i - U_f = \frac{Q^2}{2C} - \frac{Q^2}{6C} = \frac{3Q^2 - Q^2}{6C} = \frac{2Q^2}{6C} = \frac{Q^2}{3C}$.

If the capacitor connected across $AB$ is charged to $Q$ and then the switch is shifted to position $2$,calculate the heat released after that.

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