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(A, C, D) Initially, when the switch $K$ is closed, both capacitors $A$ and $B$ are in parallel with the battery of voltage $V$. The total energy stor

Vedclass · Accepted Answer

When the switch is opened, total energy stored in two capacitors is $\frac{5}{3} CV^2$

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(A, C, D) Initially, when the switch $K$ is closed, both capacitors $A$ and $B$ are in parallel with the battery of voltage $V$. The total energy stored is $U = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$.When the switch $K$ is opened, capacitor $A$ remains connected to the battery, while capacitor $B$ is isolated. The charge on $B$ remains constant at $q_B = CV$.After inserting the dielectric $(K=3)$:For capacitor $A$: The new capacitance is $C_A' = KC = 3C$. The voltage remains $V$. The energy stored is $U_A = \frac{1}{2}C_A'V^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$.For capacitor $B$: The charge $q_B = CV$ remains constant. The new capacitance is $C_B' = KC = 3C$. The energy stored is $U_B = \frac{q_B^2}{2C_B'} = \frac{(CV)^2}{2(3C)} = \frac{C^2V^2}{6C} = \frac{1}{6}CV^2$.The total energy stored is $U_{total} = U_A + U_B = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \frac{9+1}{6}CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$.Thus, statements $A$, $C$, and $D$ are correct.

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