JEE Physics 2024 Question with Solution

Given: First capacitor has capacitance $C$, potential $V$, and energy $E = \frac{1}{2}CV^2$. Second capacitor has capacitance $2C$ and potential $2V$.

Find: The value of $x$ if the loss of energy is $\frac{x}{3}E$.

Use conservation of charge after connecting the capacitors in parallel and then compare initial and final energies.

Initial energy of the first capacitor:

$$E_1 = \frac{1}{2}CV^2 = E$$

Initial energy of the second capacitor:

$$E_2 = \frac{1}{2}(2C)(2V)^2 = 4CV^2$$

So, total initial energy is

$$E_{\text{initial}} = E_1 + E_2 = \frac{1}{2}CV^2 + 4CV^2 = \frac{9}{2}CV^2$$

Initial charges are

$$Q_1 = CV$$ $$Q_2 = (2C)(2V) = 4CV$$

Hence total charge is

$$Q_{\text{total}} = Q_1 + Q_2 = 5CV$$

Total capacitance after parallel connection is

$$C_{\text{total}} = C + 2C = 3C$$

Therefore common potential is

$$V_{\text{common}} = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{5CV}{3C} = \frac{5V}{3}$$

Final energy of the system is

$$E_{\text{final}} = \frac{1}{2}(3C)\left(\frac{5V}{3}\right)^2 = \frac{25}{6}CV^2$$

Thus loss of energy is

$$\Delta E_{\text{loss}} = E_{\text{initial}} - E_{\text{final}} = \frac{9}{2}CV^2 - \frac{25}{6}CV^2 = \frac{1}{3}CV^2$$

Now, since

$$E = \frac{1}{2}CV^2$$

we get

$$\Delta E_{\text{loss}} = \frac{1}{3}CV^2 = \frac{2}{3}\left(\frac{1}{2}CV^2\right) = \frac{2}{3}E$$

Comparing with

$$\Delta E_{\text{loss}} = \frac{x}{3}E$$

we obtain

$$x = 2$$

Therefore, the correct option is B.

Given: Two isolated capacitors, one with $C, V$ and the other with $2C, 2V$.

Find: The value of $x$ in the expression $\frac{x}{3}E$ for energy loss.

The key idea is that charge is conserved, but electrostatic energy decreases after redistribution.

1. First capacitor stores

$$E = \frac{1}{2}CV^2$$

2. Second capacitor stores

$$\frac{1}{2}(2C)(2V)^2 = 4CV^2$$

3. Total initial energy:

$$\frac{1}{2}CV^2 + 4CV^2 = \frac{9}{2}CV^2$$

4. Total charge before connection:

$$CV + 4CV = 5CV$$

5. Equivalent capacitance after connection:

$$3C$$

6. Common potential:

$$\frac{5CV}{3C} = \frac{5V}{3}$$

7. Final energy:

$$\frac{1}{2}(3C)\left(\frac{5V}{3}\right)^2 = \frac{25}{6}CV^2$$

8. Energy loss:

$$\frac{9}{2}CV^2 - \frac{25}{6}CV^2 = \frac{1}{3}CV^2$$

9. Since $E = \frac{1}{2}CV^2$,

$$\frac{1}{3}CV^2 = \frac{2}{3}E$$

So the loss is $\frac{2}{3}E$, which means $x = 2$.

Therefore, the correct option is B.