1. Initial energy of the first capacitor

Given:

• First capacitor: capacitance $C$, potential $V$
• Its energy is given as $E$

Using $U = \frac{1}{2}CV^2$ so $E = \frac{1}{2}CV^2$

2. Initial energy of the second capacitor

Second capacitor has:

• Capacitance $2C$
• Potential $2V$

Its energy is $U_2 = \frac{1}{2}(2C)(2V)^2$ $U_2 = \frac{1}{2}(2C)(4V^2) = 4CV^2$

Now, since $E = \frac{1}{2}CV^2 \implies CV^2 = 2E$ therefore $U_2 = 4CV^2 = 4(2E) = 8E$

So, total initial energy is $U_i = E + 8E = 9E$

3. Initial charges on the capacitors

Charge on first capacitor: $Q_1 = CV$

Charge on second capacitor: $Q_2 = (2C)(2V) = 4CV$

Assuming like plates are connected together, total charge becomes $Q_{\text{total}} = Q_1 + Q_2 = CV + 4CV = 5CV$

4. Common final potential

After connection, equivalent capacitance is $C_{\text{eq}} = C + 2C = 3C$

Hence final common potential is $V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{5CV}{3C} = \frac{5V}{3}$

5. Final energy

Final energy stored in the combination: $U_f = \frac{1}{2}(3C)\left(\frac{5V}{3}\right)^2$ $U_f = \frac{1}{2}(3C)\cdot \frac{25V^2}{9}$ $U_f = \frac{25}{6}CV^2$

Using $CV^2 = 2E$ we get $U_f = \frac{25}{6}(2E) = \frac{25E}{3}$

6. Loss of energy

$\Delta U = U_i - U_f = 9E - \frac{25E}{3}$ $\Delta U = \frac{27E - 25E}{3} = \frac{2E}{3}$

Given loss of energy is $\frac{x}{3}E$ So, $\frac{x}{3}E = \frac{2}{3}E$ which gives $x = 2$

7. Comparison with stored answer

Stored correct answer = $2$

Our derived answer also is $2$.