A capacitor of capacitance C and potential V | vedclass

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(C) Initial energy of the first capacitor: $E_1 = \frac{1}{2} C V^2 = E$.Initial energy of the second capacitor: $E_2 = \frac{1}{2} (2C) (2V)^2 = \fra

Vedclass · Accepted Answer

$2$

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(C) Initial energy of the first capacitor: $E_1 = \frac{1}{2} C V^2 = E$.Initial energy of the second capacitor: $E_2 = \frac{1}{2} (2C) (2V)^2 = \frac{1}{2} (2C) (4V^2) = 4 C V^2 = 8E$.Total initial energy: $E_i = E_1 + E_2 = E + 8E = 9E$.Total charge: $Q_{total} = Q_1 + Q_2 = CV + (2C)(2V) = CV + 4CV = 5CV$.Total capacitance: $C_{eq} = C + 2C = 3C$.Common potential: $V_{common} = \frac{Q_{total}}{C_{eq}} = \frac{5CV}{3C} = \frac{5}{3} V$.Final energy: $E_f = \frac{1}{2} C_{eq} V_{common}^2 = \frac{1}{2} (3C) (\frac{5}{3} V)^2 = \frac{1}{2} (3C) (\frac{25}{9} V^2) = \frac{25}{6} C V^2 = \frac{25}{3} E$.Loss of energy: $\Delta E = E_i - E_f = 9E - \frac{25}{3} E = \frac{27E - 25E}{3} = \frac{2}{3} E$.Comparing with $\frac{x}{3} E$,we get $x = 2$.

$A$ capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2V$. Then the loss of energy is $\frac{x}{3} E$,where $x$ is . . . . . . .

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