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(B) $1$. Initially,the three capacitors are in series with a battery of emf $E$. The charge on each capacitor is $q = C_{eq} E = (C/3) E = CE/3$. The

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$\frac{2}{27} C E^2$

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(B) $1$. Initially,the three capacitors are in series with a battery of emf $E$. The charge on each capacitor is $q = C_{eq} E = (C/3) E = CE/3$. The initial energy stored in the system is $U_i = 3 	imes (q^2 / 2C) = 3 	imes ((CE/3)^2 / 2C) = CE^2/6$.$2$. When the battery is removed and resistors are connected as shown,the capacitors are effectively in parallel. Let the final potential difference across each capacitor be $V$. By conservation of charge,the total charge $q_{total} = CE/3$ is redistributed among the three capacitors in parallel. Thus,$3CV = CE/3$,which gives $V = E/9$.$3$. The final energy stored in the system is $U_f = 3 	imes (1/2 C V^2) = (3/2) C (E/9)^2 = (3/2) C (E^2/81) = CE^2/54$.$4$. The total heat dissipated is $\Delta H = U_i - U_f = CE^2/6 - CE^2/54 = (9CE^2 - CE^2) / 54 = 8CE^2/54 = 4CE^2/27$.$5$. Since the circuit is symmetric,the heat dissipated in each of the two resistors is equal. Therefore,the heat dissipated in each resistor is $\Delta H_{resistor} = (4CE^2/27) / 2 = 2CE^2/27$.

Three identical capacitors (each of capacitance $C$,initial charge zero) are connected in series and charged by a battery of emf $E$. After removing the battery,two resistors are connected to these capacitors as shown in the figure. The heat dissipated in each of the resistors is:

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