A circuit shown in the figure consists of a b | vedclass

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(D) In a series circuit,the charge on each capacitor is the same. Since $C_1$ and $C_2$ are in series,the charge $Q$ on both capacitors is equal.$Q_1

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$A$ and $B$ both.

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(D) In a series circuit,the charge on each capacitor is the same. Since $C_1$ and $C_2$ are in series,the charge $Q$ on both capacitors is equal.$Q_1 = Q_2 = Q$.Let the potential at the node between $C_1$ and the battery be $V_x$ and between the battery and $C_2$ be $V_y$. Given $V_A - V_B = 5\,V$,we can analyze the series combination.The equivalent capacitance $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 	imes 2}{1 + 2} = \frac{2}{3}\,\mu F$.The total charge $Q = C_{eq} 	imes V_{total} = \frac{2}{3} 	imes 5 = \frac{10}{3}\,\mu C$.Since the capacitors are in series,the charge on each is $Q = \frac{10}{3}\,\mu C$.Thus,statement $A$ is correct.Now,check energy: $E = \frac{Q^2}{2C}$.$E_1 = \frac{Q^2}{2C_1} = \frac{(10/3)^2}{2 	imes 1} = \frac{100/9}{2} = \frac{50}{9}\,\mu J$.$E_2 = \frac{Q^2}{2C_2} = \frac{(10/3)^2}{2 	imes 2} = \frac{100/9}{4} = \frac{25}{9}\,\mu J$.Since $E_1 = 2 E_2$,statement $B$ is also correct.Therefore,both $A$ and $B$ are correct.

$A$ circuit shown in the figure consists of a battery of $emf$ $10\,V$ and two capacitors $C_1$ and $C_2$ of capacitances $1.0\,\mu F$ and $2.0\,\mu F$ respectively. The potential difference $V_A - V_B$ is $5\,V$. Which of the following statements is correct?

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