Two capacitors C_1 and C_2 = 2C_1 are connect | vedclass

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(B) In the steady state,the current in the circuit becomes zero,and both capacitors are connected in parallel across the same potential difference. Le

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$\frac{Q}{3}, \frac{2Q}{3}$

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(B) In the steady state,the current in the circuit becomes zero,and both capacitors are connected in parallel across the same potential difference. Let the new charges on $C_1$ and $C_2$ be $Q_1$ and $Q_2$ respectively.Since they are in parallel,their potentials must be equal:$V_1 = V_2 \implies \frac{Q_1}{C_1} = \frac{Q_2}{C_2}$Given $C_2 = 2C_1$,we have:$\frac{Q_1}{C_1} = \frac{Q_2}{2C_1} \implies Q_2 = 2Q_1$According to the law of conservation of charge,the total charge remains constant:$Q_1 + Q_2 = Q$Substituting $Q_2 = 2Q_1$ into the conservation equation:$Q_1 + 2Q_1 = Q \implies 3Q_1 = Q \implies Q_1 = \frac{Q}{3}$Then,$Q_2 = 2Q_1 = \frac{2Q}{3}$Thus,the charges on the capacitors are $\frac{Q}{3}$ and $\frac{2Q}{3}$.

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch and a resistor as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on each capacitor will be:

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