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

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(B) In the steady state,the potential difference across both capacitors $C_1$ and $C_2$ must be equal because they are connected in parallel when the

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$Q/3, 2Q/3$

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(B) In the steady state,the potential difference across both capacitors $C_1$ and $C_2$ must be equal because they are connected in parallel when the switch is closed.Let the charges on $C_1$ and $C_2$ at steady state be $Q_1$ and $Q_2$ respectively.Since the potential difference is the same,we have $\frac{Q_1}{C_1} = \frac{Q_2}{C_2}$.Given $C_2 = 2C_1$,substituting this gives $\frac{Q_1}{C_1} = \frac{Q_2}{2C_1}$,which implies $Q_2 = 2Q_1$.According to the law of conservation of charge,the total charge remains constant: $Q = Q_1 + Q_2$.Substituting $Q_2 = 2Q_1$ into the conservation equation,we get $Q = Q_1 + 2Q_1 = 3Q_1$.Therefore,$Q_1 = Q/3$ and $Q_2 = 2Q_1 = 2Q/3$.

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch $S$ and a resistor $R$ 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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