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

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(D) Initially,capacitor $C_1$ has charge $Q$ and $C_2$ is uncharged. When the switch is closed,charge flows until both capacitors reach the same poten

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Initially,capacitor $C_1$ has charge $Q$ and $C_2$ is uncharged. When the switch is closed,charge flows until both capacitors reach the same potential $V$.In the steady state,the potential difference across both capacitors is equal:$V = \frac{Q_1}{C_1} = \frac{Q_2}{C_2}$Given $C_2 = 2C_1$,we have:$\frac{Q_1}{C_1} = \frac{Q_2}{2C_1} \implies Q_2 = 2Q_1$By the law of conservation of charge,the total charge remains constant:$Q_1 + Q_2 = Q$Substituting $Q_2 = 2Q_1$ into the 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 between them 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 the capacitors will be:

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