In the circuit shown,initially C_1 and C_2 ar | vedclass

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(B) When the switch is closed,the capacitors $C_1$ and $C_2$ are connected in series with the batteries. In a series circuit,the charge $q$ flowing th

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The charge on $C_1$ and $C_2$ are the same.

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(B) When the switch is closed,the capacitors $C_1$ and $C_2$ are connected in series with the batteries. In a series circuit,the charge $q$ flowing through each component is the same.Therefore,the charge on $C_1$ and $C_2$ is the same.Using the formula $V = q/C$,the potential drop across each capacitor is:$V_1 = q / C_1 = q / 4 \mu F$$V_2 = q / C_2 = q / 8 \mu F$Since $C_1  V_2$. Thus,the potential drop across $C_1$ is greater than that across $C_2$.

In the circuit shown,initially $C_1$ and $C_2$ are uncharged. After closing the switch:

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