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A network of four capacitors of capacity equal to $C_1 = C$, $C_2 = 2C$, $C_3 = 3C$ and $C_4 = 4C$ are connected to a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is -
($C_1$, $C_2$ and $C_3$ are in series with one another, and this branch is in parallel with $C_4$ across the battery of voltage V.)
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($C_1$, $C_2$ and $C_3$ are in series with one another, and this branch is in parallel with $C_4$ across the battery of voltage V.)
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A
$\dfrac{7}{4}$
B
$\dfrac{22}{3}$
C
$\dfrac{3}{22}$
D
$\dfrac{4}{7}$
Detailed Solution
$C_4 = 4C$ is connected directly across the battery, so $Q_4 = 4CV$.
$C_1$, $C_2$ and $C_3$ are in series: $\dfrac{1}{C_s} = \dfrac{1}{C} + \dfrac{1}{2C} + \dfrac{1}{3C} = \dfrac{6 + 3 + 2}{6C} = \dfrac{11}{6C}$
$C_s = \dfrac{6C}{11}$
Capacitors in series carry the same charge, so $Q_2 = C_sV = \dfrac{6CV}{11}$
$\dfrac{Q_2}{Q_4} = \dfrac{6CV}{11} \times \dfrac{1}{4CV}$
$\dfrac{Q_2}{Q_4} = \dfrac{6}{44} = \dfrac{3}{22}$
$C_1$, $C_2$ and $C_3$ are in series: $\dfrac{1}{C_s} = \dfrac{1}{C} + \dfrac{1}{2C} + \dfrac{1}{3C} = \dfrac{6 + 3 + 2}{6C} = \dfrac{11}{6C}$
$C_s = \dfrac{6C}{11}$
Capacitors in series carry the same charge, so $Q_2 = C_sV = \dfrac{6CV}{11}$
$\dfrac{Q_2}{Q_4} = \dfrac{6CV}{11} \times \dfrac{1}{4CV}$
$\dfrac{Q_2}{Q_4} = \dfrac{6}{44} = \dfrac{3}{22}$