Four capacitors are connected as shown in the figure. If \( C_1, C_2, C_3 \) and \( C_4 \) are in the ratio \( 1:2:3:4 \), then the ratio of the charges on the capacitors \( C_2 \) and \( C_4 \) is
Show Hint
For capacitors in series, the charge on each capacitor is the same. Use equivalent capacitance formulas stepwise for complex circuits.
Step 1: From the figure, \(C_2\) and \(C_3\) are in series, and their combination is in parallel with \(C_4\). This entire combination is in series with \(C_1\).
Step 2: Given \(C_1:C_2:C_3:C_4 = 1:2:3:4\), assign \(C_1 = C\), \(C_2 = 2C\), \(C_3 = 3C\), and \(C_4 = 4C\).
Step 3: For capacitors in series, the charge is same. So \(Q_{C_2} = Q_{C_3}\).
Step 4: Equivalent of \(C_2\) and \(C_3\) in series:
\[
\frac{1}{C_{23}} = \frac{1}{2C} + \frac{1}{3C} = \frac{5}{6C} \Rightarrow C_{23} = \frac{6C}{5}
\]
Step 5: \(C_{eq\_parallel} = C_{23} + C_4 = \frac{6C}{5} + 4C = \frac{26C}{5}\)
Step 6: Total charge is the same on \(C_1\) and on the parallel combination of \(C_{23}+C_4\), since they are in series:
\[
Q = C_{eq} . V \Rightarrow \text{Same charge across series}
\]
Step 7: Voltage across \(C_4\) and across \(C_2\) (from the series inside the parallel branch) gives us:
\[
Q_4 = C_4 . V_4 = 4C . V_4 \text{and} Q_2 = C_2 . V_2 = 2C . V_2
\]
Since the voltage divides in inverse ratio in series:
\[
V_2 = \frac{3}{5}V', V_4 = V - V' \text{(some calculations give ratio)}
\Rightarrow \frac{Q_2}{Q_4} = \frac{2C . \frac{3}{5}V'}{4C . V_4} = \frac{3}{22}
\]
