Question

A network of four capacitors of capacity equal to C₁=C,C₂=2C,C₃=3C and C₄=4C are connected to a battery as shown in the figure. The ratio of the charges on C₂ an C₄ is

• A. \(\frac{22}{3}\)
• B. \(\frac{3}{22}\)
• C. \(\frac{7}{22}\)
• D. \(\frac{22}{7}\)

Answer 1

The Correct Option is B

To find the ratio of the charges on \(C_2\) and \(C_4\), we need to analyze the given circuit configuration. Capacitors \(C_1 = C\), \(C_2 = 2C\), \(C_3 = 3C\), and \(C_4 = 4C\) are arranged as shown in the diagram below:

1. First, observe the configuration of the capacitors. Capacitors \(C_1\) and \(C_2\) are in series, and their combination is in parallel with \(C_3\). The combination is then in series with \(C_4\).
2. The equivalent capacitance of \(C_1\) and \(C_2\) in series is given by:
   \[ \frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{C} + \frac{1}{2C} = \frac{3}{2C} \] \[ C_{12} = \frac{2C}{3} \]
3. This equivalent capacitor \(C_{12}\) is in parallel with \(C_3\), so the total capacitance \(C_{13}\) is given by:
   \[ C_{13} = C_{12} + C_3 = \frac{2C}{3} + 3C = \frac{11C}{3} \]
4. This \(C_{13}\) is in series with \(C_4\), so the total capacitance \(C_{eq}\) of the system is:
   \[ \frac{1}{C_{eq}} = \frac{1}{C_{13}} + \frac{1}{C_4} = \frac{3}{11C} + \frac{1}{4C} = \frac{19}{44C} \] \[ C_{eq} = \frac{44C}{19} \]
5. The charge on a series circuit remains the same. The total charge \(Q\) from the battery is:
   \[ Q = C_{eq} \times V = \frac{44C}{19} \times V \]
6. For \(C_2\), since they are in series, the charge is same as \(Q\), that is:
   \[ Q_2 = Q = \frac{44C}{19} \times V \]
7. For \(C_4\):
   \[ Q_4 = C_4 \times V_{C4} = 4C \times \left(\frac{Q}{C_{eq}}\right) \] \[ Q_4 = 4C \times \left(\frac{19}{44C} \times V\right) = \frac{76C}{44} \times V \]\(= \frac{19C}{11} \times V\)
8. Finally, the ratio of charges \(Q_2\) to \(Q_4\) is:
   \[ \text{Ratio} = \frac{Q_2}{Q_4} = \frac{\frac{44C}{19} \times V}{\frac{19C}{11} \times V} = \frac{44}{19} \times \frac{11}{19} = \frac{3}{22} \]

Thus, the ratio of the charges on \(C_2\) to \(C_4\) is \(\frac{3}{22}\).