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

Equivalent capacitance for three capacitors C₁,C₂ and C₃ in series is given by

\(\frac{1}{C_{eq}}\)=\(\frac{1}{c_1}+\frac{1}{c_2}+\frac{1}{c_3}\)

\(\Rightarrow\) \(C_{eq}\)=\(\frac{C_1C_2C_3}{C_1C_2+C_2C_3+C_3C_1}\)

\(\Rightarrow\) \(C_{eq}\)=\(\frac{C(2C)(3C)}{C(2C)+(2C)(3C)+(3C)C}\)=\(\frac{6}{11}C\)

\(\Rightarrow\) Charge on capacitors (C₁,C₂ & C₃) in series

=\(C_{eq}\)V = \(\frac{6C}{11}V\)

Now, Charge on capacitor C₄=C₄V=4CV

\(\frac{Charge onC_2}{Charge onC_4}\)

=\(\frac{\frac{6C}{11}V}{4CV}\)

=\(\frac{6}{11}\times\frac{1}{4}\) = \(\frac{3}{22}\)

Therefore, the correct option is (B): \(\frac{3}{22}\)