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

Answer: (B)

$\frac{3}{22}$

Answer 2

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): $\frac{3}{22}$