Question

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

• A. 44658
• B. 44642
• C. 44746
• D. 22-Mar

Answer 1

Answer: (B)

44642

Answer 2

$C_1, \, C_2$ and $\, C_3$ are in series
$\frac{1}{C'} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C}$
or, $\frac{1}{C'} = \frac{ 6 + 3 + 2}{ 6 C} = \frac{11}{6C}$
or, C' = $\frac{6C}{11}$
All the capacitors in branch 1 is in series so the charge on each capacitor is Q' = $\frac{ 6}{ 11}$ CV
Also charge on capacitor $C_4$ is Q = 4 CV
$\therefore$ Ratio $= \frac{ Q \, '}{ Q} = \frac{ 6 CV}{ 11 \times 4CV} = \frac{3}{22}$.