Q.
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
Delhi UMET/DPMTDelhi UMET/DPMT 2006Electrostatic Potential and Capacitance
Solution:
$ 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 $= \frac{ Q \, '}{ Q} = \frac{ 6 CV}{ 11 \times 4CV} = \frac{3}{22}$.
