Question

A network of four $10\mu F$ capacitors is connected to a $500V$ supply, as shown in fig. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor.(Note, the charge on a capacitor is the charge on the place with higher potential, equal and opposite to the charge on the plate with lower potential).

Answer 1

$C_1=C_2=C_3=C_4=10\mu F$

$C_1,C_2$ and $C_3$ are in series

So

$\frac{1}{C^{\prime }}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\Rightarrow C^{\prime }=\frac{10}{3}\mu F$

$C_{AD}=C_{eq}=C^{\prime }+C_4$ [$\because$ They are in parallel]

$\Rightarrow C_{eq}=\frac{10}{3}+10=\frac{40}{3}\mu F$

Charge on $C_4=10\mu F\times 500v=5\times {10}^{-3}C$

Charge of $C_1,C_2$ and $C_3$ will be same as they are in series

So charge on $C_1,C_2$ and $C_3$

$=C^{\prime }\times V=\frac{10}{3}\mu F\times 500=\frac{5}{3}\times {10}^{-3}C$