Question

Three capacitors having capacitances $20\mu F,30\mu F$ and $40\mu F$ are connected in series with a $12V$ battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors?

Answer 1

$\frac{1}{C_{eq}}=\frac{1}{20}+\frac{1}{30}+\frac{1}{40}=\frac{6+4+3}{120}$

$\Rightarrow C_s=\frac{120}{13}\mu F$

$Q=\frac{120}{13}\times 12V$

Charges on each capacitors will be same, as of series combinations.

Work done $W=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{120}{13}\right)\times (12{)}^2=\frac{60}{13}\times 144$