Question

Three capacitors of $2\mu F,3\mu F$ and $6\mu F$ are joined in series and the combination is charged by means of a 24 V battery. The potential difference between the plates of the $6\mu F$ capacitor is :

• A. 4 V
• B. 6 V
• C. 8 V
• D. 10 V

Answer 1

The correct option is A 4 V

Equivalent capacitance of capacitors connected in series $\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$
$\Rightarrow C_{eq}=1\mu F$
Hence, charge flowing through the circuit $q=C_{eq}.V=1\times 24=24uF$
$\therefore$ Potential difference between the plates of $6\mu F$ capacitor
$V^{\prime }=\frac{q}{C}=\frac{24}{6}=4V$