Question

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

• A. 10 V
• B. 8 V
• C. 4 V
• D. 6 V

Answer 1

Answer: (C)

$\frac{1}{C_{e q}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
$ \Rightarrow C_{e q}=1\, \mu F$
Total charge $Q=C_{e q} \cdot V=1 \times 24=24 \mu C$
Potential diff. across $6 \mu F$ capacitor $=\frac{24}{6}=4$ volt