Q. 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 $24\ volt$ battery. The potential difference between the plates of the $6\mu F$ capacitor is
A) $4\ volt$
B) $6\ volt$
C) $8\ volt$
D) $10\ volt$
Explanation
(a) $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \Rightarrow C_{eq} = 1\ \mu\ F$
Total charge $Q = C_{eq}.\ V = 1 \times 24 = 24\ \mu C$
So p.d. across $6\ \mu F$ capacitor $= \frac{24}{6} = 4\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.