Q. Two capacitor of capacity $6\mu F$ and $12\mu F$ in series are connected by potential of 150 V. the potential of capacitor of capacity $12\mu F$ will be
A) 25 V
B) 50 V
C) 100 V
D) 150 V
Explanation
(b) Given, $C_{1} = 6\mu F,C_{2} = 12\mu F,V = 150volt$.
Total capacity, $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{6} + \frac{1}{12}\ \ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{2 + 1}{12}\frac{1}{C} = \frac{3}{12}C = 4\mu F$
Potential of $12\mu F$ capacitor
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = \frac{q}{C}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = \ \frac{4 \times 150}{12}\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V\ = \ $50 volt
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.