Q. Three capacitors 2, 3 and 6 $\mu F$ are joined with each other. What is the minimum effective capacitance?
A) $\frac{1}{2}\ \mu F$
B) 1 $\mu F$
C) 2 $\mu F$
D) 3 $\mu F$
Explanation
(b) Minimum effective capacitance
$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\ \ \ \ \ \ \ \ \ = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = 1$
$\ \ \ \ \ \ \ C = 1\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.