Q.
The equivalent capacitance between the points $A\ and\ B$ in the following circuit is
A) 1 $\mu F$
B) 2 $\mu F$
C) 4 $\mu F$
D) 8 $\mu F$
Explanation
(a) The two capacitors each of value 1.5$\mu F$ are in parallel. So, their equivalent capacitance
[image:2_image104.emf]
$\ \ \ \ \ = 1.5 + 1.5 = 3\ \mu F\ $
Now, three capacitors each of value $3\mu F$ are in series. Hence, their equivalent capacitance is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{C} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{C}\ \ = \frac{3}{3}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c = 1\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.