Q.
An infinite number of identical capacitors each of capacitance $1\mu F$ are connected as in adjoining figure. Then the equivalent capacitance between $A$ and $B$ is
A) $1\mu F$
B) $2\mu F$
C) $\frac{1}{2}\mu F$
D) $\infty$
Explanation
(b) This combination forms a G.P. $S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}\cdots\cdots$
Sum of infinite G.P. $S = \frac{a}{1 - r}$
Here $a =$ first term $= 1$ and $r =$ common ratio $= \frac{1}{2}$
$\Rightarrow S = \frac{1}{1 - \frac{1}{2}} = 2 \Rightarrow C_{eq} = 2\mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.