In This post you will get answer of Q.
Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance $C$ if the equivalent capacitance between point $A$ and $B$ is to be $1, mu F$ is :
Solution:
$frac{8 times 12}{18}=4 ,mu F$
$4 ,mu F +4 ,mu F =8 ,mu F$
$frac{8 times 1}{8+1}=frac{8}{9} ,mu F$
$frac{8 times 4}{8+4}=frac{32}{12}=frac{8}{3}, mu F$
$frac{8}{3}+frac{8}{9}=frac{24+8}{9}=frac{32}{9} ,mu C$
$frac{frac{32}{9} times C }{frac{32}{9}+ C }=1 $
$Rightarrow frac{32}{9} times C =frac{32}{9}+ C$
$Rightarrow C left(frac{32-9}{9}right)=frac{32}{9}$
$C =frac{32}{23}, mu F$
Solution:
$frac{8 times 12}{18}=4 ,mu F$
$4 ,mu F +4 ,mu F =8 ,mu F$
$frac{8 times 1}{8+1}=frac{8}{9} ,mu F$
$frac{8 times 4}{8+4}=frac{32}{12}=frac{8}{3}, mu F$
$frac{8}{3}+frac{8}{9}=frac{24+8}{9}=frac{32}{9} ,mu C$
$frac{frac{32}{9} times C }{frac{32}{9}+ C }=1 $
$Rightarrow frac{32}{9} times C =frac{32}{9}+ C$
$Rightarrow C left(frac{32-9}{9}right)=frac{32}{9}$
$C =frac{32}{23}, mu F$