Question

In the given figure, find the value of capacitance C if the equivalent capacitance between A and B is $\frac{32}{17}\mu F$. Values shown are in $\mu F$

Answer 1

The simplified circuit is

Capacitance b/w C and D

$=4\mu F+\left(\frac{12\times 6}{12+6}\right)=(4+4)\mu f=B\mu F$

Capacitance b/w E and F $=2+2=4\mu F$

Capacitance b/w G and B

$=\left(\frac{8\times 1}{8+1}\right)+\left(\frac{8\times 4}{8+4}\right)=\frac{8}{9}+\frac{8}{3}$

$=\frac{32}{9}\mu F$

Now $\frac{17}{32}=\frac{9}{32}+\frac{1}{C}$

$C=4\mu F$