Question

In the following circuit, the resultant capacitance between $A$ and $B$ is $1\mu F$. Find the value of $C$.

Answer 1

The equivalent capacitance for RS is $C_{RS}=4+\frac{6\times 12}{6+12}=4+4=8\mu F$

Now, $C_{PS}=\frac{1\times 8}{1+8}=8/9\mu F$ and $C_{PQ}=\frac{8\times 4}{8+4}=8/3\mu F$

Now $C_{PQ}$ and $C_{RS}$ are in parallel and then it is series with C.

Thus, $C_{AB}=\frac{C\left(32/9\right)}{C+32/9}=1$

or $C+32/9=32C/9$

or $9C+32=32C$ or $23C=32$

so, $C=32/23=1.4\mu F$