CGPET 2019 Solved Paper

According to the question, Given, capacitance between $A$ and $B, C_{AB}=1\ \mathrm{\mu F}$. $\therefore$ Capacitors between point $C$ and $D$ are connected in parallel. So, capacitance. $C_{CD}=2\ \mathrm{\mu F}+2\ \mathrm{\mu F}=4\ \mathrm{\mu F}$. $\therefore$ Capacitors between points $E$ and $F$ connected in series, so the capacitance between points $E$ and $F$, is $\frac{1}{C_{EF}}=\frac{1}{6}+\frac{1}{12}$ or $\frac{1}{C_{EF}}=\frac{2+1}{12}=\frac{3}{12}$ or $C_{EF}=\frac{12}{3}=4\ \mathrm{\mu F}$ Capacitors between points $G$ and $D$ are connected in series, therefore $\frac{1}{C_{GD}} = \frac{1}{8\ \mathrm{\mu F}} + \frac{1}{4\ \mathrm{\mu F}}$ ∴ $=\frac{1+2}{8}$ or $\frac{1}{C_{GD}}=\frac{3}{8}$ or $C_{GD}=\frac{8}{3}\ \mathrm{\mu F}$ $\therefore$ Capacitance between points $E$ and $F$, are connected in parallel, so $\therefore C_{EF}=4\ \mathrm{\mu F}+4\ \mathrm{\mu F}=8\ \mathrm{\mu F}$ Capacitance between points $G$ and $F$, are connected in series, therefore $\frac{1}{C_{GF}}=\frac{1}{1}+\frac{1}{8}$ or$=\frac{8+1}{8}=\frac{9}{8}$ or$C_{GF}=\frac{8}{9}\ \mathrm{\mu F}$ Now, $\therefore C_{PR} = \frac{8}{9} + \frac{8}{3} = \frac{24+8}{9}$ $=\frac{32}{9}\ \mathrm{\mu F}$Capacitors are connected in series, then the total capacitance across points $A$ and $B$, $\therefore \frac{1}{C_{AB}} = \frac{1}{C} + \frac{1}{\frac{32}{9}}$ or $1 = \frac{1}{C} + \frac{9}{32}$ Here $C_{AB}=1\ \mathrm{\mu F}$ or $\frac{1}{C}=1-\frac{9}{32}=\frac{32-9}{32}=\frac{23}{32}$ or $\frac{1}{C}=\frac{23}{32}$ $C=\frac{32}{23}\ \mathrm{\mu F}$