CGPET 2021 Solved Paper

The circuit diagram is given as From above circuit diagram, $C_1$ and $C_2$ are in series ∴ Equivalent capacitance of $C_1$ and $C_2$ is given as $C_{12} = \frac{C_1 C_2}{C_1+C_2} = \frac{2 \times 2}{2+2} = 1\,\mu\text{F}$ Now, $C_{12}$ and $C_3$ are in parallel, hence equivalent capacitance of $C_{12}$ and $C_3$, $C_{123} = C_{12}+C_3 = 1+1 = 2\,\mu\text{F}$ Now, $C_{123}$ and $C_4$ are in series, hence their equivalent capacitance, $C_{1234}= \frac{C_{123} \times C_4}{C_{123}+C_4} = \frac{2 \times 2}{2+2} = 1\,\mu\text{F}$ Again, $C_{1234}$ and $C_5$ are in parallel, hence their equivalent capacitance, $C_{12345} = C_{1234}+C_5 = 1+1 = 2\,\mu\text{F}$ Now, $C_{12345}$ and $C_6$ are in series, hence their equivalent capacitance, $C_{123456} = \frac{C_{12345} \times C_6}{C_{12345}+C_6} = \frac{2 \times 2}{2+2} = 1\,\mu\text{F}$ Now, $C_{123456}$ and $C_7$ are in parallel, hence their equivalent capacitance, $C_{1234567} = C_{123456} + C_7 = 1+1 = 2\,\mu\text{F}$ Now, $C_{1234567}$ and $C_8$ are in series, hence their equivalent capacitance, $C_{12345678} = \frac{C_{1234567} \times C_8}{C_{1234567} + C_8} = \frac{2 \times 2}{2+2} = 1\,\mu\text{F}$ ∴ $C_{AB} = C_{12345678} = 1\,\mu\text{F}$