CGPET 2021 Solved Paper

The circuit diagram is given as

From above circuit diagram,
C1 and C2 are in series
∴ Equivalent capacitance of C1 and C2 is given as
C12=

C1C2

C1+C2

=

2×2

2+2

=1µF
Now, C12 and C3 are in parallel, hence equivalent capacitance of C12 and C3,
C123=C12+C3=1+1=2µF
Now, C123 and C4 are in series, hence their equivalent capacitance,
C1234=

C123×C4

C123+C4

=

2×2

2+2

=1µF
Again, C1234 and C5 are in parallel, hence their equivalent capacitance,
C12345=C1234+C5=1+1=2µF
Now, C12345 and C6 are in series, hence their equivalent capacitance,
C123456=

C12345×C6

C12345+C6

=

2×2

2+2

=1µF
Now, C123456 and C7 are in parallel, hence their equivalent capacitance,
C1234567=C123456+C7=1+1=2µF
Now, C1234567 and C8 are in series, hence their equivalent capacitance,
C12345678=

C1234567×C8

C1234567+C8

=

2×2

2+2

=1µF
∴ CAB=C12345678=1µF