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===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===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===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===1µF ∴
CAB=C12345678=1µF