In the following circuit the resultant capacitance between A and B is 1 μF.  Then value of C is

12μF and 6μF  are in series and again are in parallel with4μF therefore resultant of these three will be =12×612+6+4=4+4=8μFThis  equivalent system is in series with 1μF Its equivalent capacitance=8×18+1=89μF                    →(1) Equivalent of 8μF , 2μF  and 2μF =4×84+8=83μF             →(2)1 and 2 are in parallel and are in series with C∴89+83=329    Ceq=1=329×C329+C ⇒C=3223μF