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Q.

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

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a

1132μF

b

2332μF

c

3223μF

d

3211μF

answer is D.

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Detailed Solution

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=8μF
This  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
 

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