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Question

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

A
3211 μF
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B
1132 μF
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C
2332 μF
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D
3223 μF
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Solution

The correct option is D 3223 μF
Since capacitor of capacitance 6 μF and 12 μF are in series and their equivalent is in parallel with 4 μF, hence their equivalent will be 4+6×126+12=8 μF
Simillarly, both 2μF capacitors are in parallel connections, therefore their equivalent capacitance will be 2+2=4 μF


Since capacitor 8 μF and 4 μF are in series, their equivalent resistance will be 8×48+4=83 μF
Simillarly equivalent of 1 μF and 8 μF will be 8×18+1=89 μF
Now, 83 μF and 89 μF will be in parallel and their equivalent will be 83+89=329 μF​​​​​​​​​​​​​​
Since total equivalent resistance is 1 μF, therefore equivalent capacitance of C and 329 μF​​​​​​​​​​​​​​ will be 1 μF.
C×32/9C+32/9=1
C=3223 μF

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