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B
7μF
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C
83μF
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D
94μF
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Solution
The correct option is A6μF The given circuit can be reduced to the above form. Since this wheatstone bridge is balanced, no current flows through the 5μF capacitor.
[6(s)12](p)[3(s)6] ⇒4(p)2=6μF Here p stands for parallel and s stands for series.