Correct Answer: Option A

Explanation

\(C_1\) = 3μF + 4μF = 7μF ( capacitance in parallel )

\(\frac{1}{ C} = \frac{ 1}{2} + \frac{1}{7} = \frac{ 7 + 2 }{14} = \frac{9}{14}\)

∴ C =  \(\frac{14}{9}\) = 1.56μF

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