Correct Answer: Option D

Explanation

The three 2µf capacitors from the left are parallel. Therefore, effective capacitance in parallel = 6µf

The 6µf, 2µf(opposite the 3µf), and the 3µf are in series.

So effective/total capacitance in series = \(\frac{1}{C_T}\) = \(\frac{1}{C_1}\) +  \(\frac{1}{C_2}\) +  \(\frac{1}{C_3}\)

=  \(\frac{1}{6}\) +  \(\frac{1}{2}\) +  \(\frac{1}{3}\) = \(\frac{1 + 3 + 2}{6}\) = \(\frac{6}{6}\)

C\(_T\) = 1µf

There is an explanation video available below.

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