Three capacitors $2,\, 3$ and $6\ \mu F$ are joined in series with each other. What is the minimum effective capacitance......$\mu \,F$
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(b) $\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}$$ = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$ $ = \frac{{3 + 2 + 1}}{6} = \frac{6}{6} = 1\,\mu F$
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