Two capacitors of $1\,\mu F$ and $2\,\mu F$ are connected in series, the resultant capacitance will be.....$\mu \,F$
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(b) $\frac{1}{{{C_{eq}}}} = \frac{1}{1} + \frac{1}{2} \Rightarrow {C_{eq}} = \frac{2}{3}\,\mu F$
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