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(B) The capacitors of $2\,\mu F$ and $3\,\mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 2\,\mu F + 3\,\mu F = 5\,\mu F$.Now,

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(B) The capacitors of $2\,\mu F$ and $3\,\mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 2\,\mu F + 3\,\mu F = 5\,\mu F$.Now,the circuit consists of three capacitors of $12\,\mu F$,$5\,\mu F$,and $20\,\mu F$ connected in series between points $P$ and $Q$.The equivalent capacitance $C_{PQ}$ is given by the formula for series combination:$\frac{1}{C_{PQ}} = \frac{1}{12} + \frac{1}{5} + \frac{1}{20}$Taking the least common multiple $(LCM)$ of $12, 5,$ and $20$,which is $60$:$\frac{1}{C_{PQ}} = \frac{5 + 12 + 3}{60} = \frac{20}{60} = \frac{1}{3}$Therefore,$C_{PQ} = 3\,\mu F$.

In the circuit diagram shown in the adjoining figure,the resultant capacitance between $P$ and $Q$ is ........ $\mu F$.

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