In the circuit diagram shown in the adjoining figure, the resultant capacitance between $P$ and $Q$ is........$\mu F$
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(b) The given circuit can be drawn as
where $C = (3 + 2)\mu F = 5\,\mu F$
$\frac{1}{{{C_{PQ}}}} = \frac{1}{5} + \frac{1}{{20}} + \frac{1}{{12}} = \frac{{20}}{{60}} = \frac{1}{3}$
$==>$ ${C_{PQ}} = 3\,\mu F$
where $C = (3 + 2)\mu F = 5\,\mu F$
$\frac{1}{{{C_{PQ}}}} = \frac{1}{5} + \frac{1}{{20}} + \frac{1}{{12}} = \frac{{20}}{{60}} = \frac{1}{3}$
$==>$ ${C_{PQ}} = 3\,\mu F$
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