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Q.
In the circuit diagram shown in the figure below, the resultant capacitance between $P$ and $Q$ is
Electrostatic Potential and Capacitance
Solution:
The given circuit can be drawn as:
$C=(3+2) \mu F =5\, \mu F$
$\frac{1}{C_{P Q}}=\frac{1}{5}+\frac{1}{20}+\frac{1}{12}$
$=\frac{20}{60}=\frac{1}{3}$
$\Rightarrow C_{P Q}=3\, \mu F$
