Their equivalent capacitance $C_{23}$ is given by:
$\frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{40 \, \mu F} + \frac{1}{60 \, \mu F} = \frac{3 + 2}{120 \, \mu F} = \frac{5}{120 \, \mu F}$
$C_{23} = \frac{120}{5} \, \mu F = 24 \, \mu F$
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