The equivalent capacitance between the points A and B in the network given below is

20 µF 20/3 µF 40/3 µF 10 µF

20 µF

The given circuit is a balanced Wheatstone bridge of capacitors, hence P.D across bridge capacitor of5 μF is zero. Thus, the circuit can be modified as \( C_{n e t}=\left(\frac{1}{30}+\frac{1}{60}\right)^{-1}=\left(\frac{1}{20}\right)^{-1}=20 \mu \mathrm{F} \)