Explanation

(d) [image:2_image131.emf]
The capacitors 2$\ \mu F$ and 2$\ \mu F$ of arm $ACD$ are in series. So, their equivalent capacitance is 1$\ \mu F$ which is in parallel with capacitor of 1$\ \mu F$ of arm $AD$.
So, equivalent capacitance now is 2$\ \mu F$.
This capacitance is now in series with 2$\ \mu F$ capacitance of arm $BD$ which equivalents to 1$\ \mu F$ is in parallel with 1$\ \mu F$ capacitance of arm $AB.$
So, final effective capacitance=2$\ \mu F$.