Find the equivalent capacitance between A and B for the following network of capacitors.

Explanation

The circuit can be redrawn as shown below:

This is a balanced Wheatstone bridge. So, the capacitor of 100 `muF` can be removed from branch ED as no charge will flow through this branch.
Equivalent capacitance of branch CEF will be:
`C_(CEF)=(1xx1)/(1+1)=(1)/(2)muF`
Equivalent capacitance of branch CDF will be:
`C_(CDF)=(1xx1)/(1+1)=(1)/(2)muF`
Equivalent capacitance between points C and F will be:
`C_(CF)=((1)/(2)+(1)/(2))=1muF`

`therefore` The equivalent capacitance between the points A and B will be:
`(1)/(C_(AB))=(1)/(1)+(1)/(1)+(1)/(1)`
`rArr C_(AB)=(1)/(3)muF`