Some capacitors each of capacitance `30 muF` are connected as shown in. Calculate the equivalent capacitance between terminals `A` and `B`
.

Answer

`22 muF`

Explanation

`Q=Q_(1)+Q_(2)`
Loop `ACDA:Q_(1)=Q_(2)+Q_(3)`
Loop `DCED:Q_(3)+Q_(1)-Q_(2)+2Q_(3)=Q_(2)-Q_(3)`
or `Q_(1)-2Q_(2)+4Q_(3)=0`
or `Q_(1)-2Q_(2)+4(Q_(1)-Q_(2))=0`
or `5Q_(1)=6Q_(2) or 6Q=11Q_(1)`
.
Loop `ADEBMA:
`CV=Q_(2)+Q_(2)-Q_(3)+Q_(1)=Q+Q_(2)-Q_(3)`
`=Q+Q_(2)-(Q_(1)-Q_(2))`
`=Q-Q_(1)+2Q_(2)`
`=Q-Q_(1)+2(Q-Q_(1))`
`=3Q-3Q_(1)`
`=3Q-3xx(6Q)/(11)=(15)/(11)Q`
`rArrC_(eq)=Q/V=(11)/(15)C`
`=(11)/(15)xx30=22 muF`.