Find the net resistance between point A and B in the circuit

Answer

`2.4 Omega`

Explanation

In closed circuit ECDE, arm ECD is parallel to arm ED so `R_(ED) = (6 xx (5+ 3))/(6 + (3+ 3)) = 3 Omega`
The resistebnce `R_(ED)` is in series `3 Omega` resistance of arm `FD `and they are in parallel its `s Omega` resistance of arm `EF` so total resistance between A and B is `R_(AB) = (4 xx (3+3))/(4+ (3+3)) = (24)/(10)`
`= 2.4 Omega`