Find the equivalent resistance of the circuit between the points A and B

Answer

(`4)/(3) Omega`

Explanation

Here, two resistance along each side triangle are in parallel , their equivalent resistance is
` = (4 xx 4)/( 4+ 4) = 2 Omega`
The given network of resistance will be reduced to the equivalent circuit .Now resistance in arm AC and CB are in series. their equivalent resistance ` = 2 Omega + 2 Omega = 4 Omega` which is in parallel with resistance of arm `AB (= 2 OMega)` Therefore equivalent resistance between A and B is
`(4 xx 2)/( 4+ 2) = (8)/(6) = (4)/(3) Omega`