Correct Answer - (i)`7.5 Omega`(ii) `6.67 Omega`
When the swich `S` open , the circuit reduce to as shown.The resistance of arm `ACB( = 5 + 10 = 15 Omega)` is in parallel to the resistance of arm `ADB ( = 10 + 5 = 15 Omega)`
Therefore , effective resistance between A and B
is `R = (15 xx 15)/(15+ 15) = 7.5 Omega`
(ii) When swich `S` is closed the circuit will be
How resistance of arm `AC(= 5 Omega)` their equivalent resistance is `R_(1) = (5 xx 10)/(5 + 10) = (10)/(5) Omega`
The resistance of arm `CB ( = 10 Omega)` is in parallel to the resistance of arm `DB = (10 xx 3)/(10 + 5) = (10)/(3) Omega`
Now `R_(1) and R_(2)` are in series on the effective resistance between A and B is
`R = R_(1) + R_(2) = (10)/(3) + (10)/(3) = (20)/(3) = 6.67 Omega`