Show how you would connect three resistors, each of resistance `6 Omega`, so that the combination has a resistance of
(i) `9 Omega`
(ii) `2 Omega`.

Explanation

To get total 9 `omega` resistance from three 6 `omega` resistors, we should connect two resistors in parallel and the third resistor in series with the resultant. The combination is given as follows :

`rarr (1)/(R_(12))=1/6+1/6+2/6=1/3`
`rarr R_(12)=3 omega`
Now `R_(12)` and 6 `omega` are connected in series, so the net resistance is given by `R=R_(12)+6 omega=3 omega+6 omega=9omega`