How can three resistors of resistances `2 Omega,3 Omega`, and `6 Omega` be connected to give a total resistance of
(a) `4 Omega`
(b) `1 Omega` ?

Explanation

To get total resistance 4 `omega`, connect 3 `omega` and 6 ft resistors in parallel and 2 `omega` resistance in series with the resultant.

Since, 3 `omega` and 6 `omega` resistors are in parallel, so the net resistance will be
`(1)/(R_(12))=(1)/(R_(1))+(1)/(R_(2))`
`rarr R_(12)=2/1 =2 omega`
Now, the resultant `R_(12)` and 2 `omega` resistors are in series. So, the net resistance `R = R_(12)+2 omega + 2 + 2 = 4omega`