How can three resistors of resistances `2Omega,3Omegaand6Omega` be connected to give a total resistance of a] `4Omega`. b] `1Omega` ?

Explanation

We can get 4 `Omega` resistance by connecting 2 `Omega` and 6 `Omega` in parallel and then this combination in series with 2`Omega` as shown in figure.

`(1)/(R_(p))=(1)/(3) +(1)/(6) = 2 Omega`
`R= 2 Omega+ 2Omega= 4 Omega`
(b) We can get `1 Omega` resistance by connecting `R_(1), R_(2)` and `R_(3)` in parallel
`(1)/(R_(p))=(1)/(2) +(1)/(3) +(1)/(6)`
`(6)/(6)`
`R_(q) = 1 Omega`