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

(a) In order to obtain a total resistance of `4 Omega` from three resistors of `2 Omega, 3 Omega` and `6 Omega`:
(i) First connect the two resistors of `3 Omega` and `6 Omega` in parallel to get a total resistance of `2 Omega`. This is because in parallel combination:
`(1)/(R) = (1)/(R_(1)) +(1)/(R_(2))`
`(1)/(R) = (1)/(3)+(1)/(6)`
`(1)/(R) =(2+1)/(6)`
`(1)/(R) = (3)/(6)`
`R = (6)/(3)`
`R = 2 Omega`

(ii) Then the parallel combination of `3 Omega` and `6 Omega` resistors (which is equivalent to `2 Omega` resistance) is connected in series with the remaining `2 Omega` resistor to get a total resistance of `4 Omega`. This is because in series combination:
`R = R_(1)+R_(2)`
`R = 2 +2`
`R = 4 Omega`
The arrangement of three resistors of `2 Omega, 3 Omega` and `6 Omega` which gives a total resistance of `4 Omega` can now be represented as follows:

Thus, we can obtain a total resistance of `4 Omega` by connecting a parallel combination of `3 Omega` and `6Omega` resistors in series with `2 Omega` resistor.
(b) In order to obtain a total resistance of `1 Omega` from three resistors of `2 Omega, 3 Omega` and `6 Omega`, all the three resistors should be connected in parallel. This is because in parallel combination:
`(1)/(R) = (1)/(R_(1)) +(1)/(R_(2))+(1)/(R_(3))`
`(1)/(R) = (1)/(2)+(1)/(3)+(1)/(6)`
`(1)/(R) = (3+2+1)/(6)`
`(1)/(R) = (6)/(6)`
`R = (6)/(6)`
`R = 1 Omega`