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

Explanation

As the total resistance (equivalent resistance) is `4Omega`, the `6Omega` resistor cannot be in series. So, it must be in parallel with some other resistors.
In parallel connection, the equivalent resistance (`4Omega`) has to be less than all the resistances. So, the resistors of `2Omega` and `3Omega` cannot be in parallel at one time with `6Omega`.

So, the resistors have to be in a mixed combination. Let us consider the combination shown in the figure. The equivalent resistance between B and C (which are in parallel)
`= (3Omega xx 6Omega)/(3Omega + 6Omega)= (18Omega)/(9Omega)= 2Omega`
The resistance between A and D = `2Omega+ 2Omega= 4Omega`.
So, the combination shown in the figure is true.

(ii) Here, `R_(1)= 2Omega, R_(2)=3Omega R_(3) = 6Omega`, and `R=1Omega`
Since the equivalent resistance of the combination is of lesser value than any of the resistors of the combination, it is clear that the resistors should be connected in parallel. It can be further confirmed by using the formula
`(1)/(R)= (1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))= (1)/(2)+(1)/(3)+(1)/(6)= (3+2+1)/(6)= (6)/(6)= 1`
i.e, R= 1 ohm
Therefore, resistors should be connected in parallel.