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

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., `6Omega+6Omega+6Omega=18Omega`, Which is not desired. If we canect the resistors in parallel, then the equivalent resistance will be `6//2=3Omega` is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistor in parallel

Two `6 Omega` resistore are connected in parallel. Their equivalent resistance will be
`1/(1/6+1/6)=(6x6)/(6+6)=3Omega`
Therefore, the total resistance is
`1/(1/6+1/6)=(6x6)/(6+6)=3Omega`.
The third `6Omega` resistor is in series with `3Omega`. Hence, the equivalent resistance of the circuit is `6Omega+3Omega=9Omega`
(b) Two resistor in series

Two `6Omega` resistors are in series. Their equivalent resistance will be the sum `6+6=12Omega`
The thirs `6Omega` resistors is in parallel with `12Omega`. Hence, equivalent resistance will be
`1/(1/12+1/6)=(12x6)/(12+6)=4Omega`
Therefore, the total resistance is `4Omega`.