Show how you would connect three resistors, each of resistance `6Omega` so that combination has a resistance of
(i) `9Omega` (ii) `4Omega`

Explanation

Here we have four options to connect the three resistors in different ways
(1) All the three resistors can be connected in series
(2) All the three can be connected in parallel
(3) Two of the three resistors can be connected in series and one in parallel and
(4) Two of the three resistors can be connected in parallel and one in series
Thus, effective resistance in the case
(1) When all the three resistors are connected in series Effective total resistance=`6Omega+6Omega+6Omega=18Omega`. This is not required
(2) when all the three are connected in parallel
Then `1/R=1/(6Omega)+1/(6Omega)+1/(6Omega)`
`=(1+1+1)/(6)Omega=3/6Omega=1/2Omega`
Thus, effective total resistance `R=2Omega`. This is also not required.
(3) When two of the three resistors are connected in parallel and one in series
When two resistors are connected in parallel
Then `1/R=1/(6Omega)=1/(6Omega)=2/6Omega=1/3Omega`
Therefore, `R=3Omega`
`=3Omega+6Omega=9Omega`
Two of the three resistors can be connected in parallel and one in series
(4) when two resistors are connected in series, then total resistance`=6Omega+6Omega=12Omega`
And one resistor is connected m series with two in parallel
then, `1/R=1/(12Omega)+1/(6Omega)=(1+2)/(12)Omega=3/2Omega=1/4Omega`
Thus, R`=4Omega`
Thus, when two resistors are connected in series and one in parallel then total effective resistance `=9Omega`
When two resistors are connected in parallel with one in series then total effective resistane `=4Omega`