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

Explanation

i] For getting a resistance of `9Omega`, the two resistors should be connected in parallel and one in series with them.
The equivalent resistance of the two resistors in parallel,
`R_(p)` is given by `1/R_(p)=1/6+1/6=2/6toR_(p)=3Omega`
Now, the equivalent resistance of `R_(P)and6Omega` in series is given by diagram.
`R=R_(p)+6=3+6=9Omega`.

ii] To get a resistance of `4Omega`, two resistors should be connected in series and one in parallel to them.
The equivalent resistance of the two resistors in series is given by
`R_(p)=6Omega+6Omega=12Omega`
Now, the equivalent resistance of `R_(p)and6Omega` in parallel is given by diagram.
`1/R=1/R_(p)+1/6=2/6+1/6=(1+2)/12=3/12=1/4`
`R=4Omega`