Show how you would connect three resistors, each of resistance `6Omega`, so that the combination has a resistance of (1) `9Omega` (ii) `4Omega`.

Explanation

(i) In order to get a resistance of `9 Omega`, We connect the given resistors (each of resistance of `6Omega`) in the following way.
Resistance between B and C
`(1)/(R)= (1)/(6)+ (1)/(6)`
`(1)/(R)= (2)/(6)= (1)/(3)`
`R=3Omega`
Resistance of the combination = `6Omega +3Omega = 9Omega`

(ii) In order to get a resistance of `4 Omega`, we connect two resistors in series and third in parallel as shown in figure. Resistors AB and BC are in series therefore,
`R_(s)= 6Omega+6Omega= 12Omega`

Now, R, is parallel with the third (`6Omega`).
`:.` Equivalent resistance of combination (`R_(rho)`) is given by
`(1)/(R_(rho))= (1)/(6)+(1)/(12)= (2+1)/(12)`
`(1)/(R_(rho))= (3)/(12)= (1)/(4)`
`R_(rho)= 4Omega`