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

(i) In order to get a resistance of `9 Omega` from three resistors of `6 Omega` each, we connect two `6 Omega` resistors in parallel and then this parallel combination is connected in series with the third `6 Omega` resistor [as shown in Figure (i)]. This is because: `(1)/(R) = (1)/(6) +(1)/(6) = (1+1)/(6) = (2)/(3)= (1)/(3)` or `R = 3 Omega`, And `3 Omega +6 Omega = 9 Omega`

(ii) In order to get a resistance of `4 Omega` from three resistors of `6 Omega` each, we connect all the three `6 Omega` resistors in parallel and then connect two such parallel combinations in series [as shown in Figure (ii)]. This is because:
`(1)/(R) = (1)/(6) +(1)/(6)+(1)/(6) = (1+1+1)/(6)= (3)/(6) = (1)/(2)` or `R = 2 Omega`, And `2 Omega +2 Omega = 4 Omega`.