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

Explanation

(i) In order to get a resistance of 9 n from three resistors, each of resistance 6 `Omega`, we connect two `6 Omega` resistors in parallel and this parallel combination is connected in series with the third `6 Omega` resistor as shown in the following figure:

This is because `1/R_P = 1/6 + 1/6 + 1/6 =(1 +1)/6 =2/6 = 1/3`
`:. R_p = 3Omega and R_p + 6 Omega = 3Omega + 6Omega = 9 Omega`
(ii) In order to get a resistance of `4 Omega` from three resistors, each of resistance `6 Omega`, we connect two `6 Omega` resistors in series and this series combination is connected in parallel to the third `6Omega` reststor as shown in the following figure :

This is because `R_s = 6 + 6 = 12 Omega`
Now, `1/R_P = 1/R_s + 1/6`
`:. R_P = (6xxR_s)/(6 + R_s) =(6 xx 12)/(6 + 12) = 72/18 = 4 Omega`