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

Show how you would connect three resistors, each of resistance 6 Omega

1.	Show how you would connect three resistors, each of resistance 6 Omega, so thatthe combination has a resistance of (i) 9 Omega, (ii) 4Omega.
Answer» Solution :(i) In ORDER to get a resistance of 9 n fromthree 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`