Show how would you join three resistors, each of resistance `9 Omega` so that the equivalent resistance of the combination is (i) `13.5 Omega (ii) 6 Omega`?

Explanation

Here resistance `R_1=R_2=R_3=9 Omega`
(i) To obtain an equivalent resistance `R_(eq)=13.5 Omega` we connect one resistors `R_1` in series to the parallel combination of `R_2 and R_3` are shown in figure (i) Then
`R_(eq)=R_1+((R_2R_3)/(R_2+R_3))=9+((9 times 9)/(9+9))`
=9+4.5=`13.5 Omega`
(ii) To obtain equivalent resistance `R_(eq)=6 Omega` we connect resistor `R_1` in parallel to the series combination of `R_2 and R_3` as shown in figure 12.29 (ii). Then
`R_(eq)=(R_1(R_2+R_3))/(R_1+(R_2+R_3))`
`=(9(9+9))/(9+(9+9))=(9 times 18)/27=6 Omega`