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?

Show how would you join three resistors, each of resistance 9 Omega so

1.	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?
Answer» SOLUTION :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`