Two resistors when connected in parallel give the resultant of 2 ohm, but when connected in series the effective resistance becomes 9 ohm ? Calculate the value of each resistance.

Two resistors when connected in parallel give the resultant of 2 ohm,

1.	Two resistors when connected in parallel give the resultant of 2 ohm, but when connected in series the effective resistance becomes 9 ohm ? Calculate the value of each resistance.
Answer» <P> Solution :Resultant resistance of parallel combination `R_(P)=2Omega` Resultant resistance of SERIES combination `R_(S)=9Omega` `(1)/(R_(P))=(1)/(R_(1))+(1)/(R_(2))` `1/2=(1)/(R_(1))+(1)/(R_(2))` `1/2=(R_(1)+R_(2))/(R_(1)R_(2))` `2(R_(1)+R_(2))=R_(1)R_(2)` . . . (1) Substitute equation (2) in equation (1) `R_(S)=R_(1)+R_(2)` `9=R_(1)+R_(2)` `R_(1)=9-R_(2)` . . (2) `2(9-R_(2)+R_(2))=(9-R_(2))R_(2)` `18=9R_(2)-R_(2)^(2)` `R_(2)^(2)-9R_(2)+18=0` `(R_(2)-3)(R_(2)-6)=0` `R_(2)=3,6` (i) If `R_(2)=3,R_(1)=9-R_(2)=9-3` `R_(1)=6` (ii) If `R_(2)=6,R_(1)=9-R_(2)=9-6` `R_(1)=3`