Two resistances when connected in parallel give resultant value of 2Omega, when connected in series the value becomes 9Omega. Calculate the value of each resistance.

Two resistances when connected in parallel give resultant value of 2Om

1.	Two resistances when connected in parallel give resultant value of 2Omega, when connected in series the value becomes 9Omega. Calculate the value of each resistance.
Answer» Solution :LET `R_(1)andR_(2)` be the two resistances. When they are in PARALLEL, resultant resistance `R_(p)` is given by `1/R_(p)=1/R_(1)+1/R_(2)` `R_(p)=(R_(1)R_(2))/((R_(1)+R_(2)))` `2=(R_(1)R_(2))/((R_(1)+R_(2)))` `2[R_(1)+R_(2)]=R_(1)R_(2)` When they are in series effective resistance `R_(s)=R_(1)+R_(2)` `9=R_(1)+R_(2)rArrR_(2)=9-R_(1)` `2xx9=R_(1)R_(2)=R_(1)[9-R_(1)]=9R_(1)-R_(1)^(2)`. `18=9R_(1)-R_(1)^(2)` `R_(1)^(2)-9R_(1)+18=0` Solving by using quadratic formula, we get `R_(1)=((9pmsqrt(81-4xx1xx18))/2)""R_(1)=((9pmsqrt9)/2)=((9pm3)/2)` `R_(1)` = 3 OHM or 6 ohm If `R_(1)` = 3 ohm, `R_(2)` = 6 ohm and vice VERSA.