Question

2 resistance when connected in parallel the resultant value is 2 ohm when connected in series the value becomes 9 ohm calculate each resistance

Answer 1

these are 6 and 3
as for // Req = R1R2/(R1+R2)
for series Req = R1+R2

Answer 2

GIVEN:-

  RESISTANCE  IN SERIES:- 9 OHM = R1+R2

  RESISTANCE IN PARALLEL:- 2 OHM= 1/R1+1/R2

PROCEDURE:-

  1/R1+1/R2=2 OHM

   R2R1/RI+R2=2 OHM

   R2R1=2(R1+R2)

   R1+R2=9(GIVEN)

   THEREFORE,

         R2R1=2(9)

         R2R1=18

         LET THE VALUE OF R2=R1-9

         (9-R1)R1=18

         9R1-(R1*R1)=18

        (R1*R1)-9R1+18=0

        (R1*R1)-6R1+3R1+18=0

        R1(R1-6)-3(RI-6)

        (R1-3) (R1-6)

ANSWER:-        

        R1=6                            OR                                      R1=  3

        R2=3                           OR                                      R2=  6