Which two resistors are connected in series with a cell of emf 2V and negligible internal resistance, a current of (2/5)A flows in the circuit. When the resistances are in parallel, the main current is (5/3)A. Calculate the resistances.

Explanation

Let `R_(1) & R_(2)` be the two resistance. When the resistances are connected in series `R_(s)=R_(1)+R_(2)` Current in the circuit is `I=(epsilon)/(R+r)i.e.,(2)/(5)=(2)/(R_(s)+0)i.e.,impliesR_(s)=R_(1)+R_(2)=5...(1)` when two resistances are connected in parallel `(1)/R_(p)=(1)/(R_(1))+(1)/(R_(2))=(R_(1)+R_(2))/(R_(1)R_(2))=(5)/(R_(1)R_(2))( :. "from"(1))R_(p)=(R_(1)R_(2))/(5)`
Current in the circuit is `I=(epsilon)/(R_(p)+r)(5)/(3)=(2)/(((R_(1)R_(2))/(5))+0)=(10)/(R_(1)R_(2))R_(1) R_(2)=6.....(2)` Using the relation
`(R_(1)-R_(2))^(2)=(R_(1)+R_(2))^(2)-4 R_(1)R_(2) (R_(1)-R_(2))^(2)=(5)^(2)-4(6)=25-24=1` `:. R_(1)-R_(2)=1.....(3)`
Adding equations (I) & (3) , weget `2R_(1)=6 or R_(1)=3 Omega" " R_(2)=5-R_(1)=5-3=2Omega`