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

E=2v,r =0 , `I_(s)=2/5`A
`I_(p)=5/3A, R_(1)=? R_(2)=?`
(a) Resistors in series :

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`I_(s)=(E)/(R_(s)+r)i.e (cancel2)^(1)/(5)=(cancel2)^(1)/(R_(1)+R_(2)+0)`
i.e `R_(1)+R_(2)=5`
(b) Resistors in parallel
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`I_(p)=(E)/(R_(p)+r)i.e 5/3 =(2)/(R_(1)R_(2))/(R_(1)+R_(2)+0)`
`therefore R_(1)+R_(2)=5 , R_(1)R_(2)=6`
and `R_(1)+R_(2)=sqrt(5)^(2)-4^(6)=1`
solving `R_(1)+R_(2)=5`
`R_(1)-R_(2)=1`
`2R_(1)=6`
`R_(1)=3 Omega` and `3-R_(1)=1`
`therefore R_(2) =2Omega`
Hence `R_(1)=3Omega R_(2) =2 Omega`