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

`I=E/(R+r)` or `R=V/I`
Case (1): `R_S=2/(2//5)=5Omega`
Case(2): `R_S=2/(5//3)=6/3 Omega`
Simplification
Arriving at `R_1 =3Omega` and `R_2=2Omega`
Given , E=2V , r=0 , `I_s=2/5A`
`I_P=5/3A R_1=? , R_2`=?
(a)Resistors in series :

`I_S=E/(R_S+r)` i.e., `2/5=2/((R_1+R_2)+0)`
i.e., `R_1+R_2=5` ...(i)
(b) Resistors in parallel :

`I_p=E/(R_p+r)` i.e., `5/3=2/((R_1R_2)/(R_1+R_2) + 0)`
i.e., `(R_1R_2)/(R_1+R_2)=6/5`
`therefore R_1+R_2= 5 , R_1R_2=6`
and `R_1+R_2=sqrt((5)^2 -4(6))=1`...(ii)
Solving `R_1+R_2=5`
`R_1-R_2=1`
`2R_1=6`
`R_1=3Omega` and `3-R_1=1`
`R_2=2W`
Hence `R_1=3Omega, R_2=2W`