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

Given : E = 2 V, r = 0.
(i)
For a series combination,
`R_(s)=R_(1)+R_(2),I=E/(R_(s)+r)`
i.e., `I=2/(R_(s))=2/5`
or `R_(s)=5Omega`
or `R_(1)+R_(2)=5` ...(1)
For a parallel combination,
`R_(p)=(R_(1)R_(2))/((R_(1)+R_(2)))=(R_(1)R_(2))/(R_(s))=(R_(1)R_(2))/5=(R_(1)R_(2))/5` ....(2)
Here `I=E/(R_(p)+r)`
i.e., `5/3=2/(R_(p))`
i.e., `R_(p)=(6/5)` using (2) we write
`(R_(1)R_(2))/5=6/5`
`thereforeR_(1)R_(2)=6` ....(3)
But `(R_(1)-R_(2))=sqrt((R_(1)+R_(2))^(2)-4(R_(1)R_(2)))`
i.e., `(R_(1)-R_(2))=sqrt((5)^(2)-4(6))=1`
Solving `R_(1)+R_(2)=5` we get `2R_(1)=6`
and `R_(1)-R_(2)=1` or `R_(1)=3Omega`
and `R_(2)=2Omega`