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

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`I_(s)=(E )/(R_(S)+r )i.e., ( cancel (2)^(1))/(5) = ( cancel (2)^(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_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_(1) R_(2) =6 `
and `R_(1) +R_(2) = sqrt((5)^2-4(6))=1....(2)`
Solving `R_1 +R_2 =5,R_1-R_2=1`
`2R_1=6 `
`R_1=3Omega and 3-R_1=1`
` R_2=2 Omega`
Hence `R_1=3Omega , R_2 = 2 Omega`