Two resistors are connected in series with 5V battery of negligible internal resistance. A current of 2 A flows through each resistor. If they are connected in parallel with the same battery a current of `(25)/(3)A` flows through combination. Calculate the value of each resistance.

Explanation

`E = 5V " " I = (E)/(R)`
`I = 2A` (series) `I = (25)/(3)A` (parallel)
Resistors in series : `R_(S) = R_(1) + R_(2)`
`I = (E)/(R_(S)) rArr R_(s) = (E)/(I) = (5)/(2)`
`R_(1) + R_(2) = 2.5` ….(1)
Resistors in parallel : `R_(p) = (R_(1)R_(2))/(R_(1) + R_(2))`
`I = (E)/(E_(p)) rArr R_(p) = (E)/(I)`
`R_(p) = (5)/(25//3) = (3)/(5)`
`R_(1)R_(2) = (3)/(5) xx (R_(1) xx R_(2)`
`R_(1)R_(2) = (3)/(5) xx (R_(1) + R_(2))`
`R_(1)R_(2) = (3)/(5) xx 2.5`
`R_(1)R_(2) = 1.5`
`(R_(1) - R_(2))^(2) = (R_(1) + R_(2))^(2) - 4R_(1)R_(2)`
`(R_(1) - R_(2))^(2) = (2.5)^(2) - 4 xx 1.5 = 6.25 - 6 = 0.25`
`(R_(1) - R_(2))^(2) = 0.25 " "R_(1) - R_(2) =0.5 Omega` ....(2)
From (1) and (2)
`R_(1) = (3)/(2) = 1.5 Omega`
`R_(1) + R_(2) = 2.5 " "1.5 + R_(2) = 2.5`
`R_(2) = 1Omega`