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

We know that ,` I = E /((R_(1) + R_(2)) +r) `
Where , r = 0
Hence , ` I = E/(R_(1) + R_(2)) or R_(1) + R_(2) = E/1 `
i.e,` R_(1) + R_(2) = 5/2 `
Using ` (R_(1) - R_(2)) = sqrt((R_(1) + R_(2))^(2) -4 R_(1)R_(2))`
` (R_(1)- R_(2)) = sqrt((5/2)^(2)- (3/2)^(2)) = 1/2`
soliving ` R_(1) + R_(2) + 5/2`
` R_(1) - R_(2) = 1/2 `
` 2 R_(1) = 3 `
` R_(1) = 3/2 Omega `
`3 /2 + R_(2) + 5/2`
` therefore R_(2) = 1 Omega or R_(2) = 1.5 Omega`
(ii) Since ` I = E/(R_(eq) + r) t=0`
` 25/3 = ( 5(R_(1)+R_(2)))/(R_(1)R_(2))`
`i.e, 25/3 = ( 5xx5)/2 1/(R_(1)R_(2))`
`i.e,R_(1) R_(2) = 3/2 `