An ideal battery sends a current of 5A in a resistor.When another resistor of value `10 Omega`is connected in parallel ,the current through the battery is increased to 6A.Find the resistance of the first resistor.

Explanation

Current through `R_(1)` in the first case, `i_(1) = 5A`
Current in the second case `i_(2)=6 A`
Effective resistance in the second case
`R=(R_(1)R_(2))/(R_(1)+R_(2)) , V=I_(1)R_(1)` and `V=I_(2) (R_(1)R_(2))/(R_(1)+R_(2))`
`I_(1)R_(1) =I_(2) (R_(1)R_(2))/(R_(1)+R_(2)) rArr I_(1) = I_(2) (R_(2)+R_(2))`
`5 = 6 xx 10/(R_(1)+R_(2)) rArr 5(R_(1) + 10)=60`
`5R_(1) + 50 = 60, 5R_(1) =10`
`R_(1) = 10/5 = 2Omega rArr R_(1) = 2 Omega`