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

ltbrrgt
Current through `R_(1)` in the first case `i_(1)=5A`
Current in the second case `i_(2)=6A`
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))impliesI_(1)=I_(2)(R_(2))/(R_(1)+R_(2))`
`5=6xx(10)/(R_(1)+10)implies5(R_(1)+10)=60`
`5R_(1)+50=60,5R_(1)=10`
`R_(1)=(10)/(5)=2OmegaimpliesR_(1)=2Omega`