An ideal battery passes a current of 5A through a resistor. When it is connected to another resistance of `10Omega` in parallel, the current is 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)=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))rArr I_(1)=I_(2)(R_(2))/(R_(1)+R_(2))`
`5=6xx(10)/(R_(1)+10)rArr 5(R_(1)+10)=60`
`5R_(1)+50=60, 5R_(1)=10`
`R_(1)=(10)/(5)=2OmegarArr R_(1)=2Omega`