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

For ideal battery internal resistance r=0
Resistance of first wire = `R_1`
Resistance of the second wire
`R_2 = 10 Omega`

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)`
`implies I_1 = I_2 (R_2)/(R_1 + R_2)`
`5 = 6 xx (10)/(R_1 + 10) implies 5(R_1 + 10) = 60`
`5R_1 + 50 = 60 , 5R_1 = 10`
`R_1 = (10)/(5) = 2 Omega implies R_1 = 2 Omega`