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

` R_1 = R, i_1 = 5A,`
` R_2 = (10R/ 10+R) , i_2 = 6A`
` Since potential is constant `
` i_1 R_1 = i_2 R_2 `
` rArr 5 xx R = (6 xx 10R / 10 + R )`
` rArr (10+R)5 = 60 ` rArr 50+ 5R = 60 `
` rArr 5R = 10 ` rArr = 2 Omega `.