The potential dofference between the terminals of a battery of emf6.0V and internal resistances`1 (Omega)`drops it 5.8 V when connected across an external resistor. Find the resistance of the external resistor.

Explanation

`E = ` 6v,` R= ` 1 Omega`
` v= 5.8 V, R=?`
` (i= E/(R+r)) = 6/(R+1) `
` V= E-iR`
` rArr 5.8 = 6-(6/R+1) xx 1 `
` rArr (6/(R+1)) = 0.2 `
` R+1 = 6/0.2 = 30 `
` R = 29 Omega` .