A cell of emf 2 volt and internal resistance `1.5Omega` is connected to the ends of 1 long wire. The resistance of wire is (0.5Omega)/(m)`. Find the value of potential gradient on the wire.

Explanation

`X=(IR)/(L)=((E)/(R+r))(R)/(L)=(2xx0.5)/(0.5+1.5)=0.5(V)/(m)`