When switch S it closed, at that moment the current in the circuit is zero. So, the induced e.m.f. across the inductance is equal to e.m.f. E of the battery. As current increases, the induced e.m.f. in the inductance decreases. If at any instant t, the current in the circuit is i, then applying Kirchhoff's law, we haveiR+e=E or e=(E−iR)So, the graph between e and i will be a straight line having negative slope.