As back e.m.f. `e = L (dI)/(dt)`, e will be maximum when `dI//dt` = max. The graph in Fig. shows that `dI//dt` is maximum for 5 x lt 6 lt 10 s.
For `0 s lt t lt 5 s, (dI)/(dt) = (1)/(5)` `:.` At `t = 3 sec , e = L (dI)/(dt) = (L)/(5)`
For `5 s lt t lt 10 s, (dI)/(dt) = (-2-1)/(5) = (-3)/(5) :.` At `7 s, e = L(dI)/(dt) = L (-(3)/(5)) = -3 e`
For `10 s lt t lt 30 s, (dI)/(dt) = (2)/(20)=(1)/(10) :.` At `t = 15 s, e = L (dI)/(dt) = (L)/(10) = (e)/(2)`
For `t gt 30 s, (dI)/(dt) = 0 :.` At `t = 40 s, e = 0`