Show that voltage in an inductor leads the current by `pi//2` rad for a pure inductor

Explanation

Consider an inductor of inductance L is connected across an AC source .
Let `V= V_(0) sinomegat` …. (1)
the self induced emf in the conductor is
`epsilon = - L (dI)/(dt)`,
According to K irchoff loop rule
`v- L (dI)/(dt) = 0`
`RightarrowV_(0)sinomegat-L(dt)/(dt)=0`
`RightarrowL (dt)/(dt)= V_(0)sinomegat.This indicates the current in an inuctor is a function of time .
`dt= v_(0)/(L)sin omegat` dt
To obtain the current at any tan t, we integrate above equation
`I = int di = (V_(0))/(L )int sin omegat dt`
`Rightarrow I = (V_(0))/(L)[-cosomegat0/(omega) + constant]`
`I = (V_(0))/(L) [(cosomegat0/(omega)+ constant]`
`I= (V_(0))/(L_(omega) (-cosomegat)`,br> If we take `(V_(0))/(Lomega)= I_(0)`, the amplitude of the current , then `I = I_(0)(-cosomegat)`
`therfore I = I_(0)sin (omegat- (pi)/(2))rightarrow (2)`
From (1) and (2) we can conclude that voltage leads current by `(pi)/(2)`,br> Phasor diagram :