Obtain an equation of current for AC voltage applied to an inductor and draw a graph of V and I.

Explanation

In figure shows an AC source connected to an inductor.

Inductor has negligible resistance. Thus, the circuit is a purely inductive AC circuit.
Let the voltage across the source be `V= V_(m) sin omega t ` . Using the Kirchoff.s loop rule,
`V - L (dI)/(dt ) = 0 ` where `- L (dI )/(dt)` is the self induced emf.
`:. V = L ( dI )/( dt ) `
` :. (dI)/(dt) = ( V )/( L )`
But `V = V_(m) sin omega t `
`:. ( dI )/( dt ) = ( V_(m) sin omega t )/( L )`
`:. dI = ( V _(m))/( L ) sin omega dt ` .....(1)
where L is the self inductance. Equation (1) indicates that current I ( t) as a function of time,must be such that its slope `(dI)/(dt)` is sine an sinusoidally varying quantity with the same phase as the source voltage and an amplitude given by `(V_(m))/( L )`.
To obtain the current integrate equations (1) with respective to time,
`:. int dI = (V_(m))/( L ) int sin omega t dt `
`:. I = ( V_(m))/( L ) xx ( cos omega t )/( omega ) +` constant ....(2)
Here, integration constant has the dimension of current and is time independent
At t=0, time I = 0
`:.` 0 `+` constant
`:.` Constant =0
`:.` From equation (2),
`:. I = - (V_(m))/( L omega ) cos omega t `
`:. I = ( V_(m))/( omega L ) sin ( omega t - ( pi )/( 2)) [ :. - cos omega t = sin ( omega t - ( pi )/( 2))]`
`omega L` is called inductive reactance denoted by `X_(L)`.
`:. X_(l) = omega L ` and its unit is Ohm.
The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit.
The inductive reactance is directly proportional to the inductance and frequency of the current.
Hence , for the source voltage and the current in an indeuctor `V = V_(m) sin omega t ` and `I = I _(m) sin( omega t - ( pi )/( 2))` . This shows the that hte current lags the voltag by `( pi )/(2)` or one -quarter `((1)/(4))` cycle.
In this case figure (a) shows that voltage and current phasors in the present case at instant `( t_(1))`.

The current phasor I is `( pi )/(2)` behind the voltage phasor V. ( Lags)
`( pi )/(2) = ( T )/( 4) :. ( pi )/( 2) = ( T )/( 2pi ) xx ( 2pi )/( 4) = (( pi )/( 2))/( omega ) = ( pi )/( 2 omega )`
When rotated with frequency `omega ` counter clockwise, they generate the voltage and current given by `V = V_(m) sin omega t ` and `I = I _(m) sin ( omega t - ( pi )/( 2))` respectively and as shown by figure (b). This graphs is for V and I versus `omega t `.