Obtain an expression for the current through an inductor when an AC emf is applied.

Explanation

Circuit consists of pure inductor of inductance L. Let an ac e.m.f `V = V_(m)` sin `omega t` is applied to it. Let I be the instantaneous current.
Back e.m.f deceloped across the inductor `= - L (di)/(dt)`
Total e.m.f `= V_(m)sin omega t - L(di)/(dt)` ....(1)
According to ohms law this must be equal to iR = 0
`V_(m)sin omega t-L(di)/(dt)=0 (because R = 0)`
`L (di)/(dt)=V_(m)sin omega t`
`di = ((V_(m))/(L))sin omega "t dt"`
Integrating we get, `int di = int (V_(m))/(L)sin omega "t dt"`
`i = (V_(m))/(omega L) cos omega t + k` ....(2)
Where k is integration constant. Its value k = 0
`i = -(V_(m))/(omega L)cos omega t = (V_(m))/(omega L)sin (omega t-(pi)/(2))`
`i=i_(0)sin(omega t-(pi)/(2))` ....(3)
`therefore i_(0)=(V_(m))/(omega L)`
This is the expression for the instantaneous current through inductor. Here current lags behind the e.m.f by `(pi)/(2)` radian (or) `90^(@)`.