Consider an inductor of inductance L is connected across an AC source,
Let v = vmsin ωt --------(1), here v - the source voltage, vm – peak voltage , ω - angular frequency of AC.
The self induced emf in the inductor is
According to Kirchoff’s loop rule,
This indicates the current in an inductor is a function of time.
To obtain the current at any instant, we integrate the above expression.
[because, we can show the integration constant over a cycle is zero]
If we take
the amplitude of the current, then i = im [-cos ωt]
Inductive reactance is given by XL = ωL = 2πνL = 2πfL
The SI unit of XL is ohm (Ω)
