Explain the term 'inductive reactance'.Show graphically the variation of inductive reactance with frequency of the applied alternating voltage.
An `a.c.` voltage `E=E_(0)sin omegat` is applied across a pure inductor `L`.Show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of `pi//2`.

Explanation

Inductive reactance :The effective resistance offered by inductance `L` is called inductive reactance.It is denoted by `X_(L)`.it `SI` unit is `ohm`.The graph between frequency `(v)` and `X_(L)` is shown in fig.
`A.C.` Source containing an inductor: Let a source of alternating `e.m.f` `E=E_(0) sin omegat` be connected to an inductor of inductance `L` only.
If `i` be the current through the circuit, then
instanteous induced `e.m.f` produced `V=L(di)/(dt)`
`di=V/L dt`
`di=V_(m)/L sin omegat dt , [because V=V_(m)sin omegat]`
Integrating both sides, we get
`intdi=V_(m)/Lint sin omegat dt`
`i=V_(m)/L((-cosomegat)/omega)=(-V_(m))/(omegaL)cos omegat`
`i=(-V_(m))/(omegaL)sin (pi/2-omegat)`
`=(-V_(m))/(omegaL)sin(omegat-pi/2)`
`i=i_(m)sin(omegat-pi/2)`...(i)
where `i=V_(m)/(omegaL)` is the amplitude of the current.The quanity `omegaL` is called the inductive reactance and is denoted by `X_(L)` i.e., `X_(L)omegaL`
The equation `(i)` shows that the current logs the voltage by `pi/2`
(or `e.m.f` lends the current by a phase angle `pi/2`)
The phasor diagram for the circuit between `vecV` and `veci` at any istant `t` is shown in Fig. (a)and the graph of `V` and `i` versus `omegat` is shown in Fig. (b)