Show mathematically that in an a.c. circuit containing only inductance, the current lags behind the e.m.f. by a phase of `pi/2`.
An a.c. voltage `E=E_0 sin omegat` is applied across an inductor L. Obtain an expression for current I.

Explanation

Consider that a pure inductor of inductance L (having no ohmic resistance) is connected to a source of an alternating emf.

Let the instantaneous value of alternating emf be given by
`E= E_0sinomegat`
If I is the current through the circuit and `(dI)/(dt)` the rate of change of current in the circuit at the instant, then
the instantaneous induced emf produced across `L = -L(dI)/(dt)`
So total instantaneous emf of the circuit = `E+(-L (dI)/(dt))`
Total instantaneous emf in the circuit = Instantaneous. potential drop across the pure inductor
`E+(-L(dI)/(dt))=0`
`E=L(dI)/(dt)`
`dI=E/Ldt`
`dI=E_0/L sinomegatdt `
Integrate both sides,
`intdI=E_0/2intsinomegatdt `
`I=E_0/2((-cosomegat)/(omega))`
`I=-E_0/(omegaL)cosomegat`
We know `-cosomegat =sin(omegat-pi//2)`
and `X_L =omegaL = ` Inductive reactance
`:. " " I=E_0/(X_L) sin(omegat - pi//2)`
Now `E_0/X_L = I_0=` Peak value of alternating current in the circuit .
So `I=I_0 sin (omegat -pi//2) " "...(ii)`
From eqns. (i) and (ii) It is clear that alternating current lags behind the alternating emf by phase angle `pi//2` as shown in fig. (b).