Prove mathematically that the average power over a complete cycle of alternating current through an ideal inductor is zero.

Explanation

Let E= Alternating emf applied to a pure inductor.
At any time applied alternating emfis given by.
`E = E_0sinomegat`
Let I = Current flowing through the pure inductor at the given time.
But we know that for pure inductor, current lags behind voltage by phase angle `pi//2`.
`I=I_0 sin(omegat -pi//2)`
`=-I_0 sin(pi//2 - omegat)`
`I =-I_0 cos omegat`
Average power over the complete cycle of a.c. is given by
`P_(av)=1/Tint_0^(T)EIdt`
`P_(av)=1/Tint_0^(T)E_0sinomegat (-I_0cosomegat)dt`
`=(-E_0I_0)/(T)int_0^Tsinomegat cos omegatdt`
`=-(E_0I_0)/(2T)int_0^T2sinomegat cos omegatdt`
`=-(E_0I_0)/(2T)int_0^Tsin2omegat dt `
`=-(E_0I_0)/(2T)|(cos2omegat)/(2omega)|_0^T`
`=-(E_0I_0)/(4omegaT)|cos2omegaT-cos0|`
`=-(E_0I_0)/(4omegaT)|cos2.((2pi)/T)T-1|`
`=-(E_0I_0)/(4omegaT)|1-1|=0`
So average power consumed for a ideal inductor during a complete cycle is zero.