Question

When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.

Answer 1

In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of π/2 rad. Here, for e = e₀ sin ωt, we have,

i = i₀ sin (ωt - π/2)

Instantaneous power, P = ei

= (e₀ sin ωt)[i₀(sin ωt cos π/2 - cos ωt sin π/2)]

= - e₀i₀ sin ωt cos ωt  as cos π/2 = 0 and sin π/2 = 1

Average power over one cycle,

`"P"_"av" = "work done in one cycle"/"time for one cycle"`

`= (int_0^"T" "P dt")/("T") = (- int_0^"T" "e"_0"i"_0 sin omega"t" cos omega"t" "dt")/("T")`

`= (- "e"_0"i"_0)/"T" int_0^"T"` sin ωt cos ωt dt

Now, `int_0^"T"` sin ωt cos ωt dt = 0

∴ P_av = 0

[Note: For reference, see the answer to Q. 6. The proof should be written as part of the answer.]