Question

Prove that an ideal capacitor in an AC circuit does not dissipate power

Answer 1

In an AC circuit containing only an ideal capacitor, the current i leads the emf e by a phase angle of π/2 rad.

Here, for e = e₀ sin wt, 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 omega"t" cos omega"t dt"`

Now, `int_0^"T" sin omega"t" cos omega"t dt" = 0`

∴ Pav = 0, i.e., the circuit does not dissipate power.

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