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Question
Prove that an ideal capacitor in an AC circuit does not dissipate power
Solution
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, Pav
`= "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.]
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