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Prove that an ideal capacitor in an a.c....

Prove that an ideal capacitor in an a.c. circuit does not dissipate power.

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Average power dissipated with a capacitor : In a circuit containing capacitor C, current leads the emf by a phase angle of `(pi)/(2)`.
`:. E=E_(0) sin omega t "then" I=I_(0) sin (omega t + (pi)/(2)) :. I=I_(0) cos omega t`
Work done in one complete cycle is
`w=underset(0)overset(T) int E I dt = underset(0)overset(T)int ( E_(0) sin omega t) ( I_(0) cos omega t) dt = E_(0) I_(0) underset(0) overset(T) int sin omega t cosomega t = E_(0)I_(0) underset(0) overset(T) int (sin 2 omegat)/(2) [:' sin 2theta = 2 sin theta cos theta]`
`=(E_(0)I_(0))/(2) [-(cos 2 omegat)/(2omega)]_(0)^(T)=-(E_(0)I_(0))/(2)[(cos 2 omega t)/(2omega)-(cos 0)/(2omega)]=-(E_(0)I_(0))/(2)[(cos 2.(2pi)/(T).T)/(2omega)-(1)/(2omega)] `
`=-(E_(0)I_(0))/(2) [(cos 4 pi)/(2omega)-(1)/(2omega)]=-(E_(0)I_(0))/(2)[(1)/(2omega)-(1)/(2omega)]=0 [:' cos 4 pi = 1]`
`:. ` Average power `=(omega)/(T)=(0)/(T)=0`
`:.` Average power supplied to an ideal capacitor by the source over a complete cycle of a.c. is zero.
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