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S.T. average power over a complete cycle in a pure inductor connected to ac is zero. |
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Answer» SOLUTION :The instantaneous power supplied to an INDUCTOR `P_(L)=iv` `=i_(m)sin(omegat-pi//2)(v_(m)SINOMEGAT)` `=-i_(m)cosomegatv_(m)sinomegat` i.e., `P_(L)=-(1)/(2)i_(m)v_(m)(sin2omegat)` where `sin2omegat=2sinomegatcosomegat` The AVERAGE power over a complete cycle is `P_(L)=(:-(i_(m)v_(m))/(2)sin2omegat:)` `=-(i_(m)v_(m))/(2)(:sin2omegat:)` = 0 `therefore (:sin2omegat:)=(:cos2omegat:)=0` Thus average power supplied to an inductor over one complete cycle is zero. |
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