Harmonics in the exciting current of transformers . r i rr -arr iZ\0 ? sin*-d0 + T 1 sin*30d0 + Ts 3ir/ cos50d0+ .... iM sincos0d 4-T, 1^ I cos3in 30d0 4- .... •Jo 5d + lHcoszd0 - iM cos^dl 20 integrating each one separately , Zrr sin*0d0 = 1*1 ( 1/2 - cos20/2 )d#= lf[( T/2 d0 - 1/2/ cos20d0 = Jo 2TT XTT Jo J0 I*0/2 4{ 1^2 sin 20 ) = 21^2 = I, fm^ajsin 30do-3#3| sin* 0d0 = JoFormula of integration Jsin* xdx = x/2 - 1/2 sinxcosxJcos2xdx - x/2 -i- 1/2 sinxcosxJl^cos^sin^d^= 0 . For the product of two sine waves is a sine wave and the integral between zero and2-rris zero .Therefore
Harmonics in the exciting current of transformers . r i rr -arr iZ\0 ? sin*-d0 + T 1 sin*30d0 + Ts 3ir/ cos50d0+ .... iM sincos0d 4-T, 1^ I cos3in 30d0 4- .... •Jo 5d + lHcoszd0 - iM cos^dl 20 integrating each one separately , Zrr sin*0d0 = 1*1 ( 1/2 - cos20/2 )d#= lf[( T/2 d0 - 1/2/ cos20d0 = Jo 2TT XTT Jo J0 I*0/2 4{ 1^2 sin 20 ) = 21^2 = I, fm^ajsin 30do-3#3| sin* 0d0 = JoFormula of integration Jsin* xdx = x/2 - 1/2 sinxcosxJcos2xdx - x/2 -i- 1/2 sinxcosxJl^cos^sin^d^= 0 . For the product of two sine waves is a sine wave and the integral between zero and2-rris zero .Therefore the total area is , iV+IVr +IV+ iVt 4-1 rr +lVr + Then the ordinate is , or or t6 KV r+IJn +I!tt + . I Tl 4-1^4-1*^4- 2t- r If +Ij 4- . 2 iT +1; +1! +. »• . • 1* which is the proof required . The same results can be obtained graphically by reversing the triple harm-onic in Plate TO . This can be done because the triple frequency harmonic is T80degrees out of phase with the primary current . Th
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Keywords: ., bookcentury1900, bookdecade1910, booksubjecttheses, bookyear1912
