Theory and calculation of alternating current phenomena . uted in (3), and transposed, give:as at the generator terminals. since Ei = e EE2 = €^e E3 = e € E - E\(l + YiZ2 + F1Z3) + E^Y^Zs + E3Y3Z, = 0e^E - ^2(1 + Y2Z3 + Y2ZO + E3Y3Zi + E\YiZ3 = 0 [ (6)E - e3(1 + F3Z1 + F3Z2) + E\YiZ, + ^2^2^! = 0 as three linear equations with the three quantities, El, El, E3. THREE-PHASE SYSTEM Substituting the abbreviations: 459 K = Ky = Ko = Kz^ - (1 + F1Z2 + Y^Za), Y^Za, YzZ^ Y,Zs, - (I + F2Z3+ F2Z1), F3Z1 F1Z2, F2Z1, - (1 + FsZi + F3Z2) 6, F2Z3 F3Z2 e\ - (1 + F2Z3 + F2Z1), F3ZX 1, FaZi, - (1 + F3Z
Theory and calculation of alternating current phenomena . uted in (3), and transposed, give:as at the generator terminals. since Ei = e EE2 = €^e E3 = e € E - E\(l + YiZ2 + F1Z3) + E^Y^Zs + E3Y3Z, = 0e^E - ^2(1 + Y2Z3 + Y2ZO + E3Y3Zi + E\YiZ3 = 0 [ (6)E - e3(1 + F3Z1 + F3Z2) + E\YiZ, + ^2^2^! = 0 as three linear equations with the three quantities, El, El, E3. THREE-PHASE SYSTEM Substituting the abbreviations: 459 K = Ky = Ko = Kz^ - (1 + F1Z2 + Y^Za), Y^Za, YzZ^ Y,Zs, - (I + F2Z3+ F2Z1), F3Z1 F1Z2, F2Z1, - (1 + FsZi + F3Z2) 6, F2Z3 F3Z2 e\ - (1 + F2Z3 + F2Z1), F3ZX 1, FaZi, - (1 + F3Z1 + F3Z2) - (I + F1Z2+ FiZ3),6, F3Z2 F1Z3, F1Z2, 6^ F3Z1 1, - (1 + F3Z1 + F3Z2). (7) - (l + FiZ2+F,Z3), F2Z3, F1Z3,F1Z2, - (1 + F2Z3 + F2ZO, eY,Z,, 1 we have: EKi?- K EK2?- K EK3 ^= K YiEKT 1 f- K Y2EKT 2 ^.- K Y3EK3 ^- k 1 Y3K3 - Y2K2E T ^- K YiKi - YJCzE ^- K J. Y2K2 — YiK\i!j (8) (9) (10) hence, E\ + E2 + ^3 = 0 I (11) 460 ALTERNATING-CURRENT PHENOMENA309. Special Cases. A. Balanced SystemY,= Y,= Y,= Y Substituting this in (6), and transposing:. 1 + 3FZ E\ = E 2 = E 3 = h = h =h = iE 1 +3 FZ 2E 1 + 3 rz E1 +3 FZ e^e - 1)EY 1 + 3 FZ(6- 1)EY 1 + 3 FZe (e - 1) EY 1 + 3FZ (12) The equations of the symmetrical balanced three-phase system. B. One Circuit Loaded, Two Unloaded Fi = F2 = 0, Ys = Y,Zi = Z2 = Z3 = Z. Substituted in equations (6): eE - Ei-\- EzYZ = 0 unloaded branches. tE - E\ -h E\YZ = 0,E - Esil +2 YZ) = 0, loaded branch. hence:E, =E 2 =E 3 = E\e-{-{l+2e)YZ} 1 + 2 FZE{e^-\-(l +2e2) YZ} 1+2 FZ E 1 +2FZ unloaded;loaded; all , and (13) of unequalphase angles. THREE-PHASE SYSTEM 461 I\ = /2 = 0 EYIz = Ii = 1 +2YZ EY1 +2FZEY f 1 + 2 FZ /a = 0 C. Two Circuits Loaded, One Unloaded Fi = F2 = F, F3 = 0,Zi = Z2 ^= Z3 = z. Substituting this in equations (6), it is e E - E\{1 + 2 FZ) + E2YZ = 0e^E - E^il + 2 FZ) + i;iFZ = 0 E - Es -\- {E\ + ^2)FZ = 0 unloaded branch. loaded branches. or, since {E\ + E2) = - E,;E- Es- EsYZ = 0, E 3 = 1 + YZ thus, je;6{i +
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