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CHAPTER 10 Magnetic Coupling. Transformers
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We must not conclude from the foregoing, however, that ALL unbalanced sets of three vectors can be expressed as the sum of just two balanced sets of vectors (one positive sequence set and one negative sequence set). This is because inspection of eqs. (448) and (449) shows that eq. (450) is valid only if " " " A0 B0 C0 0 that is, only if the sum of the vectors in the unbalanced set is equal to zero, a condition which MAY or MAY NOT be true in practical work. This indicates that a more general form of eq. (450) is needed, to cover cases in which the sum of the unbalanced vectors is not equal to zero. In this regard, a more general form of eq. (450) can be arrived at by thinking in terms of, degrees of freedom, as follows. We recall that, in general, an unbalanced set of three vectors, such as in Fig. 261, has six degrees of freedom. So far, however, in Figs. 262 and 263 we have only four degrees of freedom. Hence, in addition to Figs. 262 and 263 we must, in order to include the most general unbalanced condition, add one more set of three balanced vectors to bring the degrees of freedom up to six. This is done by de ning what is called a zero sequence set of vectors which consists of three IDENTICAL vectors, meaning that the three vectors all have the same magnitude of amplitude at the same angle k with respect to the reference axis, as shown in Fig. 264.
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Fig. 264
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Thus a zero sequence set of three vectors has two degrees of freedom and, using the subscript 0, is de ned by writing that " " " A0 B0 C0 451
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Now, while a zero-sequence set of vectors is a balanced set (in accordance with the de nition following Fig. 259), note that the vector sum is NOT zero (as it is when the vectors are 1208 apart); instead, for a zero-sequence set we have that " " " " " " A0 B0 C0 3A0 3B0 3C0 452 " Thus, in the three equations that comprise eq. (450), we now add A0 to both sides of the " "0 to both sides of the second equation, and C0 to both sides of the third rst equation, B " " " equation. Then, letting A 0 A0 A, and so on, eq. (450) becomes 9 " " " " A1 A2 A0 A > = " " " " 453 B1 B2 B0 B > ; " " " " C1 C2 C0 C " " " where A, B, and C , without subscripts, represent three components of an equivalent unbalanced set of three vectors having the required six degrees of freedom. Thus we ve " " " now expressed an unbalanced set of vectors, A, B, C , in terms of the components of three balanced sets of vectors. Now suppose the components of an unbalanced set are known, and we wish to nd the values of the three equivalent balanced sets. That is, let the PROBLEM be: GIVEN the
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CHAPTER 10 Magnetic Coupling. Transformers
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" " " values of an unbalanced set, A, B, and C , nd the values of the components of the three balanced sets whose vector sum is equal to the given unbalanced set. This would seem, o hand, to be a most di cult problem, but fortunately, because of the symmetry of balanced sets, it turns out not to be so hard after all. Let us proceed as follows. First, carefully note, again, the set of equations given just prior to eq. (450); doing this, and also keeping eq. (451) in mind, note that eq. (453) becomes (using degrees) " " " " A1 A2 A0 A " " " " A1  j120 A2  j240 A0 B " " " " A1  j240 A2  j120 A0 C Now, for convenience, let " a  j120 then, also{ " a2  j240 and hence the foregoing three equations can be written in the easier-to-handle forms " " " " A1 A2 A0 A " " " " "" aA1 a2 A2 A0 B " " "" " " a2 A1 aA2 A0 C 455 456 457 454 *
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" " Note that we now have three simultaneous equations in three unknowns, A1 , A2 , and " A0 , the values of which can be found in several ways, including the method of elimination. Let us, however, use the more straightforward method of determinants, as follows. You ll recall that the rst step in the procedure is to nd the value of the determinant formed from the coe cients of the unknowns which, as you should now verify from inspection of the above three equations, is equal to    1 1 1     " " aa a D  a a2 1  " "3 3" 2   2  a " " a 1 " " Or, since a  j120 , then a3  j360 1, the above reduces to D 3" " 1 aa 458
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To continue on, in our solution of eqs. (455) through (457), let us next nd the value of " A1 ; thus   A 1 1 "     " "  B a2 1     C a 1  A "2 a B 1 a C 1 a2 " " "a " " " " " " A1 D 3" " 1 aa
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" * Thus, a  j120 cos 120 j sin 120 0:5 j0:8660 , in rectangular form. { Using the basic relationship xy n xny .
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