. . ( 5.1):
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, ( 5.2):
B C D
L 1 L 2 L 3 P B P C P D
5.2.- L = (L 1+ L 2+ L 3) [ ] L = (P B+ P C + P D) [ ].
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S [ 2 ] Δ U [ ] [ ] I [ ] L [ ] ):
100% × S(P I × L i) 100% ×S(I i × L i) S(I i × L i) S = = = ; (5.8) D U % × U 2 × sD U % × U 2 × sD U 2 × s |
: U [ ];
D U % ;
100 % ;
s
[ ( ) 1] = [ ( 2) 1 ], . 5/5.
(5.8) :
2× 100% × S(P I × L i) 2 × S(I i × L i) S = = . (5.9) D U % × U 2 × sD U 2 × s |
D U S [] ( D):
I i × L I D U S = S D U i = S I i × R i = S . (5.10) s× S |
D U % 357 77:
) 5 %;
)
5 % + 10%;
) 10 %;
)
20 % 30 %;
) 10 % 15 %;
) , 1/3 40 % ( , , ).
( ) :
Z D U % = 100 %; (5.11) Z + Z |
: Z = (Z + Z ) [ ];
Δ U % × U
Z = - -
|
|
100 % I Ö 3 , -
[ ];
U
Z = -
I * I Ö 3
( ) [ ];
I * - ;
U ; I - [ ] [ ]
;
Δ U % - %,
;
Z - [ ],
5.7, 5.8 .