ejercicios_11_stevenson.pdf

CAPITULO II: ANALISIS DE SISTEMAS DE POTENCIA DE STEVENSON
ING. WIDMAR AGUILAR
Octubre -2022

ING. WIDMAR AGUILAR
Octubre -2022
Se sabe que: ( )
= 10∠180 ; ( )
= 50∠0 ; ( )
= 20∠90
= ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
Además: = ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
( )
= ( ) ( )
= ( )
( )
= ( ) ( )
= ( )
( )
= ( )
= ( )
Donde: a= 1∠ 120
Si: = ( )
+ ( )
+ ( )
= 10∠180 + 50∠0 + 20∠90
= −10 + 50 + 20 = 40 + 20
= 44.72 ∠ 26.57

ING. WIDMAR AGUILAR
Octubre -2022
= ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
= 10∠180 + 1∠ 240 ∗ 50∠0 + 1∠ 120 ∗ 20∠90
= −10 + 50∠ 240 + 20 ∠ 210
= −10 − 25 − 43.30 − 17.32 − 10
= −52.32 − 53.3 = 74.688∠ − 134.47
= ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
= 10∠180 + 1∠ 120 ∗ 50∠0 + 1∠ 240 ∗ 20∠90
= −10 + 50∠ 120 + 20 ∠ 330
= −10 − 25 + 43.30 + 17.32 − 10
= −17.682 + 33.3 = 37.7∠117.97

ING. WIDMAR AGUILAR
Octubre -2022
( )
= 600 ∠ − 90 ; ( )
= 250 ∠90 ; ( )
= 350 ∠90
= ( )
+ ( )
+ ( )
= 0
= 600 ∠ − 90 +250 ∠90 + 350 ∠90
= − 600 + 350 + 250= 0 (A)
Se tiene que: = ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
= ∗ 600 ∠ − 90 + ∗ 250 ∠90 +350 ∠90
= 1∠240 ∗ 600 ∠ − 90 + 1∠120 ∗ 250 ∠90 +350 ∠90
= 600∠ 150 + 250∠210 + 350∠90

ING. WIDMAR AGUILAR
Octubre -2022
= −519.615 + 300 − 216.506 − 125 + 350
= −736.1213 + 525 = 904.1568 ∠144.50
= ( )
+ ( )
+ ( )
= ( )
+ ( )
+ ( )
= ∗ 600 ∠ − 90 + ∗ 250 ∠90 +350 ∠90
= 1∠120 ∗ 600 ∠ − 90 + 1∠240 ∗ 250 ∠90 +350 ∠90
= 600∠30 + 250∠330 + 350∠90
= 519.615 + 300 + 216.506 − 125 + 350
= 736.12 + 525 = 904.156 ∠35.496
LA corriente por el neutro es:
+ + = ; = 0
= + = −736.1213 + 525 + 736.12 + 525
= 1050 ( )
Ó:
= 3 ( )
= 3 ∗ ( 350) = 1050
De:
!
( )
( )
( )
" =
#
$
1 1 1
1
1
% $ %
( )
=
#
( + + )

ING. WIDMAR AGUILAR
Octubre -2022
( )
=
#
(10∠0 + 10∠230 + 10∠130)
( )
=
#
(10 − 6.4279 − 7.660 − 6.428 + 7.660)
( )
= #
(−2.8559) = −0.952
( )
= 0.952∠180
( )
=
#
( + + )
( )
=
#
( 10∠0 + 1∠120 ∗ 10∠230 + 1∠240 ∗ 10∠130)
( )
=
#
( 10∠0 + 10∠350 + 10∠370)
( )
= #
( 10 + 9.848 − 1.7365 + 9.848 + 1.7365)
( )
= #
( 29.696) = 9.899
( )
= 9.899∠ 0
( )
= #
( + + )
( )
= #
( 10∠0 + 1∠240 ∗ 10∠230 + 1∠120 ∗ 10∠130)
( )
=
#
( 10∠0 + 10∠470 + 10∠250)
( )
= #
( 10 − 3.420 + 9.397 − 3.420 − 9.397)
( )
=
#
( 3.16) = 1.0533
( )
= 1.0533∠ 0
Para b:
( )
= ( )
=
( )
= 0.952∠180

ING. WIDMAR AGUILAR
Octubre -2022
( )
= ( )
( )
= 1∠240 ∗ 9.899∠ 0
( )
= 9.899 ∠240
( )
= ( )
( )
= 1∠120 ∗ 1.0533∠ 0
( )
= 1.0533 ∠ 120
Para c:
( )
= ( )
=
( )
= 0.952∠180
( )
= ( )
( )
= 1∠120 ∗ 9.899∠ 0
( )
= 9.899 ∠120
( )
= ( )
( )
= 1∠240 ∗ 1.0533∠ 0
( )
= 1.0533 ∠ 240

ING. WIDMAR AGUILAR
Octubre -2022
Se sabe que:
( )
=
#
[ + + ]
( )
=
#
[ 100 + 141.4∠ 225 + 100∠90]
( )
=
#
[ 100 − 100 − 100 + 100]
( )
=
#
[ 0] = 0
( )
= 0
( )
=
#
( + + )
( )
=
#
[ 100 + 1∠120 ∗ 141.4∠ 225 + 1∠240 ∗ 100∠90]
( )
=
#
[ 100 + 141.4∠ 345 + 100∠330]
( )
= #
[ 100 + 136.582 − 36.597 + 86.603 − 50]
( )
= #
[ 323.185 − 86.597] = 107.73 − 28.866
( )
= 111.53 ∠ − 15
( )
= #
( + + )
( )
=
#
( 100 + 1∠240 ∗ 141.4∠ 225 + 1∠120 ∗ 100∠90)
( )
=
#
( 100 + 141.4∠ 465 + 100∠21)
( )
=
1
3
( 100 − 36.597 + 136.582 − 86.603 − 50) =
1
3
[−23.2 + 86.582]
( )
=-7.733+j 28.886 = 29.9 ∠105

ING. WIDMAR AGUILAR
Octubre -2022
b)
( )
=
(())
*
√#
∠30
( )
=
.,# ∠- ,.#
√#
= 64.392∠15( )
( )
=
((/)
*
√#
∠ − 30
( )
=
0.0 ∠( 1.00-# )
√#
= 17.263∠75 ( )
Luego: = ( )
+ ( )
=64.392∠15 + 17.263∠75
= 62.198+j 16.666+4.4668+j16.675
= 66.666 + 33.341 = 74.54 ∠26.57

ING. WIDMAR AGUILAR
Octubre -2022
= 100 ; = 80.8 ∠ − 121.44 ; = 90∠130
Utilizando el diagrama:
( )
=
2())
*3
√#
∠ − 30
( )
=
2(/)
*3
√#
∠30
( )
=
#
[ 4 + 45 + 2
5 ]
( )
=
#
[ 100 + 1∠120 ∗ 80.8∠ − 121.44 + 1∠240 ∗ 90∠130]
( )
=
#
[ 100 + 80.8∠ − 1.44 + 90∠370]
( )
=
#
[ 100 + 80.78 − 2.031 + 88.63 + 15.63]
( )
= 89.80 + 4.53 = 89.914 ∠2.89
( )
=
1
3
[ 4 + 2
45 + 5 ]

ING. WIDMAR AGUILAR
Octubre -2022
( )
=
#
[ 100 + 1∠240 ∗ 80.8∠ − 121.44 + 1∠120 ∗ 90∠130]
( )
=
#
[ 100 + 80.8∠118.56 + 90∠250]
( )
=
#
[ 100 − 38.63 + 70.97 − 30.78 − 84.57]
( )
= 10.197 − 4.533 = 11.158 ∠ − 23.95
De; ( )
=
2())
*3
√#
∠ − 30
( )
=
60.0 1 ∠ .60-#
√#
=51.91 ∠ − 27.11 ( )
( )
= 46.21 − 23.66
( )
=
2(/)
*3
√#
∠30
( )
=
. ,6 ∠- #.0,.#
√#
=6.442 ∠6.05 ( )
( )
= 6.406 + 0.678
Se tiene que:
= ( )
+ ( )
= 46.21 − 23.66 + 6.406 + 0.678
= 52.616 − 22.982 = 57.42 ∠ − 23.6
=
2*7
8
=
,9.1 ∠- #.:
= 5.742 ∠ − 23.6 ( )

ING. WIDMAR AGUILAR
Octubre -2022
( )
= ( )
∠ − 27.11 + 240
( )
= 51.91 ∠212.89
( )
= ( )
∠6.05 + 120
( )
= 6.442 ∠126.05
= ( )
+ ( )
= 51.91 ∠212.89 + 6.442 ∠126.05
= −43.59 − 28.19 − 3.791 + 5.208= -47.381-j22.98
= 52.66∠ − 154.13
=
237
8
=
, .:: ∠- ,1. #
= 5.267 ∠ − 154.13 ( )
( )
= ( )
∠(−27.11 + 120)
( )
= 51.91 ∠92.9
( )
= ( )
∠(6.05 + 240)
( )
= 6.442 ∠246.05
= ( )
+ ( )
= 51.91 ∠92.9 + 6.442 ∠246.05
= −2.63 + 51.84 − 2.615 − 5.89= -5.24+j45.95
= 46.248∠96.5
=
2;7
8
=
1:. 16 ∠0:.,
= 4.625 ∠96.5 ( )
Además se tiene: ( )
= 0

ING. WIDMAR AGUILAR
Octubre -2022
De:
<#∅ = 3 (>) ( ) ∗
+ 3 ( ) ( ) ∗
+ 3 ( ) ( ) ∗
Como ( )
= 0
<#∅ = 3 ( ) ( ) ∗
+ 3 ( ) ( ) ∗
( )
=
( )
?
=
51.91
10
∠ − 27.11
( )
=
2(/)
*7
8
=
:.11
∠6.05 = 0.6442∠6.05
<#∅ = 3 ∗ 5.191 ∠27.11 ∗ 51.91 ∠ − 27.11 +3*6.442∠6.05 ∗
0.6442∠ − 6.05
<#∅ = 808.394∠0 + 12.45 ∠0
<#∅ = 808.394 + 12.45
<#∅ = @ = 820.84 (A)
Se puede verificar que la potencia es:

ING. WIDMAR AGUILAR
Octubre -2022
= 5.742 ∠ − 23.6 ; = 5.267 ∠ − 154.13 ∶ = 4.625 ∠96.5
P= 5.742 ∗ 10 + 5.267 ∗ 10 + 4.625 ∗ 10
P= 329.71+277.41+213.91 = 821.03 W
, D EFG HFIJ KE 5FG LKEMK5JF I JNKLL .
<#∅ = ( + ) ∗
+ ( + ) ∗
+ ( + ) ∗

ING. WIDMAR AGUILAR
Octubre -2022
= − +
=
2*-23
8∆
−
2;-2*
8∆
+
2*
8P
=
2*
8∆
+
2*
8P
-
23
8∆
−
2;
8∆
= (
8∆
+
8P
) +
2*
8∆
-
2*
8∆
−
23
8∆
−
2;
8∆
= (
#
8∆
+
8P
) −
8∆
( + + )
En forma de matriz:
= (
#
8∆
+
8P
) −
8∆
[ 1 1 1] $ %
Para :
= − +
=
23-2;
8∆
−
2*-23
8∆
+
23
8P
=
23
8∆
+
23
8P
−
2;
8∆
−
2*
8∆

ING. WIDMAR AGUILAR
Octubre -2022
= (
8∆
+
8P
) +
23
8∆
-
2;
8∆
−
2*
8∆
−
23
8∆
= (
#
8∆
+
8P
) −
8∆
( + + )
En forma de matriz:
= (
#
8∆
+
8P
) −
8∆
[ 1 1 1] $ %
Efectuando algo similar para Ic, se tiene:
= (
#
8∆
+
8P
) −
8∆
[ 1 1 1] $ %
En forma matricial:
$ % = (
#
8∆
+
8P
) $ % −
8∆
$
1 1 1
1 1 1
1 1 1
% $ %
b)
usando:
A!
(0)
(1)
(2)
" = (
#
8∆
+
8P
) !
(0)
(1)
(2)
" −
8∆
$
1 1 1
1 1 1
1 1 1
% !
(0)
(1)
(2)
"

ING. WIDMAR AGUILAR
Octubre -2022
A!
(0)
(1)
(2)
" = {(
#
8∆
+
8P
) −
8∆
$
1 1 1
1 1 1
1 1 1
%} !
(0)
(1)
(2)
"
!
(0)
(1)
(2)
" = -
{(
#
8∆
+
8P
) −
8∆
$
1 1 1
1 1 1
1 1 1
%} !
(0)
(1)
(2)
"
!
(0)
(1)
(2)
" = {(
#
8∆
+
8P
) -
−
8∆
-
$
1 1 1
1 1 1
1 1 1
% } !
(0)
(1)
(2)
"
-
=
#
$
1 1 1
1
1
% $
1 1 1
1
1
%
-
=
#
$
3 0 0
0 3 0
0 0 3
%= $
1 0 0
0 1 0
0 0 1
%
!
( )
( )
( )
" = {
⎣
⎢
⎢
⎢
⎢
⎢
⎡
3
?∆
+
1
?V
0 0
0
3
?∆
+
1
?V
0
0 0
3
?∆
+
1
?V ⎦
⎥
⎥
⎥
⎥
⎥
⎤
−
1
?∆
-
$
1 1 1
1 1 1
1 1 1
% } !
( )
( )
( )
"
#
$
1 1 1
1
1
% $
1 1 1
1 1 1
1 1 1
% =
#
$
3 3 3
1 + + 1 + + 1 + +
1 + + 1 + + 1 + +
%= $
1 1 1
0 0 0
0 0 0
%
1 + + = 0

ING. WIDMAR AGUILAR
Octubre -2022
$
1 1 1
0 0 0
0 0 0
% !
1 1 1
1 2
1 2
" = !
3 1 + + 1 + +
0 0 0
0 0 0
"
$
1 1 1
0 0 0
0 0 0
% !
1 1 1
1 2
1 2
" = $
3 0 0
0 0 0
0 0 0
%
Se tiene:
!
( )
( )
( )
" = {
⎣
⎢
⎢
⎢
⎡
#
8∆
+
8P
0 0
0
#
8∆
+
8P
0
0 0
#
8∆
+
8P ⎦
⎥
⎥
⎥
⎤
−
8∆
$
3 0 0
0 0 0
0 0 0
% } !
( )
( )
( )
"
!
( )
( )
( )
" = {
⎣
⎢
⎢
⎢
⎡
#
8∆
+
8P
0 0
0
#
8∆
+ 8P
0
0 0
#
8∆
+ 8P ⎦
⎥
⎥
⎥
⎤
− Z
3
?∆
0 0
0 0 0
0 0 0
[ } !
( )
( )
( )
"
!
( )
( )
( )
" =
⎣
⎢
⎢
⎢
⎢
⎢
⎡
1
?V
0 0
0
3
?∆
+
1
?V
0
0 0
3
?∆
+
1
?V⎦
⎥
⎥
⎥
⎥
⎥
⎤
!
( )
( )
( )
"
c) Los diagramas de secuencia son’:
secuencia cero:
Secuencia positiva:
Se tiene el paralelo de la delta y la Y.

ING. WIDMAR AGUILAR
Octubre -2022
?V =
8∆
#
Secuencia negativa:

ING. WIDMAR AGUILAR
Octubre -2022
= − +
=
2*-23
8∆
−
2;-2*
8∆
+
2*-27
8P
=
2*
8∆
+
2*
8P
-
23
8∆
−
2;
8∆
−
27
8P
= (
8∆
+
8P
) +
2*
8∆
-
2*
8∆
−
23
8∆
−
2;
8∆
−
27
8P
= (
#
8∆
+
8P
) −
8∆
( + + ) −
27
8P
= (
#
8∆
+
8P
) −
8∆
[ 1 1 1] $ % −
27
8P
Para :
= − +
=
23-2;
8∆
−
2*-23
8∆
+
23-27
8P
=
23
8∆
+
23
8P
−
2;
8∆
−
2*
8∆
−
27
8P
= (
8∆
+
8P
) +
23
8∆
-
2;
8∆
−
2*
8∆
−
23
8∆
−
27
8P
= (
#
8∆
+
8P
) −
8∆
( + + ) −
27
8P
En forma de matriz:
= (
#
8∆
+
8P
) −
8∆
[ 1 1 1] $ % −
27
8P
Efectuando algo similar para Ic, se tiene:

ING. WIDMAR AGUILAR
Octubre -2022
= (
#
8∆
+
8P
) −
8∆
[ 1 1 1] $ % −
27
8P
En forma matricial:
$ % = (
#
8∆
+
8P
) $ % −
8∆
$
1 1 1
1 1 1
1 1 1
% $ % −
8P
$
1
1
1
%
b) usando:
A!
( )
( )
( )
" = (
#
8∆
+
8P
) !
( )
( )
( )
" −
8∆
$
1 1 1
1 1 1
1 1 1
% !
( )
( )
( )
" −
8P
$
1
1
1
%
A!
(0)
(1)
(2)
" = {(
#
8∆
+
8P
) −
8∆
$
1 1 1
1 1 1
1 1 1
%} !
(0)
(1)
(2)
" −
1
?
]
1
1
1
^ G
!
(0)
(1)
(2)
" = -
{(
#
8∆
+
8P
) −
8∆
$
1 1 1
1 1 1
1 1 1
%} !
(0)
(1)
(2)
" −
- 1
?
]
1
1
1
^ G

ING. WIDMAR AGUILAR
Octubre -2022
!
(0)
(1)
(2)
" = {(
#
8∆
+
8P
) -
−
8∆
-
$
1 1 1
1 1 1
1 1 1
% } !
(0)
(1)
(2)
" −
- 1
?
]
1
1
1
^ G
-
=
#
$
1 1 1
1
1
% $
1 1 1
1 1 1
1 1 1
% =
#
$
3 3 3
1 + + 1 + + 1 + +
1 + + 1 + + 1 + +
%= $
1 1 1
0 0 0
0 0 0
%
1 + + = 0
$
1 1 1
0 0 0
0 0 0
% !
1 1 1
1 2
1 2
" = !
3 1 + + 1 + +
0 0 0
0 0 0
"
$
1 1 1
0 0 0
0 0 0
% !
1 1 1
1 2
1 2
" = $
3 0 0
0 0 0
0 0 0
%
Se tiene:
!
( )
( )
( )
" =
⎩
⎪
⎨
⎪
⎧
⎣
⎢
⎢
⎢
⎡
#
8∆
+
8P
0 0
0
#
8∆
+
8P
0
0 0
#
8∆
+
8P ⎦
⎥
⎥
⎥
⎤
−
8∆
$
3 0 0
0 0 0
0 0 0
%
⎭
⎪
⎬
⎪
⎫
!
( )
( )
( )
" −
8P
−1
$
1
1
1
%

ING. WIDMAR AGUILAR
Octubre -2022
#
$
1 1 1
1
1
% ]
1
1
1
^ =
1
3
]
1 + 1 + 1
1 + + 2
1 + 2
+
^ = ]
1
0
0
^
!
( )
( )
( )
" =
⎩
⎪
⎨
⎪
⎧
⎣
⎢
⎢
⎢
⎡
#
8∆
+
8P
0 0
0
#
8∆
+
8P
0
0 0
#
8∆
+
8P ⎦
⎥
⎥
⎥
⎤
− Z
3
?∆
0 0
0 0 0
0 0 0
[
⎭
⎪
⎬
⎪
⎫
!
( )
( )
( )
" −
8P
$
1
0
0
%
!
( )
( )
( )
" =
⎣
⎢
⎢
⎢
⎢
⎢
⎡
1
?V
0 0
0
3
?∆
+
1
?V
0
0 0
3
?∆
+
1
?V⎦
⎥
⎥
⎥
⎥
⎥
⎤
!
( )
( )
( )
" −
1
?V
$
1
0
0
%
Se sabe que: = 3 ( )
∗ ?f
= 3?f g
8P
( )
h −
27
8i
j1 +
8i
k =
#8i∗ 2(l)
*
8P
c) secuencia cero:
Secuencia positiva:

ING. WIDMAR AGUILAR
Octubre -2022
Secuencia negativa:

ING. WIDMAR AGUILAR
Octubre -2022
?> = ?m + ?n = ? + 2? + 3? − 6?
? = ?m − ?n = ? − ?
? = ?m − ?n = ? − ?
De:
$ % = $
200
200∠240
200∠120
%
A!
( )
( )
( )
" = $ %
!
( )
( )
( )
" = -
$
200
200∠240
200∠120
%
!
( )
( )
( )
" =
#
$
1 1 1
1
1
% ]
`1
1∠240
1∠120
^
!
( )
( )
( )
" =
#
$
1 + 1∠240 + 1∠120
1 + 1∠360 + 1∠360
1 + 1∠480 + 1∠240
%
!
( )
( )
( )
" =
#
$
0
1 + 1 + 1
1 − .5 + 0.867 − 0.5 − 0.867
% (KV)

ING. WIDMAR AGUILAR
Octubre -2022
!
( )
( )
( )
" = $
0
200
0
% KV
!
( )
( )
( )
" − Z
( )
p p
( )
p p
( )
p ′
[ = !
?>
( )
? ( )
? ( )
"
Además: = 0 ; = 0
( )
=
#
( + + ) =
#
( )
=
#
( + + ) =
#
( )
=
#
( + + ) =
#
!
( )
p p
( )
p p
( )
p ′
" = !
( )
( )
( )
" − !
?>
( )
? ( )
? ( )
"
!
( )
p p
( )
p p
( )
p ′
" = !
( )
( )
( )
" −
1
3
$
?>
?
?
%
Del ejemplo 11.5:
?> = 60 + 40 + 240 − 180 = 160
? = ?m − ?n = 60 − 20= j40
? = ?m − ?n = 60 − 20 = 40
!
( )
p p
( )
p p
( )
p ′
" = $
0
200
0
% 10#
−
1
3
$
160
40
40
%

ING. WIDMAR AGUILAR
Octubre -2022
= r
!
( )
p p
( )
p p
( )
p ′
" = $
0
200
0
% 10#
−
1
3
$
160 r
40 r
40 r
%
b) =
2*p7p
8
;
p p = ( )
p p + ( )
p p + ( )
p t
p p = 0 −
u : (v
#
+ 220000 −
u1 (v
#
−
u1 (v
#
p p = 200000 − 80 r
? ∗ w = 200000 − 240 w ;
r =
u1 .u6
= − 400 (A)
c)
!
( )
p p
( )
p p
( )
p ′
" = !
( )
( )
( )
" −
1
3
!
?>
(0)
? (1)
? (2)
"
!
( )
p p
( )
p p
( )
p ′
" = $
0
200000
0
% −
1
3
!
160 ∗ (− 400)
40 ∗ (− 400)
40 ∗ (− 400)
"
!
( )
p p
( )
p p
( )
p ′
" = $
−64000/3
600000 − 16000
−16000/3
%
!
( )
p p
( )
p p
( )
p ′
" =
⎣
⎢
⎢
⎢
⎡−
:1
#
,61
#
−
:
# ⎦
⎥
⎥
⎥
⎤
x1000 KV
p p = ( )
p p + ( )
p p + ( )
p t

ING. WIDMAR AGUILAR
Octubre -2022
p p = ( )
p p + ( )
p p + ( )
p t
p p = −
:1
#
+ 1∠240 ∗
,61
#
+ 1∠120 ∗ (−
:
#
)
p p = −
:1
#
− 97.33 − 168.59 + 2.67 − 4.62
p p = -115.993 -j173.21 = 208.461 ∠ − 123.81 {
p p = ( )
p p + ( )
p p + ( )
t
p p = ( )
p p + ( )
p p + ( )
p t
p p = −
:1
#
+ 1∠120 ∗
,61
#
+ 1∠240 ∗ (−
:
#
)
p p = −21.33 − 97.33 + 168.59 + 2.67 + 4.62
p p = -115.99 +j173.21 = 208.46 ∠123.81 {
d) Sin suso de componentes simétricas, se tiene:
Se sabe que:
?> = ?m + ?n = ? + 2? + 3? − 6?
? = ?m − ?n = ? − ?
? = ?m − ?n = ? − ?
ƒ„ = j60+j80-j60 = j80

ING. WIDMAR AGUILAR
Octubre -2022
ƒ… = j20+j80-j60 = j40
$
− t t
− t t
− t t
% = $
?m ?n ?n
?n ?m ?n
?n ?n ?m
% $ %
$
− t t
− t t
− t t
% = $
80 40 40
40 80 40
40 40 80
% $
− 400
0
0
%
$
− t t
− t t
− t t
% = $
− 32000
− 16000
− 16000
%
!
†‡‡p
t
t
" = $
− t t
− t t
− t t
% = $
32000
16000
16000
%
− p p = 32000
p p = 200000 − 32000 = −168000 ( )
p p= − p = 200000∠240 − 16000
p p= −100000 − 173205.08 − 16000 = −116000 − 173205.08
p p = 208461.03∠ − 123.81 ( )
p p= − t = 200000∠120 − 16000
p p= −100000 + 173205.08 − 16000 = −116000 + 173205.08
p p = 208461.03∠123.81 ( )

ING. WIDMAR AGUILAR
Octubre -2022
!
( )
( )
( )
" − Z
( )
p p
( )
p p
( )
p ′
[ = !
?>
( )
? ( )
? ( )
"
DE LA GRÁFICA: = −
= 0
( )
=
#
( + + ) =
#
( − + ) = 0
( )
=
#
( + + ) =
#
( − ) =
(*
#
(1 − )
( )
=
3
(1 + 0.5 − 0.87) =
3
(1.5 − 0.866)
( )
=
(*
#
∗ 1.73234∠ − 30
( )
=
(*
√#
∠ − 30
( )
=
1
3
( + + )
( )
=
1
3
( + + ) =
1
3
( − ) =
3
(1 − )
( )
=
3
(1 + 0.5 + 0.87) =
3
(1.5 + 0.866)
( )
=
(*
#
∗ 1.73234∠ + 30
( )
=
(*
√#
∠30

ING. WIDMAR AGUILAR
Octubre -2022
!
( )
( )
( )
" = -
$
200
200∠240
200∠120
%
!
( )
( )
( )
" =
#
$
1 1 1
1
1
% ]
`1
1∠240
1∠120
^
!
( )
( )
( )
" =
#
$
1 + 1∠240 + 1∠120
1 + 1∠360 + 1∠360
1 + 1∠480 + 1∠240
%
!
( )
( )
( )
" =
#
$
0
1 + 1 + 1
1 − .5 + 0.867 − 0.5 − 0.867
% (KV)
!
( )
( )
( )
" = $
0
200
0
% KV
!
( )
p p
( )
p p
( )
p ′
" = !
( )
( )
( )
" − !
?>
(0)
? (1)
? (2)
"
!
( )
p p
( )
p p
( )
p ′
" = $
0
200000
0
% −
⎣
⎢
⎢
⎢
⎡
0
40 ∗
√3
∠ − 30
40 ∗
√3
∠30
⎦
⎥
⎥
⎥
⎤
!
( )
p p
( )
p p
( )
p ′
" =
⎣
⎢
⎢
⎢
⎡
0
200000 − 40 ∗
√3
∠60
−40 ∗
√3
∠120
⎦
⎥
⎥
⎥
⎤
b)

ING. WIDMAR AGUILAR
Octubre -2022
Se tiene que: t t = t t − t t
t t = ( )
p p + ( )
p p+ ( )
p p
t t = ( )
p p + ( )
p p+ ( )
p p − ( ( )
p p + ( )
p p + ( )
p p)
t t = ( )
p p+ ( )
p p − ( ( )
p p + ( )
p p)
t t = ( )
p p(1 − )+ ( )
p p(1-a)
t t = ( )
p p(1 + 0.5 + 0.867) + ( )
p p(1 + 0.5 − 0.867)
t t = ( )
p p(1.5 + 0.867) + ( )
p p(1.5 − 0.867)
t t = ( )
p p√3∠30 + ( )
p p√3∠ − 30
t t = ‹ 200000 − 40 ∗
(*
√#
∠60Œ √3∠30 − 40 ∗
(*
√#
∠120 ∗ √3∠ − 30
t t = √3 ∗ 200000∠30 − 40 − 40 ∗
t t = √3 ∗ 200000∠30 − 80
†‡t•t = Ž••••• ∗ √•∠•• − ‘’• ∗ “‡
∗ 420 = Ž••••• ∗ √•∠•• − ‘’• ∗ “‡
∗ 500 = Ž••••• ∗ √•∠••
=
2*3
8
=
∗√#∠#
u,
= 692.82∠ − 60 (A)
c)
De: p p = ( )
p p + ( )
p p + ( )
p p
p p = ( )
p p + ( )
p p

ING. WIDMAR AGUILAR
Octubre -2022
p p = 1∠120 ∗ ‹200000 − 40 ∗
(*
√#
∠60Œ + 1∠240 ∗ (−40 ∗
(*
√#
∠120)
p p = 200000∠120 − 40 ∗
(*
√#
∠180 − 40
(*
√#
∠360
p p = 200000∠120 + 40 ∗
(*
√#
− 40
(*
√#
p p = 200000∠120 (V)
d)
p = − p p ; = 0
− p p = ( 20 − 30)( + )+(j30-j80)
− p p = ( 20 − 30)( − )+(j30-j80)
= + + = − + 0 = 0
− p p = 0
p p = = 200∠ 120 ( { )

ING. WIDMAR AGUILAR
Octubre -2022
Se sabe que:
!
( )
( )
( )
" = #
$
1 1 1
1
1
% $ %
( )
=
#
[ 0 + 3.75∠150 + 3.75∠30]
( )
=
#
[ −3.248 + 1.875 + 3.248 + 1.875]
( )
= #
[ 3.75] = 1.25
( )
= 1.25 ∠90 ( )
( )
=
1
3
[ 0 + 3.75∠270 + 3.75∠270] =
1
3
(− 3.75 − 3.75)
( )
= − 2.5
( )
=
1
3
[ 0 + 3.75∠390 + 3.75∠150] =
1
3
(3.25 + 1.875 − 3.25 + 1.875)
( )
= 1.25
Se obtiene voltajes, así:

ING. WIDMAR AGUILAR
Octubre -2022
( )
= − ( )
(?> + 3? ) = − 1.25( 3 ∗ 0.09 + 0.09)
( )
= 1.25( 0.36) = 0.45 M™ ( )
( )
= š − ( )
(? ) = 1 + 2.5( 0.22)
( )
= 1 − 0.55 = 0.45 M™
( )
= − ( )
(? ) = − 1.25( 0.36)
( )
= 0.45 M™
Como:
$ % = !
( )
( )
( )
"= $
1 1 1
1
1
% !
( )
( )
( )
"
$ % = $
1 1 1
1
1
% $
0.45
0.45
0.45
%
$ % = $
0.45 + 0.45 + 0.45
0.45 + 0.45 + 0.45
0.45 + 0.45 + 0.45
%=0.45$
3
1 + +
1 + +
%
$ % =0.45$
3
1 + +
1 + +
% = 0.45 $
3
0
0
%= $
1.35
0
0
% (pu)
b) › =?

ING. WIDMAR AGUILAR
Octubre -2022
› = − ( )
∗ (3? )
› = = − 1.25 ∗ ( 3 ∗ 0.09) = 0.3375 M™
c) •™K: $ % =0.45$
3
1 + +
1 + +
% = 0.45 $
3
0
0
%= $
1.35
0
0
% (pu)
y en donde se observa que: = 0 ; = 0 , EK M™žK 5FG5I™NL •™K I Ÿ II KE:
→ … … . . Ÿ II INGK − INGK JNKLL
Se conoce que:
!
( )
( )
( )
" =
#
$
1 1 1
1
1
% $ %

ING. WIDMAR AGUILAR
Octubre -2022
( )
=
#
[ 0 − 2.986 + 2.986]
( )
= 0
( )
=
1
3
[ 0 − 2.986∠120 + 2.986∠240]
=
1
3
(1.493 − 2.5859529 − 1.493 − 2.585952)
( )
= − 1.7240
( )
=
1
3
[ 0 − 2.986∠240 + 2.986∠120] =
1
3
(1.493 + 2.586 − 1.493 + 2.586)
( )
= 1.724
Se obtiene voltajes, así:
( )
= − ( )
(?> + 3? ) = −0( 3 ∗ 0.09 + 0.09)
( )
= 0
( )
= š − ( )
(? ) = 1 + 1.724( 0.22)
( )
= 1 − 0.3793 = 0.621 M™
( )
= − ( )
(? ) = − 1.724( 0.36)
( )
= 0.621 M™
Como:

ING. WIDMAR AGUILAR
Octubre -2022
$ % = !
( )
( )
( )
"= $
1 1 1
1
1
% !
( )
( )
( )
"
$ % = $
1 1 1
1
1
% $
0
0.621
0.621
%
$ % = $
0.621 + 0.621
0.621 + 0.621
0.621 + 0.621
%=0.621$
1 + 1
+
+
%
+ = 1∠ 240 + 1 ∠120 = −0.5 − 0.8667 − 0.5 + 0.8667 = −1
$ % =0.621$
2
−1
−1
% = $
1.242
−0.621
−0.621
% (pu)
b) › =?
› = − ( )
∗ (3? )
› = = 0 ∗ ( 3 ∗ 0.09) = 0 M™
c) •™K: $ % =0.61$
2
−1
−1
% = $
1.242
−0.621
−0.621
% (pu)
y en donde se observa que: = , EK M™žK 5FG5I™NL •™K I Ÿ II KE:
→ … … . . Ÿ II INGK − INGK
=

ING. WIDMAR AGUILAR
Octubre -2022
BASES: E = 10 ¢
Las corrientes bases en los lados del transformador de potencia es:
( #. £2) =
√#∗ #.
= 437.39
(::£2) =
√#∗::
= 87.48
Del problema 11.4: en el lado Y:
( )
=
1
3
[ + + ]
( )
¤ = 0
( )
=
#
( + + )
( )
= 111.53 ∠ − 15
( )
= #
( + + )
( )
= 29.9 ∠105
= ( )
+ ( )
= 107.73 − 28.87 − 7.73 + 28.82
“‡ = ¥•• ¦
§ = ( )
+ ( )
= 111.53 ∠225 + 29.9 ∠225
= −78.86 − 78.86 − 21.14 − 21.14
= −100 −j100 = ¥¨¥. ¨Ž ∠ − ¥•©
= ( )
+ ( )
ª = 111.53 ∠105 + 29.9 ∠345
ª = −28.866 + 107.73 + 28.88 − 7.74
ª = 100= ¥••∠«•

ING. WIDMAR AGUILAR
Octubre -2022
En por unidad se tiene: lado de 66 KV
( )
= 0 M™
( )
=
.,# ∠- ,
69.16
= 1.275 ∠ − 15 M™
( )
=
0.0 ∠ ,
69.16
= 0.3418 ∠105 M™
En el lado delta del transformador:
( )
= ( )
¤∠ − 30
( )
¤ =
.,# ∠- ,-#
69.16
= 1.275 ∠ − 45 M™
( )
¤ =
0.0 ∠ ,.#
69.16
= 0.3418 ∠135 M™
¤ = ( )
+
(1)
+
(2)
= 0 + 1.275 ∠ − 45 + 0.342 ∠135
= 0.902 − 0.902 − 0.242 + 0.242
¤ = 0.66 − 0.66 = 0.934∠ − 45 M™
¤ = 0.934∡ − 45
§ = ( )
+ 2 (1)
+
(2)
= 0 + 1.275 ∠195 + 0.342 ∠255
§ = −1.232 − 0.33 − 0.0885 − 0.33 = −1.3205 − 0.66
§ = 1.476∠ − 26.56 pu

ING. WIDMAR AGUILAR
Octubre -2022
ª = ( )
+
(1)
+ 2 (2)
= 0 + 1.275 ∠75 + 0.342 ∠375
ª = 0.33 + 1.232 + 0.33 + 0.0885 = 0.66 + 1.3205
ª = 1.476∠63.44 pu
¤ = 437.39 ∗ 0.934 = 408.52
§ = 437.39 ∗ 1.476 = 645.59
ª = 437.39 ∗ 0.934 = 645.59
b)
L› =
#.
--
√®
= 0.346
()
(/
=
¯°
= 2.89
¤§ = 2.89 ∗
= 2.89 ∗ 100 = 289∠0
= 289
ª¤ = 2.89 ∗
ª¤ = 2.89 ∗ 100∠90 = 289
De: ¤ = ¤§− ª¤
¤ = 289 − 289 = 408.7 ∠ − 45 ( )
¤ = 408.7 ( )

ING. WIDMAR AGUILAR
Octubre -2022
Y / ∆ − − − −: ( )
¤ − − − − − žKI GJ KG 30 ±L žFE ( )
a)

ING. WIDMAR AGUILAR
Octubre -2022
b)
Y / ∆ − − − −: ( )
− − − − − Ÿ™KL žžK Ÿ EK KG 90 ±L žFE 5FG ( )
¤

ING. WIDMAR AGUILAR
Octubre -2022
p = ?
p = ?
p = ?
$
p
p
p
% = $
? 0 0
0 ? 0
0 0 ?
% $ %
$
1 1 1
1
1
% !
p
( ) p
( ) p
" = $
? 0 0
0 ? 0
0 0 ?
% $
1 1 1
1
1
% ! ( )
( )
"

ING. WIDMAR AGUILAR
Octubre -2022
!
( ) p
( ) p
( ) p
" =
#
$
1 1 1
1
1
% $
? 0 0
0 ? 0
0 0 ?
% !
+ ( )
+ ( )
+ ( )
+ ( )
+ ( )
+ ( )
"
!
p
( ) p
( ) p
" =
#
!
? ? ?
? ? ?
? ? ?
" !
( + ( )
+ ( )
)
( + ( )
+ ( )
)
( + ( )
+ ( )
)
"
Como el neutro no esta aterrizado se tiene que:
( )
= ( )
= ( )
= 0
!
p
( ) p
( ) p
" =
#
!
? ? ?
? ? ?
? ? ?
" !
( ( )
+ ( )
)
( ( )
+ ( )
)
( ( )
+ ( )
)
"
!
p
( ) p
( ) p
" =
#
Z
? ² ( )
+ ( )
³ + ? ( ( )
+ ( )
) + ? ( ( )
+ ( )
)
? ² ( )
+ ( )
³ + ? ² ( )
+ ( )
³ + ? ( ( )
+ ( )
)
? ² ( )
+ ( )
³ + ? ² ( )
+ ( )
³ + ? ( ( )
+ ( )
[
!
p
( ) p
( ) p
" =
⎣
⎢
⎢
⎢
⎡ #
( )
( ? + ? + ? ) + #
( )
(? + ? + ? )
#
( )
( ? + #
? + #
? ) + #
( )
(? + ? + 1
? )
#
( )
( ? + 1
? + ? ) +
#
( )
(? + #
? + #
? ⎦
⎥
⎥
⎥
⎤
!
p
( ) p
( ) p
" =
⎣
⎢
⎢
⎢
⎡ #
( )
( ? + ? + ? ) + #
( )
(? + ? + ? )
#
( )
( ? + ? + ? ) + #
( )
(? + ? + ? )
#
( )
( ? + ? + ? ) +
#
( )
(? + ? + ? ⎦
⎥
⎥
⎥
⎤
1
= 1∠480 = −05 + 0.866
a= 1∠120 = −0.5 + 0.866
= 1∠240 = −05 − 0.866
!
? + ? + ?
? + ? + ?
? + ? + ?
" = $
60
10 − 10 + 17.32 − 15 − 25.98
10 − 10 − 17.32 − 15 + 25.98
% =$
60
−15 − 8.66
−15 + 8.66
%

ING. WIDMAR AGUILAR
Octubre -2022
!
? + ? + ?
? + ? + ?
? + ? + ?
" = $
60
17.32∠ − 150
17.32∠150
%
!
p
( ) p
( ) p
" =
⎣
⎢
⎢
⎢
⎢
⎡
1
3
( )
( 17.32∠150) +
1
3
( )
(17.32∠ − 150)
1
3
( )
( 60) +
1
3
( )
(17.32∠150)
1
3
( )
( 17.32∠ − 150) +
1
3
( )
(60) ⎦
⎥
⎥
⎥
⎥
⎤
Como ( )
= 0
]
(1)
′
(2)
′
^ = g
20 5.773∠150
5.773∠ − 150 20
h ]
( )
( )
^
]
( )
( )
^ = g
20 5.773∠150
5.773∠ − 150 20
h
-
]
(1)
′
(2)
′
^
Se tiene que:
( )
=
1
3
[ + + ]
( )
= #
[57.74 + 57.74∠360 + 57.74∠360]
( )
= #
[57.74 + 57.74 + 57.74] = 57.74
]
( )
( )
^ = g
20 5.773∠150
5.773∠ − 150 20
h
-
g
57.74
0
h
g
20 5.773∠150
5.773∠ − 150 20
h
-
=
1 -##.##
g
20 − 5.773∠150
−5.773∠ − 150 20
h

ING. WIDMAR AGUILAR
Octubre -2022
g
20 5.773∠150
5.773∠ − 150 20
h
-
=
#::.:9
g
20 − 5.773∠150
−5.773∠ − 150 20
h
g
20 5.773∠150
5.773∠ − 150 20
h
-
= g
0.0545 − 0.0157∠150
−0.0157∠ − 150 0.0545
h
]
( )
( )
^ = g 0.0545 − 0.0157∠150
−0.0157∠ − 150 0.0545
h g
57.74
0
h
]
( )
( )
^ = g 3.146
0.906∠30
h (A)
La corriente en la fase A:
= (0)
+ (1)
+ (2)
= 3.146 + 0.906∠30 = 3.146 + 0.7846 + 0.453
“‡ = •. «•¥ + ‘•. ¨©•= 3.957∠ ´. ©µ ¶·
De: !
p
( ) p
( ) p
" =
#
!
? ? ?
? ? ?
? ? ?
" !
( ( )
+ ( )
)
( ( )
+ ( )
)
( ( )
+ ( )
)
"
p = 1/3[? ² ( )
+ ( )
³ + ? ( ( )
+ ( )
) + ? ( ( )
+ ( )
)]
( )
+ ( )
= 3.957∠ 6.57
( )
+ ( )
= 3.146∠240 + 0.906∠150
( )
+ ( )
= -1573-j2.7245-0.7846+j0.453=--2.3576-j2.272
( )
+ ( )
= 3.275 ∠ − 136.06
( )
+ ( )
= 3.146∠120 + 0.906∠270 = −1.573 + 2.7245 − 0.906
( )
+ ( )
= −1.573 + 1.8185 = 2.404∠130.85
p =
#
[10 ∗ .957∠ 6.57 + 20 ∗ 3.275 ∠ − 136.06+30*2.404∠130.85]
p =
#
[9.507 + 1.095 − 47.164 − 45.45 − 47.17 + 54.55]

ING. WIDMAR AGUILAR
Octubre -2022
p =
#
(−84.83 + 11) = 28.28 ∠179.3 ( )
= p + p
]
(1)
′
(2)
′
^ = g
20 5.773∠150
5.773∠ − 150 20
h ]
( )
( )
^
p = 20 ∗ 3.146 + 5.773∠150 ∗ .906∠30
p = 62.92-5.23 = 57.69 V
= p + p = −28.28 + 3.67 + 57.69
= 29.42 + 3.67 = 29.65 ∠7.11 ( )
Base: <§ = 50 ¢ ; { § = 13.8 {
Los valores en pu en la base dada son:

ING. WIDMAR AGUILAR
Octubre -2022
G1: ¸1 = 0.2 ∗ ‹
,
Œ = 0.5 M™
¸ = 0.05 ∗
,
= 0.125 M™
G2: ¸2 = 0.2 ∗ ‹
,
#
Œ = 0.333 M™
¸ = 0.05 ∗
,
#
= 0.0833 M™
G3: ¸3 = 0.2 ∗ (20/22) ‹
,
#
Œ = 0.275 M™
¸ = 0.05 ∗ (20/22) ‹
,
#
Œ = 0.0689 M™
T1: ¹› = 0.1 ∗ ‹
,
,
Œ = 0.2 M™
T2: ¹› = 0.1 ∗ ‹
,
#
Œ = 0.167 M™
T3: ¹›# = 0.1 ∗ ‹
,
#,
Œ = 0.143 M™
L1 y L2: ?§ =
£2/
º
=
/
,
= 968 Ω
?¼ =
6
0:6
= 0.0826 M™
?r =
0:6
= 0.103 M™
? ¼ =
0:6
= 0.217 M™
? ¼ =
,
0:6
= 0.258 M™
El diagrama de secuencia cero es:

ING. WIDMAR AGUILAR
Octubre -2022
El diagrama de secuencia negativa es:

ING. WIDMAR AGUILAR
Octubre -2022
G1:
¸1 = 0.2 ∗ (
6
) ‹
,
Œ = 0.405 M™
¸f = 0.05 ∗ (
6
) ‹
,
Œ = 0.10125 M™
¸ = 0.08 ∗ (
6
) ‹
,
Œ = 0.162 M™
G2:
¸2 = 0.2 ∗ (
6
) ‹
,
Œ = 0.405 M™
¸f = 0.05 ∗ (
6
) ‹
,
Œ = 0.10125 M™
¸ = 0.08 ∗ (
6
) ‹
,
Œ = 0.162 M™
M3:
¸3 = 0.2 ∗ ‹
,
#
Œ = 0.333 M™
¸f = 0.05 ∗ ‹
,
#
Œ = 0.0833 M™
¸ = 0.08 ∗ ‹
,
#
Œ = 0.133 M™

ING. WIDMAR AGUILAR
Octubre -2022
T1:
¹› = 0.1 ∗ ‹
,
Œ = 0.25 M™
T2:
¹› = 0.1 ∗ ‹
,
Œ = 0.25 M™
T3:
¹›# = 0.1 ∗ ‹
,
Œ = 0.25 M™
T4:
¹›1 = 0.1 ∗ ‹
,
,
Œ = 0.333 M™
T5:
¹›, = 0.1 ∗ ‹
,
,
Œ = 0.333 M™
T6:
¹›: = 0.1 ∗ ‹
,
Œ = 0.25 M™
L1 y L2, L3: ?§ =
£2/
º
=
#6/
,
= 381 Ω
?¼ =
1
#6
= 0.105 M™
?r = ?r# = #6
= 0.053 M™
?¼ ( ) =
120
381
= 0.315 M™
?r = ?r# =
:
#6
= 0.16 M™

ING. WIDMAR AGUILAR
Octubre -2022
b) Secuencia negativa

ING. WIDMAR AGUILAR
Octubre -2022
Secuencia cero:
a) T3 sin aterrizar:
El equivalente de thevenin visto desde C es:
?›½ =
(u .1 9)(u .1 ,)
u( .1 9. .1 ,)
= 0.2105
b)
con el neutro aterrizado en T3:

ING. WIDMAR AGUILAR
Octubre -2022
8¾¿
=
8)
+
8/
+
8®
=
8)8/.8)8®.8/8®
8)8/8®
?ÀÁ =
8)8/8®
8)8/.8)8®.8/8®
?›½ =
( 0.417)( 0.425)( 0.4186)
( 0.417)( 0.425) + ( 0.417)( 0.4186) + ( 0.4186)( 0.425)
?›½ =
− 0.074186
−0.17723 − 0.1746 − 0.17791
=
0.074186
0.53093
?›½ = 0.1394

ING. WIDMAR AGUILAR
Octubre -2022
Ejemplo 1)
Se puede colocar la carga en conexión estrella, para ello:
?V =
8∆
#
= 10 ∠ 40
$ % = $
277
260∠ − 120
295∠115
%
( )
=
#
[ + + ] =
#
[ 277 − 130 − 225.17 − 124.67 + 267.36]

ING. WIDMAR AGUILAR
Octubre -2022
( )
=
#
( 22.33 + 42.19) = 7.443 + 14.063
( )
= 15.91 ∠62.11
( )
=
1
3
[ + + ] =
1
3
[ 277 + 260∠ 0 + 295∠ 355]
( )
= #
( 277 + 260 + 293.88 − 25.72) = 276.933 − 8.573
( )
= 277.066 ∠ − 1.77
( )
=
1
3
[ + + ] =
1
3
[ 277 + 260∠ 120 + 295∠ 235]
( )
=
#
( 277 − 130 + 225.167 − 169.205 − 241.65) = −7.402 − 5.494
( )
= 9.218 ∠ − 143.42
Como el neutro de la carga está aislado, se tiene:
( )
= 0 = ( )
( )
= ( )
=
2())
*7
∠6,. ∠1
=
99. :: ∠- .99
. 69 .u .00: .9.::.u:.1#
( )
= ( )
=
99. :: ∠- .99
.9# ∠1#.90
= 25.82 ∠ − 45.56 ( )
( )
= ( )
=
2(/)
*7
∠6,. ∠1
=
0. 6 ∠- 1#.1
. 69 .u .00: .9.::.u:.1#
( )
= ( )
=
0. 6 ∠- 1#.1
.9# ∠1#.90
= 0.859 ∠ − 187.21 ( )
Las corrientes en las líneas son:
= (0)
+ (1)
+ ( )
= 0 + 25.82 ∠ − 45.56 + 0.859 ∠ − 187.21
= 18.078 − 18.435 − 0.8522 + 0.108 = 17.23 − 18.327
= 25.155∠ − 46.77
= ( )
+ ( )
+ ( )

ING. WIDMAR AGUILAR
Octubre -2022
= 0+ 25.82 ∠194.44 + 0.859 ∠ − 67.21
= −25 − 6.44 + 0.333 − 0.792 = −24.667 − 7.232
= 25.705∠ − 163.66
= ( )
+ ( )
+ ( )
= 0+ 25.82 ∠194.44 + 0.859 ∠ − 67.21
= −25 − 6.44 + 0.333 − 0.792 = −24.667 − 7.232
= 25.705∠ − 163.66
= ( )
+ ( )
+ ( )
= 0+ 25.82 ∠74.44 + 0.859 ∠52.79
= 6.944 + 24.869 + .519 + 0.684 = 7.463 + 25.553
= 26.598∠73.71
Ejemplo2)
Ejemplo 8.6 es el ejercicio anterior:

ING. WIDMAR AGUILAR
Octubre -2022
Bases: <§ = 0.075 ¢ ;
?§ =
. 6/
. 9,
= 0.576 Ω
?¼ =
∠6,
.,9:
= 1.736 ∠85 M™
? ¯f (Àm›¯À¼¼ ) =
∠1
.,9:
= 17.361 ∠40
Del ejemplo anterior:
( )
=
,.0
ÂÃl
√®
∠62.11 = 0.0574∠62.11 M™
( )
=
99. :: ∠- .99
ÂÃl
√®
∠ − 1.77= 1 ∠ − 1.77
( )
=
0. 6 ∠- 1#.1
ÂÃl
√®
=0.033∠ − 143.42 pu
Se tiene: las corrientes de secuencia de la fase a: ( )
= 0

ING. WIDMAR AGUILAR
Octubre -2022
( )
=
∠- .99
u . . .9#: ∠6,.
Ä/.lÃ®
®
∠1
(la carga en Y)
( )
=
∠- .99
u . . .9#: ∠6,. 9.#: ∠1
( )
=
∠- .99
u . . . , #.u .9#. #.#.u . :
=
∠- .99
6.9∠11
( )
= 0.0535∠ − 45.77 pu
( )
=
. ##∠- 1#.1
u . . .9#: ∠6,. 9.#: ∠1
(la carga en Y)
( )
=
. ##∠- 1#.1
u . . .9#: ∠6,. 9.#: ∠1
( )
=
. ##∠- 1#.1
u . . . , #.u .9#. #.#.u . :
=
. ##∠- 1#.1
6.9∠11
( )
= 0.001765∠ − 187.42
La corriente en a fuente de la fase a, es:
= ( )
+ ( )
+ ( )
= 0 + 0.0535∠ − 45.77 + 0.001765∠ − 187.42
= 0.0374 − 0.0383 − 0.00175 + 0.00023 = 0.0357 − 0.038
= 0.0521∠ − 46.78 M™
La corriente base en el lado de la fuente es:
§ =
9,
√#∗16
= 90.211
= 0.0521 ∗ 90.211∠ − 46.78 ( )
= 4.7 ∠ − 46.78 ( )

ING. WIDMAR AGUILAR
Octubre -2022
Ejemplo 11.2)
Tomando como base a 2300 KV
=
61
#
= 0.8 M™ =
9:
#
= 1.2 pu =
#
#
= 1 M™
Si se considera que el ángulo de Vac es de 180 grados, se determina los otros ángulos
de los fasores
1.2 = 0.8 + 1 − 2(1)(0.8)5FEÅ

ING. WIDMAR AGUILAR
Octubre -2022
5FEÅ = 0.125 ; Å = 82.82
0.8 = 1.2 + 1 − 2(1)(1.2)5FEÆ
5FEÆ = 0.75 ; Æ = 41.4
Se tiene:
= 0.8∠ 82.82 ; = 1.2 ∠ − 41.4 ; = 1 ∠180
Las componentes simétricas se obtienen de:
( )
=
#
[ 4 + 45 + 5 ]
( )
=
#
[ 0.8∠ 82.82 + 1.2 ∠ − 41.4 + 1 ∠180]
( )
=
#
[0.1 + 0.794 + 0.9 − 0.794 − 1]
( )
=
#
(0)=0
( )
=
1
3
[ 4 + 45 + 2
5 ]
( )
=
#
[ 0.8∠ 82.82 + 1.2 ∠78.6 + 1 ∠420]
( )
=
#
[0.1 + 0.794 + 0.237 + 1.176 + 0.5 + 0.866]
( )
=
#
(0.837+j2.836)= 0.985 ∠73.56
( )
=
1
3
[ 4 + 2
45 + 5 ]
( )
=
#
[ 0.8∠ 82.82 + 1.2 ∠198.6 + 1 ∠300]
( )
=
#
[0.1 + 0.794 − 1.137 − 0.383 + 0.5 − 0.866]
( )
=
#
(-0.537-j0.455)= 0.233 ∠ − 139.73

ING. WIDMAR AGUILAR
Octubre -2022
De: El voltaje de secuencia cero de la fase ‘a’ es cero ( neutro aislado)
( )
= ( )
∠ − 30
( )
= 0.985 ∠73.56 − 30
( )
= 0.985 ∠43.56
( )
=0.233 ∠ − 139.73+30
( )
= 0.233 ∠ − 109.73
( )
=
.06,
∠43.56
( )
= 0.985 ∠43.56 pu
( )
=
. ##
∠ − 109.73
( )
= 0.233 ∠ − 109.73 pu
= ( )
+ ( )
+ ( )
= ( )
+ ( )
Ejemplo 11.4)

ING. WIDMAR AGUILAR
Octubre -2022
Se tiene:
$ %= $
21 0 0
0 21 0
0 0 21
% $ %
!
( )
( )
( )
"= $
21 0 0
0 21 0
0 0 21
% !
( )
( )
( )
"
!
( )
( )
( )
"= $
21 0 0
0 21 0
0 0 21
% -
!
( )
( )
( )
"
-
=
#
$
1 1 1
1
1
% $
1 1 1
1
1
%
-
=
#
$
3 1 + + 1 + +
1 + + 1 + #
+ #
1 + +
1 + + 1 + 1
+ 1 + #
+ #
%
-
=
#
$
3 0 0
0 1 + 1 + 1 1 + +
1 + + 1 + + 1 + 1 + 1
%
-
=
#
$
3 0 0
0 3 0
0 0 3
%= $
1 0 0
0 1 0
0 0 1
%

ING. WIDMAR AGUILAR
Octubre -2022
!
( )
( )
( )
"= $
21 0 0
0 21 0
0 0 21
% !
( )
( )
( )
"
Se conoce que ( )
= 0
→ ( )
= 0
]
( )
( )
^= Ç
21 0
0 21
È = ]
( )
( )
^
?É =
#
= 7 Ω
Se tiene que:
Los circuitos de secuencia son:
b) Con las mutuas:
$ %= $
21 6 6
6 21 6
6 6 21
% $ %

ING. WIDMAR AGUILAR
Octubre -2022
!
( )
( )
( )
"= $
21 6 6
6 21 6
6 6 21
% !
( )
( )
( )
"
Utilizando las propiedades de matrices:
$
21 6 6
6 21 6
6 6 21
% = 15 $
1 0 0
0 1 0
0 0 1
% + 6 $
1 1 1
1 1 1
1 1 1
%
!
( )
( )
( )
"= { 15 $
1 0 0
0 1 0
0 0 1
% + 6 $
1 1 1
1 1 1
1 1 1
%} !
( )
( )
( )
"
!
( )
( )
( )
"= { 15 -
$
1 0 0
0 1 0
0 0 1
% + 6 -
$
1 1 1
1 1 1
1 1 1
%} !
( )
( )
( )
"
-
=
#
$
1 1 1
1
1
% $
1 1 1
1
1
%
-
=
#
$
3 1 + + 1 + +
1 + + 1 + #
+ #
1 + +
1 + + 1 + 1
+ 1 + #
+ #
%
-
=
#
$
3 0 0
0 1 + 1 + 1 1 + +
1 + + 1 + + 1 + 1 + 1
%
-
=
#
$
3 0 0
0 3 0
0 0 3
%= $
1 0 0
0 1 0
0 0 1
%

ING. WIDMAR AGUILAR
Octubre -2022
!
( )
( )
( )
"= { 15 $
1 0 0
0 1 0
0 0 1
% $
1 0 0
0 1 0
0 0 1
% +
6
#
$
1 1 1
1
1
% $
1 1 1
1 1 1
1 1 1
% } !
( )
( )
( )
"
Usando wólfram para las multiplicaciones, se tiene:
!
( )
( )
( )
"= { 15 $
1 0 0
0 1 0
0 0 1
% $
1 0 0
0 1 0
0 0 1
% +
6
#
$
3 3 3
0 0 0
0 0 0
% } !
( )
( )
( )
"
!
( )
( )
( )
"= { 15 $
1 0 0
0 1 0
0 0 1
% + 6 $
1 1 1
0 0 0
0 0 0
% } !
( )
( )
( )
"

ING. WIDMAR AGUILAR
Octubre -2022
!
( )
( )
( )
"= { 15 $
1 0 0
0 1 0
0 0 1
% + 6 $
3 0 0
0 0 0
0 0 0
%} !
( )
( )
( )
"
!
( )
( )
( )
"= $
33 0 0
0 15 0
0 0 15
% !
( )
( )
( )
"
Las impedancias de secuencia negativa y positiva son:
? (V) =
,
#
= 5
? (V) =
15
3
= 5

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