2n cos e + 3 2n cos e cos e 3 2 sin 0 3 sin 0 3 2n sin 0+ 3 K, 1 1 1 cose sine 1 (K,)- =| cos| 0 sin 0- 3 1 3 %3D cos 0+ sin 0+ 3 3 Lis + Lms Lms - Lms 2 L, = 1 - Lms s + Lms 1 Lms -- --Lms Lis + Lms Lis + LM K,L,(K,)-' = Lis + LM Lis
2n cos e + 3 2n cos e cos e 3 2 sin 0 3 sin 0 3 2n sin 0+ 3 K, 1 1 1 cose sine 1 (K,)- =| cos| 0 sin 0- 3 1 3 %3D cos 0+ sin 0+ 3 3 Lis + Lms Lms - Lms 2 L, = 1 - Lms s + Lms 1 Lms -- --Lms Lis + Lms Lis + LM K,L,(K,)-' = Lis + LM Lis
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Related questions
Question
mutual leakage terms for flux linkage
Ks, Ks^-1 , Ls
If I multiply Ks by it inverse I should get the intentity matrix, then If I muliply by Ls matrix I should get the Ls matrix right.
I did the matrix but It doenst not become the indenty matrix multpy by Ls and I also I have no Idea how the anwer would Lis at the last term of the last matrix on the first picture.
![sin e
cos (e + 2*)
sin (e + 2")
cos (e) cos (e - 4)
cos e
1
L_is + L_ms -1/2L_ms -1/2L_ms
* -1/2L_ms L_is + L_ms -1/2L_ms *
-1/2L_ms
3
2 A
cos (e
sin (e
sin (e) sin (e
3
3
-1/2L_ms L_is + L_ms
cos (e - 2*) sin (e
1
1/2
1/2
1/2
3
3
1
cos sin e
e|L m, cos (e)L m),
1
t{21]= {{ cos? e? (L_is + L_rm),
+ eL_rm
3
3
1
cos sin e
2
2 JT
2 JT
+ eL rm
1
+ eL_rm, sin?
3
(L_is + L_rm),
sin
3
3
1
2 JT
L_is + L_rm
1
L_rm,
4
sin
cos
L_rm,
4
3
3
[23]:= MatrixForm[%21]
B1/MatrixForm=
cos? e? (L_is + L_rm)
2 A
cos sin e (-
+ e) L_rm
cos
2
L rm
3
2 A
+ e) L_rm sin? (- 2* + e) (4 + e) (L
1
cos sine
(-
(L_is + L_rm)
sin (24 + e) L_rm
3
3
3
2
3
cos (- + e) L_rm
! sin (
2 A
1
L_is+L_rm
e) L_rm
4
3
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb010926-0d81-478d-9736-679f94816a5d%2F7b4ee2e6-d7b5-4de0-9e83-70133fc460d1%2Fe0nsosr_processed.png&w=3840&q=75)
Transcribed Image Text:sin e
cos (e + 2*)
sin (e + 2")
cos (e) cos (e - 4)
cos e
1
L_is + L_ms -1/2L_ms -1/2L_ms
* -1/2L_ms L_is + L_ms -1/2L_ms *
-1/2L_ms
3
2 A
cos (e
sin (e
sin (e) sin (e
3
3
-1/2L_ms L_is + L_ms
cos (e - 2*) sin (e
1
1/2
1/2
1/2
3
3
1
cos sin e
e|L m, cos (e)L m),
1
t{21]= {{ cos? e? (L_is + L_rm),
+ eL_rm
3
3
1
cos sin e
2
2 JT
2 JT
+ eL rm
1
+ eL_rm, sin?
3
(L_is + L_rm),
sin
3
3
1
2 JT
L_is + L_rm
1
L_rm,
4
sin
cos
L_rm,
4
3
3
[23]:= MatrixForm[%21]
B1/MatrixForm=
cos? e? (L_is + L_rm)
2 A
cos sin e (-
+ e) L_rm
cos
2
L rm
3
2 A
+ e) L_rm sin? (- 2* + e) (4 + e) (L
1
cos sine
(-
(L_is + L_rm)
sin (24 + e) L_rm
3
3
3
2
3
cos (- + e) L_rm
! sin (
2 A
1
L_is+L_rm
e) L_rm
4
3
2
![2n
cos e
cos 0
3
cos
3
sin(0-
K,
sin e
sin 0+
3
1
1
2
2
2
cose
sine
1
(K,)- =| cos 0 –-
sin 0–
3
1
%3D
3
2n
cos 0+-
sin 0 +-
1
Lis + Lms
Lms
2
Lms
L, = -. 4 + L.
1
-L,
'ms
1
- Lms
1
Ls + Lms
2
Lis + LM
K,L,(K,)-!
Li, + LM
Lis
Lms
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb010926-0d81-478d-9736-679f94816a5d%2F7b4ee2e6-d7b5-4de0-9e83-70133fc460d1%2Fkcldym5c_processed.png&w=3840&q=75)
Transcribed Image Text:2n
cos e
cos 0
3
cos
3
sin(0-
K,
sin e
sin 0+
3
1
1
2
2
2
cose
sine
1
(K,)- =| cos 0 –-
sin 0–
3
1
%3D
3
2n
cos 0+-
sin 0 +-
1
Lis + Lms
Lms
2
Lms
L, = -. 4 + L.
1
-L,
'ms
1
- Lms
1
Ls + Lms
2
Lis + LM
K,L,(K,)-!
Li, + LM
Lis
Lms
2
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