mutual leakage terms for flux linkage Ks, Ks^-1 , LsIf 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 ecos (e + 2*)sin (e + 2")cos (e) cos (e - 4)cos e1L_is + L_ms -1/2L_ms -1/2L_ms* -1/2L_ms L_is + L_ms -1/2L_ms *-1/2L_ms32 Acos (esin (esin (e) sin (e33-1/2L_ms L_is + L_mscos (e - 2*) sin (e11/21/21/2331cos sin ee|L m, cos (e)L m),1t{21]= {{ cos? e? (L_is + L_rm),+ eL_rm331cos sin e22 JT2 JT+ eL rm1+ eL_rm, sin?3(L_is + L_rm),sin3312 JTL_is + L_rm1L_rm,4sincosL_rm,433[23]:= MatrixForm[%21]B1/MatrixForm=cos? e? (L_is + L_rm)2 Acos sin e (-+ e) L_rmcos2L rm32 A+ e) L_rm sin? (- 2* + e) (4 + e) (L1cos sine(-(L_is + L_rm)sin (24 + e) L_rm33323cos (- + e) L_rm! sin (2 A1L_is+L_rme) L_rm432 2ncos ecos 03cos3sin(0-K,sin esin 0+311222cosesine1(K,)- =| cos 0 –-sin 0–31%3D32ncos 0+-sin 0 +-1Lis + LmsLms2LmsL, = -. 4 + L.1-L,'ms1- Lms1Ls + Lms2Lis + LMK,L,(K,)-!Li, + LMLisLms2

When a matrix is multiplied with its inverse matrix, the result is an identity matrix. It is…

Answered: 2n cos e + 3 2n cos e cos e 3 2 sin 0 3…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

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

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

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

Similar questions