# Mutual Inductance - Mesh analysis

Discussion in 'Homework Help' started by ydgmms, Dec 4, 2007.

1. ### ydgmms Thread Starter Member

Oct 6, 2007
10
0
I'm desperately trying to figure out this problem with ZERO luck.

I'm thinking its either my mesh equations or my calculating Vo after I obtain I2.

Mesh1 I get: 1*J1 + j6*J1 + j2*J2 = 10 angle 0
Mesh2 I get: j2*J1 + j4*J2 + 4*J2 - j3*J2 = 0

In Matrix(scilab input format, cuz I dont know how to do a matrix here)
a=[ 1+6j , 2j ; 2j , 4+j ]
b=[10;0]

So then, to calculate J1 and J2, I do

c=a/det(a)
c =

0.2416534 - 0.0206677i 0.0794913 + 0.0063593i
0.0794913 + 0.0063593i 0.0524642 - 0.1558029i

d=c*b
d =

2.4165342 - 0.2066773i
0.7949126 + 0.0635930i

Now I'm stuck here, getting my J2 into Vo. I try V=IR, so
d(2,1)*-3j =

ans =

0.1907790 - 2.3847377i

Which is not Vo according to the book...

Anyone?

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Your problem is that the calculation c=a/det(a) does not give the inverse of a. Doesn't Scilab have a function like c = Inv(a) or c = Inverse(a)?

3. ### ydgmms Thread Starter Member

Oct 6, 2007
10
0
The example problem I was trying to follow, had the coeffecient matrix over its determinate. ?

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
If you will scan or photograph the page(s) from the example you're trying to follow and post them, I'll explain what they're doing.

In the meantime, if you will substitute c = inverse(a), (or whatever the correct syntax in Scilab is) instead of c = a/det(a), you will get the correct answer.

5. ### ydgmms Thread Starter Member

Oct 6, 2007
10
0
You are absolutely correct sir. I tried it with the way I SHOULD have the entire time. and within a minute I had the correct answer. Pissed me off to no end, i spent 2 hours last nite and i dunno how long monday and sunday nites.

i did c=a\b, then d=c(2,1)*-3j and got the proper Vo

Code ( (Unknown Language)):
1.
2. -->a
3.  a  =
4.
5.     1. + 6.i    2.i
6.     2.i         4. + i
7.
8. -->b
9.  b  =
10.
11.     10.
12.     0.
13.
14. -->c=a\b
15.  c  =
16.
17.     0.5246423 - 1.5580286i
18.   - 0.7949126 - 0.0635930i
19.
20. -->c(2,1)*-3*%i
21.  ans  =
22.
23.   - 0.1907790 + 2.3847377i
24.
25. -->
26.
And when I convert ans into a phasor, I get the proper Vo.

Thanks.. I dunno what the example problem and I were doing.

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6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
See the attachment for an explanation.

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7. ### ydgmms Thread Starter Member

Oct 6, 2007
10
0

Huh..interesting. I've never tried to figure out the inverse of a matrix by hand. Thats what threw me off. Never been taught it, always just been told 'use a calculator or MATLAB(scilab for us poor college students)'. So I thought they were doing something different when they had the matrix over its determinate.

I had noticed the sign change of the -j5 values. But thought it may be a misprint, as during the course we've seen a few in the book.

Oh well. Thanks for the explanation. I think I'll stick to scilab and how I normally solve coeffecient matrices.