Electrical Resistance - Left Division Method

I'm having some trouble with an assingment using the left divison method. The examples I've been given in my text books are rather confusing, and I need some help.
I have been given three equations of Kirchoff's Voltage Law:
v - R2i2 - R4i4= 0
-R2i2 + R1i1 + R3i3 = 0
-R4i4 - R3i3 + R5i5 = 0
Now I know that I need to rearrange these so that they are "in order":
R2i2 + R4i4 = v
R1i1 - R2i2 + R3i3 = 0
-R3i3 - R4i4 + R5i5 = 0
I've also been given 4 equations for the conservation of charge at my nodes:
i6 = i1 + i2
i4 = i2 + i3
i1 = i3 + i5
i6 = i4 + i5
I don't understand what to do. Why can't I just put the given equations into a matrix and do the divison? That would look something like:
R=[1 5 2 10 5]*1000;
v1=100;
A1=[0 R(2) 0 R(4) 0 0];
A2=[R(1) -R(2) R(3) 0 0 0];
A3=[0 0 -R(3) -R(4) R(5) 0];
A=[A1;A2;A3];
b=[v1;0;0];
current=A\b

回答 (1 件)

Sean de Wolski
Sean de Wolski 2012 年 4 月 9 日

0 投票

Then look at:
doc mldivide

4 件のコメント

Joseph
Joseph 2012 年 4 月 9 日
There's nothing in the FAQ relating to Left Division that I saw ("Find" didn't help either).
mldivide did say that "has at most K nonzero components per column... the effective rank, K". When I tried to expand the number of terms per row in my matrix using the current equations, MATLAB was producing an error.
It could be that I'm making an error there. Any idea if the first set of values could come out to:
A1=[-R(2) R(2) R(2) R(4) R(2) 0];
Sean de Wolski
Sean de Wolski 2012 年 4 月 9 日
The point of the FAQ link is to show you how to get rid of many variables named things like r1i2, and how to build a matrix of values. Since you have constraints, you'll have to enforce these as well.
Joseph
Joseph 2012 年 4 月 9 日
Ah ha, when the images load, it scrolls me saving file conventions. You have no idea how confused I was- thought you were just showing me to the FAQ page. Silly images loading!
Sean de Wolski
Sean de Wolski 2012 年 4 月 9 日
I agree that's very annoying!

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