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Solving linear systems with a function

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Luke Radcliff
Luke Radcliff 2016 年 7 月 10 日
コメント済み: Akash Chauhan 2021 年 1 月 18 日
have a circuit that has 5 resistors and 1 applied voltage. Kirchoff's voltage law was applied to 3 loops that gave me 3 linear equations.
v - R2*i2 - R4*i4 = 0
-R2*i2 + R1*i1 + R3*i3 = 0
-R4*i4 - R3*i3 + R5*i5 = 0
from that law Is known:
6 = i1 + i2
4 = i2 + i3
1 = i3 + i5
6 = i4 + i5
using this I made a function that just finds the current in i4 with a given set of values.
values to use: R1 = 1, R2 = 4, R3 = 5, R4 = 1, R5 = 5, v = 100, measured in ohms and volts
function I made to solve this.
function [i4] = I(R1, R2, R3, R4, R5, v)
format shortg;
A = [0 -R2 0 -R4 0 0
R1 -R2 R3 0 0 0
0 0 -R3 -R4 R5 0];
b = [-v; 0; 0;];
x = A\b;
i2 = x(2,:);
i3 = x(3,:);
i4 = i2 + i3;
return
This system is overdetermined, using gauss-jordan in matlab I got x, which give all 6 currents values. With this function i4 = 45, the answer is i4 = 27.638. What do I have wrong in my function or need to add?

採用された回答

Star Strider
Star Strider 2016 年 7 月 10 日
I get the same result you do, coding your matrices myself. I would have to see your circuit. (I usually use the node voltage approach, since it’s easier for me.)
  14 件のコメント
Hayriye Esin Basaran
Hayriye Esin Basaran 2020 年 12 月 16 日
編集済み: Hayriye Esin Basaran 2020 年 12 月 16 日
nction [i,i1,i2,i3,i4,i5,i6]=kirchoff(R1,R2,R3,R4,R5,V)
A=[0 -R2 0 -R4 0 0; R1 -R2 R3 0 0 0; 0 0 -R3 -R4 R5 0];
B=[-V;0;0];
i=pinv(A)*B;
i1 = i(3,1)+i(5,1);
i2 = i(4,1)-i(3,1);
i3 = i(1,1)-i(5,1);
i4 = i(2,1)+i(3,1);
i5 = i(1,1)-i(3,1);
i6 = i(4,1)+i(5,1);
Command Window:
>> [i,i1,i2,i3,i4,i5,i6]=kirchoff(1,5,2,10,5,100)
Hayriye Esin Basaran
Hayriye Esin Basaran 2020 年 12 月 16 日
I want to get the same result with the current results I found by typing i = pinv (A) * B and the new current equations I added below.
i1 = i(3,1)+i(5,1);
i2 = i(4,1)-i(3,1);
i3 = i(1,1)-i(5,1);
i4 = i(2,1)+i(3,1);
i5 = i(1,1)-i(3,1);
i6 = i(4,1)+i(5,1);

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その他の回答 (1 件)

Andrei Bobrov
Andrei Bobrov 2016 年 7 月 10 日
編集済み: Andrei Bobrov 2016 年 7 月 11 日
R1 = 1, R2 = 4, R3 = 5, R4 = 1, R5 = 5, v = 100
R = [R1;R2;R3;R4;R5;0];
E = [zeros(5,1);v];
J = zeros(6,1);
D = [[1 -1 1 0 0 0];[0 0 -1 -1 1 0];[0 1 0 1 0 1]];
RR = D*diag(R)*D';
EE = D*(E - J.*R);
Ik = RR\EE;
I = D'*Ik + J;
  1 件のコメント
Akash Chauhan
Akash Chauhan 2021 年 1 月 18 日
can u please explain this to me ?

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