Solving linear systems with a function

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?

Answer 1

[Star Strider]

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.)

[Luke Radcliff]

I attached the circuit for you

[Luke Radcliff]

Thanks for trying, I have solved it.

[Star Strider]

I apologise for the delay. It was too late last night for me to work on this, and I got a late start today.

My nodal analysis labeled the node defined by the connection of ‘R2’, ‘R3’, and ‘R4’ as ‘VA’, and the node defined by the connection of ‘R1’, ‘R3’, and ‘R5’ as ‘VB’ (the apex of the triangle). I defined ‘IT’ as the total current supplied by the source. I did all the calculations with the Symbolic Math Toolbox, largely because I’m lazy today. The currents correspond to those in the circuit schematic diagram you supplied. The nodes were not labeled, so I created my own labels.

The Nodal Equations:

syms R1 R2 R3 R4 R5 V VA VB IT
R1 = 1; R2 = 4; R3 = 5; R4 = 1; R5 = 5; V = 100;

I1 = (V-VB)/R1;
I2 = (V-VA)/R2;
I3 = (VB-VA)/R3;
I4 = VA/R4;
I5 = VB/R5;

Eq1 = I1+I2 == IT;
Eq2 = -I1+(I5+I3) == 0;
Eq3 = I2+I3-I4 == 0;

S = solve(Eq1, Eq2, Eq3);
V_A = vpa(S.VA, 7)
V_B = vpa(S.VB, 7)
I_T = vpa(S.IT, 7)

The Voltages and Total Current:

V_A =
27.63819

V_B =
75.37688

I_T =
42.71357

That calculates to a total network resistance of 2.3412 Ohms (seems reasonable), in the event you want to do a Norton or Thévenin equivalent circuit of it. With the solved voltages, the currents are straightforward calculations.

I haven’t done circuit analysis in at least a few months (since I don’t usually need to), so thanks for the opportunity to come back up to speed on it! It was fun!

[Marc]

Looks good to me... I wish I could give Star Strider some gold for the effort after Luke got a bit frisky.... Much better man than I would have been.

[Star Strider]

@Marc —

Thank you!

[Luke Radcliff]

A bit frisky.... wait what? haha looks good star thanks for your help.

[Star Strider]

My pleasure!

[Hayriye Esin Basaran]

how can i do this using first method. i want to calculate currents

[Star Strider]

Hayriye Esin Basaran —

Solve for the voltages, then plug those back into the current equations to solve for the currents. That’s how I always do it.

[Hayriye Esin Basaran]

function [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);

[Hayriye Esin Basaran]

i do like that but i dont understand what is the wrong. i calculate the currents but after i1,i2... the result is not same. i am confused.

[Star Strider]

Hayriye Esin Basaran — I cannot follow what you are doing.

[Hayriye Esin Basaran]

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]

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);

Answer 2

[Andrei Bobrov]

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;

[Akash Chauhan]

can u please explain this to me ?