Question

R611 + R1(11 – 12) + R2(11 – 13) = V1 R312 + R4(12 – 13) + Ri(12 – I1) = V2 R513 + R4(13 – 12) + R2(13 – 11) = V3.

using MatLab

Answer 1

a)

I1*(R6+R1+R2) +I2*(-R1)+I3*(-R2) = V1

I1*(-R1)+I2*(R4+R3+R1)+I3*(-R4) = V2

I1*(-R2) + I2*(-R4) +I3*(R2+R4+R5) =V3

Hence the matrix can be formed as-

R6 + R1 + R2 -R1 -R2 -R1 R4 + R3 + R1 -R4 -R2 -R4 R5 + R4 + R2 * 11 12 13 V1 V2 V3

b)

code-

% %Part (a)
R1=10;
R2=20;
R3=5;
R4=15;
R5=30;
R6=25;
A=[R6+R1+R2 -R1 -R3 ;-R1 R3+R4+R1 -R4 ;-R2 -R4 R5+R4+R2]
[L ,U ,P] = lu(A)
A1 = U*P*L;

%Part (b)
V1 =50;
V2 =0;
i=1;

for V3b=1:1:100

B = [V1 ;V2 ;V3b];
Y = inv(L) *B;
X = inv(U) *Y ; %Solving AX =B where X = current vector[I1 ;I2 ;I3]
I2b(1,i) = X(2,1);
i=i+1;
end
disp('I2b=')
for i =1:100
disp(I2b(1,i)) %The required vector I2 from Part c (Inverse)
end %The required vector I2 from Part b (LU decomposition)

%Part c
j=1;

for V3c =1:1:100
  
B1 = [V1 ;V2 ;V3c];
I = inv(A)*B1;
I2c(1,j) =I(2,1);
j=j+1;
end

disp('I2c=')
for j =1:100
disp(I2c(1,j)); %The required vector I2 from Part c (Inverse)
end

disp('I2c-I2b=')
for i=1:100
A3(1,i) = I2c(1,i)-I2b(1,i);

end
A3

Result-