Analysis of Mesh Circuit Using Linear Algebra Submitted By: Md. Anowar Hossain Researcher No: 75654 Submitted To: Dr. Yacine Benhadid 1
Abstract An interesting application of linear algebra can be found in electrical engineering and specifically electrical networks. I have found that linear algebra is a useful tool in analyzing electric circuits in terms of organization and saving time. In this report, I have included examples how to analyze a circuit using linear algebra and solved it by using Gaussian Elimination, LU Decomposition and SOR (Successive Over Relaxation) method and determined Eigenvalues and Eigenvectors using Shifted Inverse Power method, Deflation method and Jacobi method. Background A simple electric circuit consists of voltage sources, wire with current running through it, and various types of resistors that cause a voltage drop in the circuit. In order to tell how much the voltage drops across a resistor, one must use Ohm s law. This law simply states that V = IR Where, V = Voltage across the resistor (volts), I = Current through the resistor (amperes) and R = Resistance of the resistor (ohms). When analyzing an electric circuit, there are many techniques that may be used. One simple technique uses the idea of Kirchhoff s voltage law (KVL). Kirchhoff s voltage law states that the algebraic sum of the voltages around any closed path in a circuit is identically zero for all time. In order to use Kirchhoff s voltage law, the method of mesh current analysis must be used to examine the circuit. In order to analyze the circuit, Kirchhoff s voltage law must be applied to each mesh. It is important to note that if a resistor is shared by two meshes, the voltage across this resistor is (current in mesh being analyzed current in other mesh) x (resistance).
After summing the voltages around a circuit with 5 meshes, there will be 5 equations with 5 unknowns. One way to solve this would be to solve each equation in terms of one variable and then substitute into the others. This method would work, but would be far too time consuming. This is where linear algebra comes in. Using Ohm s law, each equation can be written in the form RI = V (similar to Ax = b). Putting these equations into matrices and solve for each current by using the matrix equation RI = V. By solving the augmented matrix, values of I 1, I 2, I 3, I 4 and I 5 can be found. The matrix equation can be very useful when large circuits with many loops are to be analyzed. In order to analyze the simple circuit below (find the amount of current flowing in each loop), the technique of mesh analysis will be combined with the use of linear algebra. Model Circuit
The equations in the model are given by: Mesh 1: I 1 R 1 + (I 1 I 2 ) R 6 + (I 1 I 4 ) R 7 = V Mesh 2: (I 2 - I 1 ) R 6 + I 2 R 2 + (I 2 - I 3 ) R 5 = 0 Mesh 3: I 3 R 3 + I 3 R 4 + (I 3 I 2 ) R 5 = 0 Mesh 4: (I 4 I 1 ) R 7 + (I 4 I 5 ) R 4 + I 4 R 10 + I 4 R 8 = 0 Mesh 5: (I 5 I 4 ) R 4 + I 5 R 11 = 0 V = 120V, R1 = 2, R2 = 5, R3 = 2, R4 = 4, R5 = 3, R6 = 1 R7 = 3 R8 = 6 R9 = 5 R10 = 2 R11 = 8 Step 1: The first step in solving this problem is to write the sum of voltages around each loop. (In this case, I arbitrarily choose current direction as clockwise). The voltage sums of each loop must equal zero, thus giving the following equations: Mesh 1: 6I 1 I 2 + 0I 3 3I 4 + 0I 5 = 120 Mesh 2: -I 1 + 9I 2-3I 3 + 0I 4 + 0I 5 = 0 Mesh 3: 0I 1 3I 2 + 9I 3 + 0I 4 + 0I 5 = 0 Mesh 4: -3I 1 + 0I 2 + 0I 3 + 15I 4-4I 5 = 0 Mesh 5: 0I 1 + 0I 2 + 0I 3 5I 4 + 13I 5 = 0 Step 2: Putting the five equations into matrix form Ax = b. In this case, values of resistance will make up the coefficient matrix, current will be the unknown x, and voltages will be the product of resistances and currents. The three equations from above will have the form RI = V. Once the mesh equations are put into matrices, they will look like: 6-1 0-3 0 I 1 120-1 9-3 0 0 I 2 0 0-3 9 0 0 I 3 = 0-3 0 0 15-4 I 4 0 0 0 0-5 13 I 5 0
Solving the system using Gaussian Elimination: 6-1 0-3 0 120-1 9-3 0 0 0 Ab = 0-3 9 0 0 0-3 0 0 15-4 0 0 0 0-5 13 0 >> Ab=[6-1 0-3 0 120;-1 9-3 0 0 0;0-3 9 0 0 0;-3 0 0 15-4 0;0 0 0-5 13 0]; >> gauss(ab) Which gives upper triangular matrix, 6.0000-1.0000 0-3.0000 0 120.0000 0 8.8333-3.0000-0.5000 0 20.0000 M = 0 0 7.9811-0.1698 0 6.7925 0 0 0 13.4681-4.0000 61.2766 0 0 0 0 11.5150 22.7488 Using backward substitution, gives the solution: I 1 = 23.0484 A, I 2 = 2.8811 A, I 3 = 0.9604 A, I 4 = 5.1365 A, I 5 = 1.9756 A
Solving the system using LU Decomposition 6-1 0-3 0-1 9-3 0 0 A = 0-3 9 0 0-3 0 0 15-4 0 0 0-5 13 >>A=[6-1 0-3 0;-1 9-3 0 0;0-3 9 0 0;-3 0 0 15-4;0 0 0-5 13]; >>LU(A) 1.0000 0 0 0 0-0.1667 1.0000 0 0 0 0-0.3396 1.0000 0 0 = L -0.5000-0.0566-0.0213 1.0000 0 0 0 0-0.3712 1.0000 6.0000-1.0000 0-3.0000 0 0 8.8333-3.0000-0.5000 0 0 0 7.9811-0.1698 0 = U 0 0 0 13.4681-4.0000 0 0 0 0 11.5150
Using Ly=b and Ux=y 1.0000 0 0 0 0 y 1 120-0.1667 1.0000 0 0 0 y 2 0 0-0.3396 1.0000 0 0 y 3 = 0-0.5000-0.0566-0.0213 1.0000 0 y 4 0 0 0 0-0.3712 1.0000 y 5 0 >>b=[120;0;0;0;0]; >>y=inv(l)*b Gives the solution, y 1 = 120 y 2 = 20 y 3 = 6.7925 y 4 = 61.2766 y 5 = 22.7488 6.0000-1.0000 0-3.0000 0 I 1 120 0 8.8333-3.0000-0.5000 0 I 2 20 0 0 7.9811-0.1698 0 I 3 = 6.7925 0 0 0 13.4681-4.0000 I 4 61.2766 0 0 0 0 11.5150 I 5 22.7488
>>I=inv(U)*y Gives the solution, I 1 = 23.0484 A I 2 = 2.8811 A I 3 = 0.9604 A I 4 = 5.1365 A I 5 = 1.9756 A Solving the System using SOR method 6-1 0-3 0-1 9-3 0 0 A = 0-3 9 0 0-3 0 0 15-4 0 0 0-5 13 120 b = 0 0 0 0 >> A=[6-1 0-3 0;-1 9-3 0 0;0-3 9 0 0;-3 0 0 15-4;0 0 0-5 13]; >> b=[120;0;0;0;0]; >> N=100; >> sor(a, b, N) It gives the solution on 7 iterations (W = 1.0628)- I 1 = 23.0484 A I 2 = 2.8811 A I 3 = 0.9604 A I 4 = 5.1365 A I 5 = 1.9756 A
Finding Eigenvalues and Eigenvectors: Eigenvalues of matrix A are: 19.0196 4.4967 6.2619 10.1315 12.0903 Power Method: Using an initial vector x 0 = [1,1,1,1,1] T, the approximation of the dominant Eigenvalue of matrix A after 20 iterations is 19.0196 and the corresponding eigenvector is [-0.1760, 0.0193, -0.0058, 0.7572, -0.6288] T >> A=[6-1 0-3 0;-1 9-3 0 0;0-3 9 0 0;-3 0 0 15-4;0 0 0-5 13]; >> x=[1,1,1,1,1]'; >> tol=1e-10; >> maxiter=20; >> [lambda,v,relerr,niter] = Powermethod(A,x,tol,maxiter) Inverse Power Method: Using an initial vector x 0 = [0,1,2,3,4] T, the approximation of the dominant Eigenvalue of matrix A -1 after 10 iterations is 4.4980 and the corresponding eigenvector is [0.8294, 0.3550, 0.2453, 0.3055, 0.1805] T. >> A=[6-1 0-3 0;-1 9-3 0 0;0-3 9 0 0;-3 0 0 15-4;0 0 0-5 13]; >> x=[0,1,2,3,4]'; >> tol=1e-10; >> maxiter=10; >> [lambda,v,relerr,niter] = InvPowerMethod(A,x,tol,maxiter) Shifted Inverse Power Method: Using an initial vector x 0 = [1,1,1,1,1] T and µ=6, the approximation of the dominant Eigenvalue of matrix B -1 = (A-µI) -1 after 15 iterations is 19.0196 and the corresponding eigenvector is [-0.1759, 0.0193, -0.0058, 0.7571, - 0.6289] T.
Matlab command: >>A=[6-1 0-3 0;-1 9-3 0 0;0-3 9 0 0;-3 0 0 15-4;0 0 0-5 13]; >> s=19; >> x=[1,1,1,1,1]'; >> tol=1e-10; >> maxiter=15; >> [lambda,v,relerr,niter] = ShiftInvPowerMethod(A,s,x,tol,maxiter) Jacobi Method: Jacobi method gives the matrix 4.4956 0.0402-0.0066-0.0230 0.1396 0.0174 12.0898 0.0865 0.0414 0.0041 D = -0.0026 0.0745 6.2630 0.0001-0.0001 0.2906 0.1229-0.1803 19.0188-0.0084-0.0455 0.0218 0.0946-0.9131 10.1328 Its diagonal elements are: 4.4956 12.0898 6.2630 19.0188 10.1328 Which are the Eigenvalues of the original matrix A. >>A=[6-1 0-3 0;-1 9-3 0 0;0-3 9 0 0;-3 0 0 15-4;0 0 0-5 13]; >> JOBM(A)
Deflation Method: Using the largest Eigenvalue 19.0196 and the corresponding Eigenvector x 1 = [-0.1760, 0.0193, -0.0058, 0.7572, -0.6288] T, Deflation method gives the following Eigenvalues: 19.0196 4.4979 6.2617 12.0905 10.1335 and the corresponding eigenvectors are: [-0.1759, 0.0193, -0.0058, 0.7571, -0.6289] T [0.8414, 0.3359, 0.2238, 0.3097, 0.1821] T [-0.3386, 0.6168, 0.6759, -0.1761, -0.1306] T [-0.3264, -0.0478, 0.1269, 0.4655, 0.8114] T [-0.1259, 0.7074, -0.6867, 0.0198, 0.1088] T >> A=[6-1 0-3 0;-1 9-3 0 0;0-3 9 0 0;-3 0 0 15-4;0 0 0-5 13]; >> Lamda=19.0196; >> XA=[-0.1760, 0.0193, -0.0058, 0.7572, -0.6288]'; >> DEFLATED(A,Lamda,XA) >>[v,d]=eig(a) Evaluation I think that having knowledge of matrices and linear algebra is extremely valuable in many fields of study. This project was a good chance to find out how useful linear algebra is in many different areas of math and science. The reason I choose only to discuss linear algebra in relation to Kirchhoff s voltage law and mesh current analysis is because I became extremely familiar with mesh current analysis. I wanted to see how linear algebra could relate to and expand my knowledge of a subject I already knew about. I thought this would be more valuable to me, than to learn about a completely new topic that I knew nothing of, and then write a report on that.