Consider the circuit depicted in the figure. Please use the two following facts: First, the voltage drop across a resistor is V = IR. Second, the sum of all voltage drops in a closed-loop sum to zero, The currents I1, I2, and I3 are determined from the 3 3 system.

where R1 = 20, R2 = 10, R3 = 25, R4 = 10, R5 = 30, R6 = 40, V2 = 0, V3 = 200, and V1 will be variable. In this form, the associated matrix is strictly diagonal dominant.

Part 1.

Vary V1 from 0 to 100 in steps of 2 (i.e. V1 = 0, 2, 4, ..., 100) and calculate I1, I2, and I3 as a function of increasing V1 by solving the system with the standard backslash. Save your results in a matrix of 3 columns and 51 rows where the first, second and third column are I1, I2 and I3 respectively.

Part 2.

Repeat part 1, but now solve it with three additional methods: LU decomposition, Jacobi Iterations, and Gauss-Seidel Iterations. For the iteration methods, begin with the guess (I1, I2, I3) = (0, 0, 0). This will give you three additional matrices of size 3 columns by 51 rows for the LU, Jacobi and Gauss-Seidel respectively.

Part 3.

For the two iteration methods, what is the average number of iterations required to solve the given equation with accuracy 10^−6 . The accuracy constraint should be based upon looking at the norm of the difference between successive iterations. Thus it is required that  

Save the two answers (for Jacobi first and Gauss-Seidel second) as a row vector with two components.

 

Transcribed Image Text:

R6I1 + R(I1 I2) + R2(I1 I3) = V R3I2 + R4(I2 I3) + R1(I2 I1) = V R513 + R4(I3 I2) + R2(I3 I1) = V3 R6I1 + R(I1 I2) + R2(I1 I3) = V R3I2 + R4(I2 I3) + R1(I2 I1) = V R513 + R4(I3 I2) + R2(I3 I1) = V3