Consider the circuit diagram below: The voltage drop across a resistor is V = IR, The sum of all voltage drops in a closed loop sum to zero (Kirchoff's Law). The previous two facts allow us to construct the following system of equations: R6I1 + R1(I1 -I2) + R2(I1 - I3) = V1, R3I2 + R4(I2 -I3) + R1(I2 - I1) = V2, R5I3 + R4(I3 - I2) + R2(I3 - I1) = V3. Let the resistances be given by R1 = 30, R2 = 25, = 20, R4 = 5, R5 = 15, R6 = 10, and let V2 = 10 and V3 = 40. We will be varying V1 throughout this exercise, and solving for the currents, I1, I2, and I3. Write the equations in matrix form Ax = b, and determine the matrices P, L, and U using the lu command so that PA = LU. ANSWERS: Save the matrices PA and LU in A22.dat and A23.dat, respectively. Vary V1 from 50 to 100 in steps of 5 (i.e., V1 = 50,55,60, , 95,100) and calculate I1,I2 and I3 as a function of the increasing V1 by solving the system above using LU-decomposition (the two-step procedure is summarized in the next exercise). ANSWERS: Save your results as a matrix of 3 rows and 11 columns in A24.dat (where the first, second, and third rows are the I1, I2, and I3 values, respectively). For the same V1 values from 50 to 100 in steps of 5, create a matrix B that is 3 by 11, one column for every new value of V1 (the second and third rows are the constant V2 and V3 values). Try solving the system AX = B in one fell swoop by using the backslash command: X=A\B;. ANSWERS: Save the resulting X in A25.dat.
