Question

Question 1 (15 Marks): The electrical circuit shown consists of resistors and voltage sources. We can determine the current in each resistor, using the mesh current method based on Kirchhoffs voltage law. (Kirchhoff's voltage law states that the sum of the voltage around a closed circuit is zero) 2 Σ vsource-IR = 0 i3 IH - Note: if two currents passes through one resistor we should use difference or summation between them based on direction of the currents i.e ii-j^,. i1+i3 Determine the voltage drops on the resistances VRI,VR2, VR3 VR4, VR5 using two methods (a) Symbolic Math in MATLAIB (b) Using block algebraic constrain in Simulink (c) Draw the solutions that comes from both MATLAB and Simulink in one figure and highlight the simulink with * and MATLAB with O [6 marks] [6 marks] [3 marks]

Answer 1

3 10V 3 1. 13 Applyin kvi for mesha, 1o./L → (i 3211-12五-20, --10 r me 1812-1241-61チー30 →的 Applying kvLtor mesh Applying kyL าขึ้r nesh@ and T usinq MATLAE NOTE: vottage actoss each resstaf- mayj be difer frm your resu based m your » etation, but magntu -de same.

MATLAB CODE:

clc
clear all
close all

syms I1 I2 I3 I4
eq1 = 32*I1-12*I2-20*I3 == -10;
eq2 = 18*I2-12*I1-6*I4 == 30;
eq3 = 28*I3-20*I1 == -30;
eq4 = 16*I4-6*I2 == 0;
eqs = [eq1,eq2,eq3,eq4];
[A,B] = equationsToMatrix(eqs);
X = inv(A)*B;
I1 = X(1);I2 = X(2);I3 = X(3);I4 = X(4);
VR1 = (I3-I1)*20;
VR2 = (I2-I1)*12;
VR3 = (-I3)*8;
VR4 = (I4-I2)*6;
VR5 = (-I4)*10;
fprintf('VR1 = %d Volts\n',VR1);
fprintf('VR2 = %d Volts\n',VR2);
fprintf('VR3 = %d Volts\n',VR3);
fprintf('VR4 = %d Volts\n',VR4);
fprintf('VR5 = %d Volts\n',VR5);

OUTPUT:

VR1-16 Volts VR226 Volts VR314 Volts VR4-4 Volts VR5-4 Volts