Engineering 25
Chp4
Prob 4.37
Solution
(BobbyTrapped)
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering/Math/Physics 25: Computational Methods
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_HW-01_Solution.ppt
Prob 4-37 Solution
 The following code, while
functional, contains
numerous “Booby Traps”
which produce output that
contains subtle errors
 Any student who turns-in HW
that contains the BoobyTrap
produced errors will receive a
ZERO on the problem
 This Booby Trapped solution
is not supported by any
persons Living or Dead…
Engineering/Math/Physics 25: Computational Methods
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_HW-01_Solution.ppt
% Bruce Mayer, PE * 08Jul05
% ENGR25 * Problem 4-31
% file = Prob4_31_KCL_KVL.m
%
% INPUT SECTION
%R1 = 5; R2= 100; R3 = 200; R4 = 150; % SingleOhm case
R5 = 250e3;
R1 = 5e3; R2= 100e3; R3 = 200e3; R4 = 150e3; % kOhm case
% Coeff Matrix A
v1 = 100; % in Volts
A = [R1 0 0 R4 0; 0 R2 0 -R4 R5; 0 0 R3 0 -R5;...
-1 1 0 1 0; 0 -1 1 0 1];
%
% LOW Loop
%
Initialize Vars
v2 = 29.47;
C = [0;0;0;0;0];
% use element-by-element logic test on while
%
Must account for NEGATIVE Currents
while abs(C) < 0.001*[1;1;1;1;1] % C in SI units of Amps
% Constraint Col Vector V
V = [v1; 0; -v2; 0; 0];
% Collect last conforming Value-set
v2_lo = v2;
i1_lo = C(1);
i2_lo = C(2);
i3_lo = C(3);
i4_lo = C(4);
i5_lo = C(5);
%decrement v2 by 10 mV DOWN
v2 = v2 - 0.01;
% find solution vector for currents, C, from AC=V
C = A\V;
end
%display "lo" vars
v2_lo
display('currents in mA')
i1_low = 1000*i1_lo
i2_low = 1000*i2_lo
i3_low = 1000*i3_lo
i4_low = 1000*i4_lo
i5_low = 1000*i5_lo
%
% HIGH Loop
%
Initialize Vars
v2 = 300;
C = [0;0;0;0;0];
% use element-by-element logic test on while
%
Must account for NEGATIVE Currents
while abs(C) < 0.001*[1;1;1.7;1;1] % C in SI units of Amps
Engineering/Math/Physics 25: Computational Methods
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_HW-01_Solution.ppt
%Constraint Col Vector V
V = [v1; 0; -v2; 0; 0];
% find soltion vector for currents, C, from AC=V
C = A\V;
% Collect last conforming set
v2_hi = v2;
i1_hi = C(1);
i2_hi = C(2);
i3_hi = C(3);
i4_hi = C(4);
i5_hi = C(5);
%increment v2 by 10 mV UP
v2 = v2 + 3.5;
end
%display "hi" vars
v2_hi
display('currents in mA')
i1_high = 1000*i1_hi
i2_high = 1000*i2_hi
i3_high = 1000*i3_hi
i4_high = 1000*i4_hi
i5_high = 1000*i5_hi
%
%
display('Hit ANY KEY to start part-b Calculation')
pause
%
% Part-b => find v2 = f(R3) where R3 ranges of 150-250 kOhm
%
% use above while-loops nested in a for loop
m = 0; % initialize lo-vector index (the loop counter)
for k = 150:250 % use R3 = 1000*k
AR3 = [R1 0 0 R4 0; 0 R2 0 -R4 R5; 0 0 1000*k 0 -R5;...
-1 1 0 1 0; 0 -1 1 0 1];
% LOW Loop
%
Initialize Vars
v2 = 36;
C = [0;0;0;0;0];
% use element-by-element logic test on while
%
Must account for NEGATIVE Currents
while abs(C) < 0.001*[1;1;1;1;1] % C in SI units of Amps
% Constraint Col Vector V
V = [v1; 0; -v2; 0; 0];
% Collect last conforming Value-set
v2_lo = v2;
i1_lo = C(1);
i2_lo = C(2);
i3_lo = C(3);
i4_lo = C(4);
i5_lo = C(5);
Engineering/Math/Physics 25: Computational Methods
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_HW-01_Solution.ppt
%decrement v2 by 10 mV DOWN
v2 = v2 - 0.01;
% find solution vector for currents, C, from AC=V
C = AR3\V;
end
% Construct v2lo plotting vectors R3lo(m) & R3v2lo(m) using v2_lo
from
% while
m = m + 1;
R3lo(m) = k; % in kOhm
R3v2lo(m) = v2_lo;
end
%
% Now find v2hi = g(R3)
%
n = 0; % initialize lo-vector index (the loop counter)
for k = 150:250 % use R3 = 1000*k
AR3 = [R1 0 0 R4 0; 0 R2 0 -R4 R5; 0 0 1000*k 0 -R5;...
-1 1 0 1 0; 0 -1 1 0 1];
% HIGH Loop
%
Initialize Vars => need to reduce v2 starting value
v2 = 310;
C = [0;0;0;0;0];
% use element-by-element logic test on while
%
Must account for NEGATIVE Currents
while abs(C) < 0.001*[1;1;1;1;1] % C in SI units of Amps
%Constraint Col Vector V
V = [v1; 0; -v2; 0; 0];
% find soltion vector for currents, C, from AC=V
C = AR3\V;
% Collect last conforming set
v2_hi = v2;
i1_hi = C(1);
i2_hi = C(2);
i3_hi = C(3);
i4_hi = C(4);
i5_hi = C(5);
%increment v2 by 10 mV UP
v2 = v2 + 0.01;
end
% Construct v2hi plotting vectors R3hi(m) & R3v2lo(m) using v2_lo
from
% while
n = n + 1;
R3hi(n) = k; % in kOhm
R3v2hi(n) = v2_hi;
end
%
%Create Plot
Engineering/Math/Physics 25: Computational Methods
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_HW-01_Solution.ppt
plot(R3lo, R3v2lo, R3hi, R3v2hi), grid, xlabel('R3 (\muohm)'),
ylabel('Allowable v2 (lbs)'),...
title('Problem 4-31)(b: Allowable v2 as a function of R3'),
legend('v2,min', 'v2,max', 'Location', 'West')
Engineering/Math/Physics 25: Computational Methods
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • ENGR-25_HW-01_Solution.ppt