slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege PowerPoint Presentation
Download Presentation
Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Loading in 2 Seconds...

play fullscreen
1 / 9

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege - PowerPoint PPT Presentation


  • 49 Views
  • Uploaded on

Engr/Math/Physics 25. MidTerm Exam Review. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. Error in Text Book All R’s Should be in k Ω Plot at right shows the large currents Generated by this Error. Problem 4-31. In NO case are ALL Currents  1mA.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege' - lamya


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Engr/Math/Physics 25

MidTerm ExamReview

Bruce Mayer, PE

Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

problem 4 31
Error in Text Book

All R’s Should be in kΩ

Plot at right shows the large currents Generated by this Error

Problem 4-31

In NO case are ALL Currents  1mA

I5 0 In all Cases

Prob4_31_KVL_KCL_Plot.m

prob 4 31
Prob 4-31
  • v1 = 100 V
  • All Resistances kΩ
  • v2 Variable
  • Max Resistor Current = 1 mA
plot v2 over 1 400v
Plot v2 over 1-400V

Green Zone

Prob4_31_KVL_KCL_Calc.m

all done for today
All Done for Today

This SpaceForRent

slide6

Engr/Math/Physics 25

Appendix

Time For

Live Demo

Bruce Mayer, PE

Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

prob4 31 kvl kcl plot m 1
Prob4_31_KVL_KCL_plot.m - 1

% 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

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];

%

% Make Loop with v2 as counter in units of Volts

for v2 =1:400 % units of volts

%Constraint Vector V

V = [v1; 0; -v2; 0; 0];

% find soltion vector for currents, C

C = A\V;

% Build plotting vectors for current

vplot(v2) = v2;

i1(v2) = C(1);

i2(v2) = C(2);

i3(v2) = C(3);

i4(v2) = C(4);

i5(v2) = C(5);

end

% PLOT SECTION

plot(vplot,1000*i1,vplot,1000*i2, vplot,1000*i3, vplot,1000*i4, vplot,1000*i5 ),...

ylabel('Resitor Current(mA)'),xlabel('Supply-2 Potential (V)'),...

title('Resistor Network currents'), grid, legend('i1', 'i2', 'i3', 'i4', 'i5')

prob4 31 kvl kcl calc m 1
Prob4_31_KVL_KCL_Calc.m - 1

% 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 = 40;

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]

% Constraint Col Vector V

V = [v1; 0; -v2; 0; 0];

% find solution vector for currents, C

C = A\V;

% 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);

%increment v2 by 10 mV DOWN

v2 = v2 - 0.01;

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

%

prob4 31 kvl kcl calc m 2
Prob4_31_KVL_KCL_Calc.m - 2

% 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;1;1]

%Constraint Vector V

V = [v1; 0; -v2; 0; 0];

% find soltion vector for currents, C

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 + 0.01;

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