Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Engr/Math/Physics 25 MidTerm ExamReview Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

Possible Confusion in Text Book All R’s Should be in kΩ Plot at right shows the large currents Generated by Not Using kΩ Problem 4-37 In NO case are ALL Currents  1mA I5 0 In all Cases Prob4_31_KVL_KCL_Plot.m

Prob4-37 • v1 = 100 V • All Resistances kΩ • v2 Variable • Max Resistor Current = 1 mA

Analyze this Circuit using Methods from ENGR43 P4-37 Electrical Engineering • Specifically use • Kirchoff’s Voltage Law (KVL) • Ohm’s Law (V = IR) • Kirchoff’s Current Law (KCL) • These Laws yield Eqns for the currents

P4-37 Applied Math • The 5 eqns for the 5 currents can be cast into Matrix form:

4-37 MATLAB Numerical Processor • Recall: • Use MATLAB’s LEFT Division to Find the solution vector C • PLOT Results to Analyze Circuit BEHAVIOR • Remember – When in Doubt PLOT

Plot v2 over 1-400V Green Zone Prob4_31_KVL_KCL_Calc.m

4-37(b)  v2min(R3), v2max(R3) • Now Allow R3 to vary: 150 kΩ250 kΩ 150 k  250 k • Use solution to part (a) as basis • Vary R3 with For-Loop, then Chk v2 as in part (a)

Plot v2 Max & Min as f(R3) • Use Solution from part(a) in FOR loop that Varies R3 to produce Plot Part (a) Case Part (a) Case

All Done for Today This SpaceForRent

Engr/Math/Physics 25 Appendix Time For Live Demo Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

Prob4_37_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 % 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_37_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

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu