[email protected] • ENGR-43_Chp3_Nodal_Analysis.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
Chp3 Chp3 Nodal Nodal
Analysis & Analysis & MATLABMATLAB
[email protected] • ENGR-43_Chp3_Nodal_Analysis.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Chp3_Nodal_Analysis.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Chp3_Nodal_Analysis.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Chp3_Nodal_Analysis.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Last of The Hand-SolnLast of The Hand-Soln
[email protected] • ENGR-43_Chp3_Nodal_Analysis.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The .m Script FileThe .m Script File% ENGR43_Chp3_Nodal_Analysis_MATLAB_Tutorial_0602.m% Bruce Mayer, PE% ENGR43 * 05Feb05%% The Coeffiecent Matrix A (unitless)A = [1 0 0 0; 0 0 -1 1; -4 7 -2 0; 30 -25 -4 -10]%% The Constraint Vector BB = [40; 60; 0; -200] % in Volts%% Solve by MATLAB "Back Division"V = A\B;disp('The Solution Vector, [V1; V2; V3; V4] in Volts = ')disp(V)%% Power to I-Src is V4*10mAdisp('Power Supplied by 10 mA current source in mW =')disp(V(4)*10)
[email protected] • ENGR-43_Chp3_Nodal_Analysis.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The SolutionThe Solution From
ENGR43_Chp3_Nodal_Analysis_MATLAB_Tutorial_0602.m
A = 1 0 0 0 0 0 -1 1 -4 7 -2 0 30 -25 -4 -10
B = 40 60 0 -200
The Solution Vector, [V1; V2; V3; V4] in Volts = 40.0000 25.9459 10.8108 70.8108
Power Supplied by 10 mA current source in mW = 708.1081
