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// Examle 3.24
// From the diagram (3.42) Node voltages are
// Have { va-vb+0vc = 6 }.......................(1
// Apply KCL at Super node
// will get { 0.33va+0.25vb-0.25vc = 2 }.......(2
// Apply KCL at node c
// will get { 0va-0.25vb+4.5vc = -7 }..........(3
// By using matrix form will get A*X = B formate
delta=[1 -1 0 ; 0.33 0.25 -0.25 ; 0 -0.25 0.45]; // value of A
d=det(delta); // Determinant of A
delta1=[1 6 0 ; 0.33 2 -0.25 ; 0 -7 0.45]; // value of A1 (when 2nd colomn is replace by B)
d1=det(delta1); // Determinant of A1
delta2=[1 -1 6 ; 0.33 0.25 2 ; 0 -0.25 -7]; // value of A2 (when 3rd colomn is replace by B)
d2=det(delta2); // Determinant of A2
Vb=d1/d; // Voltage at node-b
Vc=d2/d; // Voltage at node-c
I=(Vb-Vc)/4; // Current through 4 ohm resistor (I)
disp(' Current through 4 ohm resistor = '+string(I)+' Amp');
// p 82 3.24
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