% Prova d'esame Albanese, de Magistris del 01-02-2016

clear
format short g
format compact

[ST,I] = dbstack;
%fprintf(2,['$$$$ ' num2str(100*rand) ' $$$$\n'])
fprintf(2,'$$$$ %0.5g $$$$ %s\n', 100*rand,ST.file)

%% Problema 1
R1 = 2;
R2 = 2;
R3 = 3;
R4 = 5;
E  = 10;
J1 = 2;
J2 = 5;
%
% Metodo 1: Norton
disp('<a href=""> Traccia A, Esercizio 1, Metodo 1: Norton eliminando R2 e J2 </a>')
% elimino R2 e J2 e calcolo Jcc con riferimento verso destra
% Jcc + IR1cc = J1 quindi mi serve IR1cc
vR1cc = (J1+E/R4) / (1/R1+1/R3+1/R4);
iR1cc = vR1cc/R1;
Jcc = J1 - iR1cc
R0 = R1 + paral(R3,R4)
% collego R2 e J2 al generatore equivalente di Norton
J = Jcc+J2;
iR2 = J*R0/(R0+R2)
% verifica Thevenin
vR10 = R1*J1;
vR30 = E*R3/(R3+R4);
E0 = vR10 - vR30
Jcc_chk = E0/R0
%
% Metodo 2: potenziali nodali
disp('<a href=""> Traccia A, Esercizio 1, Metodo 2: potenziali nodali </a>')
% | u1/R1 + (u1-u2)/R2              = J1+J2   | (1/R1+1/R2)*u1 - 1/R2*u2       = J1+J2
% | (u2-u1)/R2 + u2/R3 + (u2-u3)/R4 = -J2     | -1/R2*u1 + (1/R2+1/R3+1/R4)*u2 = -J2 + E/R4
% | u3 = E                                    |
G = [1/R1+1/R2, -1/R2;  -1/R2, 1/R2+1/R3+1/R4]
d = [J1+J2; -J2+E/R4]
u = G\d
iR2 = (u(1)-u(2))/R2

%% Problema 2