% Newton Raphson - Problem 1


close;
clear;
clc;


P2 = 1.8  % P p.u.
Q2 = 0.6  % Q p.u.
Z= i*0.1; % impedance

Y = [1/Z , -1/Z ;
     -1/Z , 1/Z]; % admittance matrix (2x2)

Gik = real(Y); % takes real component of Y matrix (ie Gik) into 2x2 matrix
Bik = imag(Y); % take imaginary cmponents of Y matrix (ie Bik) into 2x2 matrix

x = [0;1];  % initial conditions


for n = 1:50 % n = # iterations


f = [Bik(2,1)*x(2)*sin(x(1)) + P2;

     -Bik(2,1)*x(2)*cos(x(1)) - Bik(2,2)*x(2)^2 + Q2];


J = [Bik(2,1)*x(2)*cos(x(1)) , Bik(2,1)*sin(x(1));
     Bik(2,1)*x(2)*sin(x(1)) , -Bik(2,1)*cos(x(1))-2*Bik(2,2)*x(2)];


x = x - inv(J)*f;

end

k1 = x(1,1); % this is the angle in radians
k2 = x(2,1); % this is the voltage (V2)
k3 = k1*(180/pi); % radians to degree


fprintf('V =');
disp(x(2,1)); % displays voltage
fprintf('theta =');
disp(k3); % Displays angle (in degrees)