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+//Chapter 5
+//Example 5.3
+//page 135
+//problem with mixed end condition
+clc;clear;
+Vr=3000; //receiving end voltage
+pfs=0.8; //sending end power factor
+Ps=2000*10^3; //sending end active power
+z=0.4+%i*0.4; //series impedance
+Ss=Ps/pfs; //sending end VA 
+Qs=Ss*sqrt(1-pfs^2); //sending end reacive power
+
+//by substituting all the values to the equation (iii)
+deff('[y]=fx(I)',"y=(Vr^2)*(I^2)+2*Vr*(I^2)*(real(z)*((Ps-real(z)*(I^2))/Vr)+imag(z)*((Qs-imag(z)*(I^2))/Vr))+(abs(z))^2*(I^4)-(Ss^2)");
+I=fsolve(100,fx);
+
+pfR=(Ps-real(z)*(I^2))/(Vr*I); //Cos(phi_r)
+Pr=Vr*I*pfR;
+Vs=(Ps/(I*pfs));
+
+printf('\nLoad Current |I|= %0.2f A',I);
+printf('\nLoad Pr=%d W',Pr);
+printf('\nReceiving end power factor=%0.2f',pfR);
+printf('\nSupply Voltage=%0.2fV',Vs);