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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 29
+
+disp("CHAPTER 7");
+disp("EXAMPLE 29");
+
+//VARIABLE INITIALIZATION
+va=200000;                             //apparent power
+v1=4000;                              //primary voltage in Volts
+v2=1000;                               //secondary voltage in Volts
+f=50;
+//loads
+pf=1;
+eff=0.97;                               // at full load and at 60% of full load
+nlpf=0.5;                               //no load pf
+lpf=0.8                                 //lagging pf
+reg=0.05;                               //%regulation at 0.8 pf
+//
+//SOLUTION  
+loss=(1-eff)*va/eff;                    //=Pc+Pcu losses
+//simultaneous equation to be solved
+//eq 1: Pc+Pcu=loss;
+//fractipon of copper/ ohmic losses
+f=(0.6)^2;                              // 60% of full load
+//the 2nd equation is Pc+f*Pcu=loss
+//now the matrix
+M=[1,1;1,f];
+A=[loss,loss*0.6];
+Mi=inv(M);
+Ans=A*inv(M);
+Pc=Ans(1,1);
+Pcu=Ans(1,2);
+//disp(sprintf("The Pc is %f",Pc));
+//disp(sprintf("The Pcu is %f",Pcu));
+//LV side
+R_e2=Pcu/va;
+//from %reg find X_e2
+phi=acos(lpf);
+X_e2=(reg-R_e2*cos(phi))/sin(phi);
+//in oms units
+R_e2=R_e2*v2^2/va;                       // in ohms
+X_e2=X_e2*v2^2/va;                       // in ohms
+disp(sprintf("The Re2 is %f Ω",R_e2));
+disp(sprintf("The Xe2 is %f Ω",X_e2));
+//
+Rc=v2^2/Pc; 
+Ie2=Pc/(v2*0.25);
+Ic=Pc/v2;
+Iphi=sqrt(Ie2^2-Ic^2);
+Xphi=v2/Iphi;
+disp(sprintf("The Rc is %f Ω",Rc));
+disp(sprintf("The Ie2 is %f A",Ie2)); 
+disp(sprintf("The Ic is %f A",Ic));
+disp(sprintf("The Iphi is %f A",Iphi)); 
+disp(sprintf("The Xphi is %f Ω",Xphi)); 
+disp(" "); 
+// 
+//END