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+//CHAPTER 7- SINGLE PHASE TRANSFORMER
+//Example 36
+
+disp("CHAPTER 7");
+disp("EXAMPLE 36");
+
+//100kVA 50 Hz 440/1100 V single phase transformer
+//VARIABLE INITIALIZATION
+va=100000;                               //apparent power
+v1=440;                                  //primary voltage in Volts
+v2=11000;                                //secondary voltage in Volts
+f=50;                                   // frequency
+//loads
+pf=1;                                    //power factor unity
+eff1=0.985;                              // at full load at 0.8pf
+eff2=0.99;                               //at half full load at unity pf
+pf1=0.8;                                 // power factor
+pf2=1;                                   //power factor
+//
+//SOLUTION  
+loss1=(1-eff1)*va*pf1/eff1;                    //=Pc+Pcu losses
+loss2=(1-eff2)*va*(1/2)*pf2/eff2;                    //=Pc+Pcu losses
+//simultaneous equation to be solved
+//eq 1: Pc+Pcu=loss;
+//fractipon of copper/ ohmic losses
+f=(1/2)^2;                              // 60% of full load
+//the 2nd equation is Pc+f*Pcu=loss
+//now the matrix
+M=[1,1;1,f];                            //Pc+Pcu=loss1; Pc+(1/2)^2*Pcu=loss2: 1,1,; 1,f
+A=[loss1,loss2];
+Mi=inv(M);
+Ans=A*inv(M);
+Pc=Ans(1,1);
+Pcu=Ans(1,2);
+disp(sprintf("The Pc is %.1f W",Pc));
+disp(sprintf("The Pcu is %.1f W",Pcu));
+//
+//maximumefficiency at farction x times the full load;and then f.Pcu=Pc
+x=sqrt(Pc/Pcu);
+disp(sprintf("The maximum efficiency would occur at a load of  %.0f kVA",x*va/1000));
+I1=va/v1;
+I1maxEff=I1*x;
+disp(sprintf("The current at maximum efficeincy is %.0f A",I1maxEff)); 
+disp(" "); 
+// 
+//END