//CHAPTER 7- SINGLE PHASE TRANSFORMER //Example 4 disp("CHAPTER 7"); disp("EXAMPLE 4"); //VARIABLE INITIALIZATION va=10*1000; //apparent power N1=50; //number of turns on primary side N2=10; //number of turns on secondary side v1=440; //primary voltage in Volts f=50; //in Hertz //SOLUTION //solution (a) v2=v1*(N2/N1); disp(sprintf("(a) The secondary voltage on no load is %d V",v2)); //solution (b) I1=va/v1; disp(sprintf("(b) The full load primary current is %f A",I1)); I2=va/v2; disp(sprintf("The full load secondary current is %f A",I2)); //solution (c) phi_m=v2/(4.44*N1*N2); disp(sprintf("(c) The maximum value of the flux is %f mWb",phi_m*1000)); //END