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//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
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