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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD 
+
+// CHAPTER : 3 : TRANSFORMERS 
+
+// EXAMPLE : 3.13
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 1500;               // 1st Test on Transformer at f1 Frequency and Vo1 voltage, Speed in RPM
+Vo1 = 250;               // 1st Test on Transformer at f1 Frequency and Vo1 voltage, Voltage in Volts
+Wo1= 55;                 // 1st Test on Transformer at f1 Frequency and Vo1 voltage, Power in Watts
+N2 = 1200;               // 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Speed in RPM
+Vo2 = 200;               // 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Voltage in Volts
+Wo2 = 40;                // 2nd Test on Transformer at f2 Frequency and Vo2 voltage, Power in Watts
+p = 4;                   // Number of poles of single phase alternator
+
+
+// CALCULATIONS
+
+f1 = (N1*p)/120;              // 1st case Supply Frequency in Hertz
+Ratio1 = Vo1/f1;              // 1st case Ratio of the Volatge and Frequency in Volts/Hertz
+f2 = (N2*p)/120;              // 2nd case Supply Frequency in Hertz
+Ratio2 = Vo2/f2;              // 2nd case Ratio of the Volatge and Frequency in Volts/Hertz
+
+c = Wo1/f1;         // No-load corrected losses Eq 1 in Watts
+d = Wo2/f2;         // No-load corrected losses Eq 2 in watts
+
+x = [ 1 f1 ; 1 f2 ];             // No-load corrected losses Eq 1 in watts
+y = [ c ; d ];                   // No-load corrected losses Eq 2 in watts
+
+E = x\y;                        // Solution of constants A in Watts/Hertz and B in watts/Hertz-Sqare in matrix form
+A = E(1,1);                      // Solution of constant A in Watts/Hertz
+B = E(2,1);                      // Solution of constant B in watts/Hertz-Sqare
+Ph1 = f1*A;                      // Hysteresis loss at 50 Hertz in Watts
+Pe1 = (f1^2)*B;                  // Eddy current loss at 50 Hertz in Watts
+Ph2 = f2*A;                      // Hysteresis loss at 40 Hertz in Watts
+Pe2 = (f2^2)*B;                  // Eddy current loss at 40 Hertz in Watts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.13 : SOLUTION :-") ;
+printf("\n (a.1) Hysteresis loss at %.f Hertz , Ph = %.f W \n ",f1,Ph1);
+printf("\n (a.2) Eddy current loss at %.f Hertz , Pe = %.f W \n",f1,Pe1);
+printf("\n (b.1) Hysteresis loss at %.f Hertz , Ph = %.f W \n ",f2,Ph2);
+printf("\n (b.2) Eddy current loss at %.f Hertz , Pe = %.f W \n",f2,Pe2);
+printf("\n\n    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n      WRONGLY PRINTED ANSWERS ARE :- (a) Hysteresis loss at %.f Hertz , Ph = 25 W instead of %.f W \n ",f2,Ph2);