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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD 
+
+// CHAPTER : 3 : TRANSFORMERS 
+
+// EXAMPLE : 3.12
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+N1 = 1000;               // 1st Test at No-load condition f1 Frequency, Speed in RPM
+Vo1 = 250;               // 1st Test at No-load condition f1 Frequency, Voltage in Volts
+Io1 = 0.5;               // 1st Test at No-load condition f1 Frequency, Current in Amphere
+Wo1= 230;                // 1st Test at No-load condition f1 Frequency, Power in Watts
+
+N2 = 900;                // 2nd Test at No-load condition f2 Frequency, Speed in RPM
+Vo2 = 225;               // 2nd Test at No-load condition f2 Frequency, Voltage in Volts
+Io2= 0.5;                // 2nd Test at No-load condition f2 Frequency, Current in Amphere
+Wo2 = 200;               // 2nd Test at No-load condition f2 Frequency, Power in Watts
+p = 6;                  // Number of poles of single phase alternator
+N = 220;                // Number of the turns of single phase alternator
+R = 0.66;               // Resistance of the single phase alternator in Ohms
+
+
+// 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-(Io1^2)*R)/f1;         // No-load corrected losses Eq 1 in Watts
+d = (Wo2-(Io2^2)*R)/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
+Ph = f1*A;                      // Hysteresis loss at 50 Hertz in Watts
+Pe = (f1^2)*B;                  // Eddy current loss at 50 Hertz in Watts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.12 : SOLUTION :-") ;
+printf("\n (a) Hysteresis loss at %.f Hertz , Ph = %.3f W \n ",f1,Ph);
+printf("\n (b) Eddy current loss at %.f Hertz , Pe = % .2f W \n",f1,Pe);
+
+
+