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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 3 : TRANSFORMERS
+
+// EXAMPLE : 3.23
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+S = 100; // Rating of the 3-Phase Transformer in kVA
+VH = 11; // HV side voltage in kilo-Volts
+VL = 440; // LV side voltage in Volts
+Vl = 400; // Line voltage in Volts
+ZA = 0.6; // Line impedance in line A in Ohms
+ZB = 0.6*(0.8 + 0.6 * %i); // Line impedance in line B in Ohms
+ZC = 0.6*(0.5 - 0.866 * %i); // Line impedance in line C in Ohms
+
+
+// CALCULATIONS
+
+Vp = Vl/sqrt(3); // Phase voltage in Volts
+VAB = Vl * exp( %i * 0 * %pi/180); // Line Voltage across line A and B in Volts
+VBC = Vl * exp( %i * (-120) * %pi/180); // Line Voltage across line B and C in Volts
+VCA = Vl * exp( %i * 120 * %pi/180); // Line Voltage across line C and A in Volts
+VAN = (Vl/sqrt(3)) * exp( %i * (-30) * %pi/180); // Phase Voltage across line A and Neutral in Volts
+VBN = (Vl/sqrt(3)) * exp( %i * (-150) * %pi/180); // Phase Voltage across line B and Neutral in Volts
+VCN = (Vl/sqrt(3)) * exp( %i * (90) * %pi/180); // Phase Voltage across line C and Neutral in Volts
+IA = VAN/ZA; // Line current in line A in Amphere
+IB = VBN/ZB; // Line current in line B in Amphere
+IC = VCN/ZC; // Line current in line C in Amphere
+IN = IA + IB + IC ; // Current in the Neutral in Amphere
+Y = (1/ZA)+(1/ZB)+(1/ZC); // Net Admittance in mho
+VN = IN/Y; // Neutral Potential in Volts
+VDA = VAN - VN; // Voltage drops across the ZA in Volts
+VDB = VBN - VN; // Voltage drops across the ZB in Volts
+VDC = VCN - VN; // Voltage drops across the ZC in Volts
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 3.23 : SOLUTION :-") ;
+printf("\n (a.1) Line current in line A , IA = %.f<%.f A \n ",abs(IA),atand(imag(IA),real(IA)));
+printf("\n (a.2) Line current in line B , IB = %.f<%.2f A \n",abs(IB),atand(imag(IB),real(IB)));
+printf("\n (a.3) Line current in line C , IC = %.f<%.f A \n ",abs(IC),atand(imag(IC),real(IC)));
+printf("\n (b.1) Phase Voltage across line A and Neutral , VAN = %.f<%.f V \n",abs(VAN),atand(imag(VAN),real(VAN)));
+printf("\n (b.2) Phase Voltage across line B and Neutral , VBN = %.f<%.f V \n ",abs(VBN),atand(imag(VBN),real(VBN)));
+printf("\n (b.3) Phase Voltage across line C and Neutral , VCN = %.f<%.f V \n",abs(VCN),atand(imag(VCN),real(VCN)));
+printf("\n (c) Neutral Potential , VN = %.1f<%.2f V \n ",abs(VN),atand(imag(VN),real(VN)));
+printf("\n\n [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n WRONGLY PRINTED ANSWERS ARE :- (a) IC = 385<-90.1 V instead of %.f<%.f A \n ",abs(IC),atand(imag(IC),real(IC)));
+printf("\n (b) VN = 230.5<78.17 V instead of %.1f<%.2f V \n ",abs(VN),atand(imag(VN),real(VN)) );
+printf("\n From Calculation of the IC, rest all the Calculated values in the TEXT BOOK is WRONG because of the IC value is WRONGLY calculated and the same used for the further Calculation part \n")