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+
+// ELECTRICAL MACHINES
+// R.K.Srivastava
+// First Impression 2011
+// CENGAGE LEARNING INDIA PVT. LTD
+
+// CHAPTER : 5 : INDUCTION MACHINES
+
+// EXAMPLE : 5.24
+
+clear ; clc ; close ; // Clear the work space and console
+
+
+// GIVEN DATA
+
+m = 3;                          // Total Number of phase in Induction Motor
+f = 50;                         // Frequency in Hertz
+V = 440;                        // Operating voltage of the Induction Motor in Volts
+R1 = 0.2;                      // Circuit Parameter in Ohms
+R2 = 0.4;                      // Circuit Parameter in Ohms
+X1 = 1.0;                      // Circuit Parameter in Ohms
+X2 = 1.5;                      // Circuit Parameter in Ohms
+Rc = 150;                      // Circuit Parameter in Ohms
+Xm = 30;                       // Circuit Parameter in Ohms
+
+
+// CALCULATIONS
+
+V1 = V/sqrt(3);                                     // Rated Voltage in Volts
+Zdol = (R1+%i*X1)+(Rc*%i*Xm*(R2+%i*X2))/(Rc*%i*Xm+Rc*(R2+%i*X2)+(%i*Xm)*(R2+%i*X2));                                           // Equivalent impedance per phase in DOL starter in Ohms
+I = V1/Zdol;                                        // Starting Current in DOL starter in Amphere
+
+// For Case(a) A per Phase resistance of 0.5 Ohms is added in Series with the stator circuit
+
+Zsr =  (0.5+R1+%i*X1)+((Rc*%i*Xm*(R2+%i*X2))/((Rc*%i*Xm+Rc*(R2+%i*X2)+(%i*Xm)*(R2+%i*X2))));                                     // Total impedance seen from the terminals in Ohms
+Isr = V1/Zsr;                                         // Starting Current in DOL starter in Amphere
+
+// For Case(b) A per Phase resistance of 0.5 Ohms is added in Series with the rotor circuit here assumed that stator to rotor turn ratio is 1.0
+
+Zrr =  (R1+%i*X1)+((Rc*%i*Xm*(0.5+R2+%i*X2))/(Rc*%i*Xm+Rc*(0.5+R2+%i*X2)+(%i*Xm)*(0.5+R2+%i*X2)));                            // Total impedance seen from the terminals in Ohms
+Irr = V1/Zrr;                                      // Starting Current in DOL starter in Amphere
+
+// For Case(c) When applied Voltage reduced to 50%
+
+I_c = (0.5*V1)/Zdol;                                // Starting Current in DOL starter in Amphere
+
+// For Case(d) When Motor is supplied by reduced Voltage of 44V ( Voltage is reduced by 10%) and the reduced frequency is 5Hz
+
+f_n = 5;                                          // Reduced Frequency in Hertz
+X1_n = (f_n/f)*X1;                                // Changed Circuit Parameter in Ohms
+X2_n = (f_n/f)*X2;                                // Changed Circuit Parameter in Ohms
+Xm_n = (f_n/f)*Xm;                                // Changed Circuit Parameter in Ohms
+Zdol_n = (R1+%i*X1_n)+((Rc*%i*Xm_n*(R2+%i*X2_n))/(Rc*%i*Xm_n+Rc*(R2+%i*X2_n)+(%i*Xm_n)*(R2+%i*X2_n)));                         // Equivalent impedance per phase in DOL starter in Ohms
+I_n = (V1*0.1)/Zdol_n;                            // Starting Current in DOL starter in Amphere
+Ratio = abs(I_n)/abs(I);                          // Ratio of the Starting Current witha the rated Voltage and frequency to the reduced Voltage and frequency
+
+
+// DISPLAY RESULTS
+
+disp("EXAMPLE : 5.24 : SOLUTION :-");
+printf("\n Normal Initial Starting Current in DOL starter, I = %.1f <%.1f A \n",abs(I),atand(imag(I),real(I)))
+printf("\n For Case(a) A per Phase resistance of 0.5 Ohms is added in Series with the stator circuit \n")
+printf("\n              Initial Starting Current in DOL starter, I = %.1f <%.2f A \n",abs(Isr),atand(imag(Isr),real(Isr)))
+printf("\n  For Case(b) A per Phase resistance of 0.5 Ohms is added in Series with the rotor circuit \n")
+printf("\n              Initial Starting Current in DOL starter, I = %.2f <%.1f A \n",abs(Irr),atand(imag(Irr),real(Irr)))
+printf("\n  For Case(c) When applied Voltage reduced to 50 percentage \n")
+printf("\n              Initial Starting Current in DOL starter, I = %.2f <%.1f A \n",abs(I_c),atand(imag(I_c),real(I_c)))
+printf("\n  For Case(d) When Motor is supplied by reduced Voltage of 44V ( Voltage is reduced by 10 percenatge ) and the reduced frequency is 5Hz \n")
+printf("\n              Initial Starting Current in DOL starter, I = %.1f <%.1f A \n",abs(I_n),atand(imag(I_n),real(I_n)))
+printf("\n By reducing volatge as well as the frequency, the peak starting current at the instant os starting is reduced by a fector of %.4f of the starting current with the reted volatge and frequency \n",Ratio)
+printf("\n\n    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY ( I verified by manual calculation )]\n" );
+printf("\n      WRONGLY PRINTED ANSWERS ARE :- For Case(d) When Motor is supplied by reduced Voltage of 44V ( Voltage is reduced by 10 percenatge ) and the reduced frequency is 5Hz,  I = 24.1 < 25.6 A instead of %.1f < (%.2f) A \n ",abs(I_n),atand(imag(I_n),real(I_n)));
+printf("\n  Ratio of the Starting Current with the rated Voltage and frequency to the reduced Voltage and frequency, Ratio = 0.2518 instead of %.4f \n ",Ratio);
+