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+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-15
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P_o = 2275 ; // Original kVA 
+Q_o = 1410 ; // Original kvar 
+S_f_conjugate = 3333.3 ; // final kVA of the load
+S_o_conjugate = P_o + %i*Q_o ; // Load of the alternator in kVA
+S_o_conjugate_m = abs(S_o_conjugate);//S_o_conjugate_m = magnitude of S_o_conjugate in kVA
+S_o_conjugate_a = atan(imag(S_o_conjugate) /real(S_o_conjugate))*180/%pi;
+//S_o_conjugate_a=phase angle of S_o_conjugate in degrees
+
+disp("Example 8-15");
+printf(" \n  Power tabulation grid : \n ");
+printf(" \n  \t\t P \t\t ±jQ \t\t S* ");
+printf(" \n  \t\t(kW) \t\t(kvar) \t\t(kVA) \t\t cosӨ ");
+printf(" \n  ________________________________________________________________________");
+printf(" \n  Original : \t%d \t\t j%.f \t\t %.1f \t%.2f lag",real(S_o_conjugate) ,imag(S_o_conjugate) ,S_o_conjugate_m,cosd(S_o_conjugate_a));
+printf(" \n  Added : \t0.8x \t\t j0.6x \t\t x \t\t0.80 lag" );
+printf(" \n  Final :    (%d + 0.8x) \tj(%.f + 0.6x)   %.1f \t0.841 lag\n",real(S_o_conjugate) ,imag(S_o_conjugate),S_f_conjugate );
+
+// Calculations
+// case a
+// Assume x is the additional kVA load. Then real and quadrature powers are 0.8x and j0.6x
+// respectively,as shown. Adding each column vertically and using the Pythagorean theorem,
+// we may write (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this eqution yields
+// the quadratic x^2 + 5352x -3947163 = 0. Applying the quadratic yields the added kVA load:
+x = poly(0,'x'); // Defining a polynomial with variable 'x' with  root at 0
+p = -3947163 + 5352*x + x^2 
+a = 1 ; // coefficient of x^2
+b = 5332 ; // coefficient of x
+c = -3947163 ; // constant
+
+// Roots of p
+x1 = ( -b + sqrt (b^2 -4*a*c ) ) /(2* a);
+x2=( -b - sqrt (b^2 -4*a*c ) ) /(2* a);
+
+// case b
+P_a = 0.8*x1 ; // Added active power of the additional load in kW
+Q_a = 0.6*x1 ; // Added reactive power of the additional load in kvar
+
+// case c
+P_f = P_o + P_a ; // Final active power of the additional load in kW
+Q_f = Q_o + Q_a ; // Final reactive power of the additional load in kvar 
+
+// case d
+PF = P_f / S_f_conjugate ; // Final power factor
+// Validity check
+S_conjugate_f = P_f + %i*Q_f ; // Final kVA of the load 
+S_conjugate_f_m = abs(S_conjugate_f);//S_conjugate_f_m = magnitude of S_conjugate_f in kVA
+S_conjugate_f_a = atan(imag(S_conjugate_f) /real(S_conjugate_f))*180/%pi;
+//S_conjugate_f_a=phase angle of S_conjugate_f in degrees
+
+// Display the results
+
+disp(" Solution : ")
+
+printf(" \n a: The given data is shown in the above power tabulation grid.Assume");
+printf(" \n    x is the additional kVA load. Then real and quadrature powers are");
+printf(" \n    0.8x and j0.6x respectively,as shown.Adding each column vertically");
+printf(" \n    and using the Pythagorean theorem, we may write");
+printf(" \n    (2275 + 0.8x)^2 + (1410 + 0.6x)^2 = (3333.3)^2, and solving this");
+printf(" \n    equation yields the quadratic as follows : \n");
+printf(" \n    x^2 + 5332x -3947163 = 0. \n ")
+printf(" \n    Applying the quadratic yields the added kVA load:");
+printf(" \n    Roots of quadratic Eqn p are \n ");
+printf(" \n    x1 = %.2f \n    x2 = %.2f ", x1, x2 );
+printf(" \n    Consider +ve value of x for added kVA so");
+printf(" \n    x = S*a = %.2f kVA \n ", x1 );
+
+printf(" \n b: P_a = %.1f kW \n ", P_a );
+printf(" \n    Q_a in kvar = \n");disp(%i*Q_a);
+
+printf(" \n c: P_f = %.1f kW \n ", P_f );
+printf(" \n    Q_f in kvar = \n");disp(%i*Q_f);
+
+printf(" \n d: PF = cosθ_f = %.3f lagging \n ", PF );
+printf(" \n    Validity check\n    S*f = ");disp(S_conjugate_f);
+printf(" \n    S*f = %.1f <%.2f kVA \n",S_conjugate_f_m,S_conjugate_f_a);
+printf(" \n    PF = cos(%.1f) = %.3f lagging",S_conjugate_f_a ,cosd(S_conjugate_f_a));