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+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION 
+
+// PART II : TRANSMISSION AND DISTRIBUTION
+// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
+
+// EXAMPLE : 11.13 :
+// Page number 337-338
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l = 20.0                 // Length of cable(km)
+r = 0.248                // Resistance(ohm/km)
+x = 0.50*10**-3          // Inductance(H/m)
+V_gen = 6600.0           // Generation voltage(V)
+f = 50.0                 // Frequency(Hz)
+V = 33000.0              // Transmission voltage(V)
+rating = 10.0            // Transformer rating(MVA)
+loss_cu = 100.0          // Copper loss at full load(kW)
+x_tr = 2.5/100           // Transformer reactance
+load = 7.5               // Load to be transmitted(MW)
+PF = 0.71                // Lagging power factor
+
+// Calculations
+R = l*r                                                         // Resistance of the cable(ohm)
+I_fl = rating*10**6/(3**0.5*V)                                  // Transformer current at full load(A)
+R_eq = loss_cu*1000/(3*I_fl**2)                                 // Equivalent resistance per phase of transformer(ohm)
+R_total_hv = R+2.0*R_eq                                         // Total resistance per conductor in terms of hv side(ohm)
+X = 2.0*%pi*f*l*x                                               // Reactance of cable per conductor(ohm)
+per_X_tr = V/3**0.5*x_tr/I_fl                                   // % reactance of transformer(ohm)
+X_total_hv = X+2.0*per_X_tr                                     // Total reactance per conductor in terms of hv side(ohm)
+I = load*10**6/(3**0.5*V*PF)                                    // Line current at receiving end(A)
+IR = I*R_total_hv                                               // IR drop(V)
+IX = I*X_total_hv                                               // IX drop(V)
+E_r = V/3**0.5                                                  // Phase voltage at station B(V)
+cos_phi_r = PF
+sin_phi_r = (1-PF**2)**0.5
+E_s = ((E_r*cos_phi_r+IR)**2+(E_r*sin_phi_r+IX)**2)**0.5/1000   // Sending end voltage(kV)
+E_s_ll = 3**0.5*E_s                                             // Sending end line voltage(kV)
+V_booster = 3**0.5*(E_s-E_r/1000)                               // Booster voltage between lines(kV)
+tan_phi_s = (E_r*sin_phi_r+IX)/(E_r*cos_phi_r+IR)               // tanΦ_s
+phi_s = atand(tan_phi_s)                                        // Φ_s(°)
+cos_phi_s = cosd(phi_s)                                         // cosΦ_s
+P_s = 3.0*E_s*I*cos_phi_s                                       // Power at sending end(kW)
+loss = P_s-load*1000                                            // Loss(kW)
+loss_per = loss/(load*1000)*100                                 // loss percentage
+
+// Results
+disp("PART II - EXAMPLE : 11.13 : SOLUTION :-")
+printf("\nLoss in the interconnector as a percentage of power received = %.3f percent", loss_per)
+printf("\nRequired voltage of the booster = %.3f kV (in terms of H.V) \n", V_booster)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
+printf("\n      kVA rating of booster is not calculated in textbook and here")