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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 18: POWER DISTRIBUTION SYSTEMS
+
+// EXAMPLE : 18.8 :
+// Page number 442-443
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V = 440.0            // Voltage between outer(V)
+I_pos = 210.0        // Ligting load current on positive side(A)
+I_neg = 337.0        // Ligting load current on negative side(A)
+I_power = 400.0      // Power load current(A)
+P_loss = 1.5         // Loss in each balancer machine(kW)
+
+// Calculations
+P = I_power*V/1000.0                          // Power(kW)
+load_pos = I_pos*V*0.5/1000.0                 // Load on positive side(kW)
+load_neg = I_neg*V*0.5/1000.0                 // Load on negative side(kW)
+loss_total = 2*P_loss                         // Total loss on rotary balancer set(kW)
+load_main = P+load_pos+load_neg+loss_total    // Load on main machine(kW)
+I = load_main*1000/V                          // Current(A)
+I_M = I-610.0                                 // Current through balancer machine(A)
+I_G = 127.0-I_M                               // Current through generator(A)
+output_G = I_G*V*0.5/1000.0                   // Output of generator(kW)
+input_M = I_M*V*0.5/1000.0                    // Input to balancer machine(kW)
+
+// Results
+disp("PART II - EXAMPLE : 18.8 : SOLUTION :-")
+printf("\nLoad on the main machine = %.2f kW", load_main)
+printf("\nOutput of generator = %.2f kW", output_G)
+printf("\nInput to balancer machine = %.2f kW", input_M)