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+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.9
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+P_i = 350 // iron loss of transformer in W
+P_cu = 650 // copper loss of transformer in W
+kVA = 30 // kVA ratingss of transformer
+pf = 0.6 // power factor
+
+// caclulations
+P_tloss = (P_i+P_cu)*10^-3 // total full load loss in kW
+P_out = kVA*pf // o/p power at full load in kW
+P_in = P_out+P_tloss // i/p power at full load
+n_1 = (P_out/P_in)*100 // efficiency at full load
+kVA_out = kVA*sqrt(P_i/P_cu) // o/p kVA corresponding to maximum efficiency
+P_01 = kVA_out*pf // o/p power in W
+P_tloss1 = 2*P_i // maximum efficiency iron loss=copper loss in W
+P_in1 = P_01+P_tloss1*10^-3 // i/p power in kW
+n_2 = (P_01/P_in1)*100 // efficiency
+
+// display the result
+disp("Example 3.9 solution");
+printf(" \n Efficiency at full load \n n_1 = %.2f percent \n", n_1);
+printf(" \n Out put power \n P_01 = %.1f kW \n", P_01);
+printf(" \n Efficiency \n n_2 = %.2f percent \n", n_2);
+