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+// FUNDAMENTALS OF ELECTICAL MACHINES 
+// M.A.SALAM 
+// NAROSA PUBLISHING HOUSE 
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.8
+clc;clear; // clears the console and command history 
+
+// Given data
+P_i = 1   // iron loss of transformer in kW
+P_cu = 2  // copper loss of transformer in kW
+kVA = 200 // kVA ratingss of transformer
+pf = 0.95 // power factor
+
+// caclulations 
+P_cu1 = (3/4)^2*P_cu   // copper loss at 1/2th of full load in kW
+P_cu2 = (1/2)^2*P_cu   // copper loss at 1/2th of full load in kW
+P_01 = (3/4)*kVA*P_i   // o/p power at 3/4 full load and unity power factor in kW
+P_in1 = P_01+P_i+P_cu1 // i/p power at 3/4 full load and unity power factor in kW
+n_1 = (P_01/P_in1)*100 // efficiency at 3/4 full load and unity power factor
+P_02 = (1/2)*kVA*pf   // o/p power factor at1/2 full load and 0.95 power factor in kW
+P_in2 = P_02+P_i+P_cu2 // i/p power at 1/2 full load and 0.95 power factor in kW
+n_2 = (P_02/P_in2)*100 // efficiency at 1/2 full load and 0.95 power factor
+
+// display the result 
+disp("Example 3.8 solution");
+printf(" \n Efficiency at 3/4 full load and unity power factor \n n_1 = %.2f percent \n", n_1);
+printf(" \n Efficiency at 1/2 full load and 0.95 power factor \n n_2 = %.2f percent \n", n_2);
+