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+//Chapter 4
+//Example 4.2
+//page 104
+// To draw the per unit reactance diagram when pu values are specified based on euipment rating
+clear;clc;
+mvab=30; kvb=11; //MVA base and KVA base are selected in the circuit of generator 1
+
+gen1_mva=30; gen1_kv=10.5; gen1_x=0.435;  //Generator No.1 details
+gen2_mva=15; gen2_kv=6.6; gen2_x=0.413;  //Generator No.2 details
+gen3_mva=25; gen3_kv=6.6; gen3_x=0.3214;  //Generator No.3 details
+
+t1_mva=15; t1_hv=33; t1_lv=11; t1_x=0.209; //Transformer T1 details
+t2_mva=15; t2_hv=33; t2_lv=6.2; t2_x=0.220; //Transformer T1 details
+
+tl_x=20.5; //Transmission line recatance
+
+//Loads are neglected as said in the problem
+
+tl_pu=(tl_x*mvab)/t1_hv^2;
+t1_pu=t1_x*(mvab/t1_mva);
+t2_pu=t2_x*(mvab/t2_mva);
+gen1_pu=gen1_x*(mvab/gen1_mva)*(gen1_kv/kvb)^2;
+gen2_kv_base=t2_lv;
+gen2_pu=gen2_x*(mvab/gen2_mva)*(gen2_kv/gen2_kv_base)^2;
+gen3_kv_base=t2_lv;
+gen3_pu=gen3_x*(mvab/gen3_mva)*(gen3_kv/gen3_kv_base)^2;
+
+//diplaying the results on console
+
+printf('Per unit impedance of the components of the given power system are as follows :\n\n');
+
+printf('Transmission line: %0.3f \n\n',tl_pu);
+
+printf('Transformer T1: %0.3f \n\n',t1_pu);
+
+printf('Transformer T2: %0.3f \n\n',t2_pu);
+
+printf('Generator 1: %0.3f \n\n',gen1_pu);
+
+printf('Generator 2: %0.3f \n\n',gen2_pu);
+
+printf('Generator 3: %0.3f \n\n',gen3_pu);
+