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+//Chapter 10
+//Example 10.2
+//page no 390
+//To draw sequence networks of the system
+clear;clc;
+
+//selecting generator rating as base in generator circuit
+
+mvab=25;
+kvGb=11;  //base voltage for generator
+kvTLb=kvGb*(121/10.8); //base voltage for TL
+kvMb=kvTLb*(10.8/121); //base voltage for motors
+
+xG=%i*0.2;
+xT=%i*0.1;
+xTL=100;
+xM=%i*0.25;
+
+mvaG=25;
+mvaT=30;
+mvaM1=15;
+mvaM2=7.5;
+
+kvM=10;
+
+//converting all the reactances to PUs
+
+xT=xT*(mvab/mvaT)*(10.8/kvGb)^2;
+xTL=xTL*(mvab/(kvTLb)^2);
+xM1=xM*(mvab/mvaM1)*(kvM/kvMb)^2;
+xM2=xM*(mvab/mvaM2)*(kvM/kvMb)^2;
+
+//displaying the results
+
+printf('\n\nTransmission line voltage base = %0.1f kV',kvTLb);
+printf('\n\Motor voltage base = %d kV',kvMb);
+printf('\n\nTransformer reactance = %0.4f pu',abs(imag(xT)));
+printf('\nLine reactance = %0.3f pu',abs(xTL));
+printf('\nReactance of motor 1 = %0.3f pu',abs(imag(xM1)));
+printf('\nReactance of motor 2 = %0.3f pu\n\n',abs(imag(xM2)));
+
+disp('Positive and Negative sequence diagram has been drawn using XCOS,simulation has not been done as it is not being asked in the problem');