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+//Chapter 9
+//Example 9.1
+//page 335
+//To calculate fault current 
+clear;clc;
+//selecting base KVA and MVA
+mvab=100;
+Gmva=10;
+T1mva=10; T2mva=5;
+Gkvb=11; //generator kV base
+OHLkvb=33; //overhead line kV base
+Ckvb=6.6;// cable kB base
+xg1=%i*0.15; xg2=%i*0.125; xt1=%i*0.10; xt2=%i*0.08;
+xOHL=0.27+%i*0.36 ; xcab= 0.135+%i*0.08;
+
+//clculating PU impedances
+
+xg1=(xg1*mvab)/Gmva;
+xg2=(xg2*mvab)/Gmva;
+xt1=(xt1*mvab)/T1mva;
+xt2=(xt2*mvab)/T2mva;
+xOHL=(30*xOHL*mvab)/(OHLkvb^2);
+xcab=(3*xcab*mvab)/(Ckvb^2);
+//displaying results
+printf('\n Reactance of G1= j%0.1f pu \n',abs(imag(xg1)));
+printf(' Reactance of G2= j%0.1f pu\n',abs(imag(xg2)));
+printf(' Reactance of T1= j%0.1f pu\n',abs(imag(xt1)));
+printf(' Reactance of T2= j%0.1f pu\n',abs(imag(xt2)));
+printf(' Overhead line impedance=(%0.3f + j%0.3f) pu\n',real(xOHL),abs(imag(xOHL)));
+printf(' Cable impedance= (%0.3f + j%0.3f) pu\n',real(xcab),abs(imag(xcab)));
+
+// Impedance diagram is as shown in the figure9.7 in the textbook
+// A XCOS simulation for this proble is done to explain the subtransient,transient and steady state periods of a symmetrical short circuit
+xtotal=((xg1*xg2)/(xg1+xg2)+xt1+xt2+xOHL+xcab);
+Isc_pu=(1/xtotal);
+Ibase=(mvab/(sqrt(3)*Ckvb))*1000;
+Isc=Isc_pu*Ibase;
+x_F_to_bus=(xt1+xt2+xOHL+xcab);
+v_11b=x_F_to_bus*Isc_pu*11;
+//displaying results
+printf('\nTotal impedance= %0.1f < %0.2f deg pu \n',abs(xtotal),atand(imag(xtotal)/real(xtotal)));
+printf('Short circuit current= %d A\n',abs(Isc));
+printf('Voltage at 11kV bus=%0.2f kV\n',abs(v_11b));
+
+