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+//Chapter 9
+//Example 9.2
+//page 337
+//To calculate subtransient and momentary current 
+clear;clc;
+mvab=25;
+Gmva=25;
+T1mva=25; T2mva=25;
+Gkvb=11; //generator kV base
+OHLkvb=66; //overhead line kV base
+Mkvb=6.6; //motor kV base
+Mmva=5; //motor mva
+
+XdG=%i*0.2; //Generator's subtransient reactance
+XdM=%i*0.25; //Motor's subtransient reactance
+XdM2=%i*0.3; //Motor's transient reactance
+Xt1=%i*0.1; // step up transformer's reactance
+Xt2=%i*0.1;//step down transformer's reactance
+Xtl=%i*0.15 ;//trnasmission line's reactance
+
+//per unit calculation
+XdM=(XdM*mvab)/Mmva ;//perunit impedance of each motor
+printf('\nSubtransient reactance of each motor = j%0.2f pu\n',abs(XdM));
+
+//(a)subtransient current in the fault
+Isc=(3*(1/XdM))+(1/(XdG+Xt1+Xt2+Xtl));
+Ibase=(mvab*1000)/(sqrt(3)*Mkvb);
+Isc=Isc*Ibase;
+printf('\nSubtransient current in the fault =%0.1fA\n',abs(Isc));
+
+//(b)subtransient current in the breaker B
+IscB=(2*(1/XdM))+(1/(XdG+Xt1+Xt2+Xtl));
+IscB=IscB*Ibase;
+printf('\nSubtransient current in breaker B=%0.1fA\n',abs(IscB));
+
+//(c) to find the momentary current through breaker B
+ImomB=1.6*IscB;
+printf('\nMomentary current through the breaker B=%dA\n',abs(ImomB));
+
+//(d) to compute current to be interrupted by breaker in 5 cycles
+XdM2=(XdM2*mvab)/Mmva ;//perunit transient impedance of each motor
+IscB=(2*(1/XdM2))+(1/(XdG+Xt1+Xt2+Xtl));
+IscB=IscB*Ibase;
+ImomB=1.1*IscB;
+printf('\nCurrent to be interrupted by breaker B in five cycles=%dA\n',abs(ImomB));
+
+
+