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+//Taking Base value MVA and KVA
+clear;clc;
+global mvab
+mvab=645; //Base MVA in 3-phase
+kvb=24; //Base KV,line-to-line
+vt=24/kvb; //Terminal voltage
+xs=1.2;
+xs=(xs*mvab)/kvb^2; // xs converted to its pu
+
+//since the generator is operating at full load & 0.9pf
+pf_angle=acos(0.9);
+Ia=1*(cos(pf_angle)-%i*sin(pf_angle)); //load current
+//to find excitation emf
+ef=vt+%i*xs*Ia;
+ef=abs(ef);
+P=1*0.9; //at Full load
+
+/////// writing an inline function /////////////////
+function [pf,lead_lag,Q]=excitation_change(P,ef,vt,xs)
+sin_delta=(P*xs)/(ef*vt);
+delta=asind(sin_delta);
+ef0=ef*(cosd(delta)+(%i*sind(delta)));
+Ia=(ef0-vt)/(%i*xs);
+Ia_mag=abs(Ia);Ia_ang=atand(imag(Ia)/real(Ia)); // Magnitude and angle of Ia
+pf=cosd(abs(Ia_ang));
+if(Ia_ang>0) then lead_lag='leading';
+  elseif (Ia_ang==0) then lead_lag='unity pf' 
+    else lead_lag='lagging';
+  end
+Q=vt*Ia_mag*sind(abs(Ia_ang));
+Q=abs(Q)*mvab;
+endfunction
+////////////////////////////////////////////////////////
+
+
+// First Case when Ef is increased by 20% at same real load now
+ ef1=ef*1.2;
+[pf1,lead_lag1,Q1]=excitation_change(P,ef1,vt,xs);
+disp("Case (i): When Ef is increased by 20% ");
+printf('\n\tPower factor pf= %0.2f %s \n',pf1,lead_lag1);
+printf('\tReactive power drawn by the load = %0.1f MVAR \n',Q1);
+
+//Second Case when Ef is decreased by 20% at same real load now
+ ef2=ef*0.8;
+[pf2,lead_lag2,Q2]=excitation_change(P,ef2,vt,xs);
+disp("Case (ii): When Ef is decreased by 20% ");
+printf('\n\tPower factor pf= %0.2f %s \n',pf2,lead_lag2);
+printf('\tReactive power drawn by the load = %0.1f MVAR \n',Q2);
+
+disp('The answers given here are exact values.Textbook answers has an approximation of upto 2 decimal places on Xs,Ia,pf. ');