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+clc;
+Vt=100;//terminnal voltage
+P=2;//no of poles
+Z=1000;//no of conductors
+A=2;//no of parallel paths for armature conductors
+Ra_=2*10e-3;//resistance of each armature
+Ra=500*Ra_*(1/2);//total armature resistance
+//Let If be field current
+//Ea=Vt+(Il+If)*0.5
+//Ea1=100+(10+If)*0.5,because at 1055 rpm Il=10.
+//Ea2=100+(20+If)*0.5,because at 1105 rpm Il=20.
+//But, Ea=k1*If*speed
+//Therefore,((If*1055)/(If*1105))=((100+(10+If)*0.5)/(100+(20+If)*0.5)),which gives-
+If=1;//field current
+Ea1=100+(10+1)*0.5;//at 1055 rpm
+N=1055;//speed of rotor
+phi=(Ea1*60*A)/(Z*N*P);
+Rf=Vt/If;//field circuit resistance
+printf('Field circuit resistance is %f ohm.\n',Rf);
+printf('Flux per pole is %f Wb.',phi);