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+clc
+//Variable Initialisation
+Ea=230//Input Voltage of motor in volts
+Ia=30//Armature Current in Ampere
+Ia2=60//Armature Current in Ampere
+N1=1000//Rated Speed of Motor in rpm
+N2=800//Rated Speed of Motor in rpm
+Ra=0.7//Armature resistance in ohm
+d2=0.6//Duty Ratio
+d3=0.9//Duty Ratio
+d4=0.9//Duty Ratio
+//Solution
+Eb1=Ea-(Ia*Ra)
+Eb2=(N2/N1)*Eb1
+E01=Eb2-(Ia*Ra)
+d1=E01/Ea
+E02=d2*Ea
+Eb3=E02+(Ia*Ra)
+N3=(Eb3/Eb1)*N1
+E03=d3*Ea
+Eb4=E03+(Ia2*Ra)
+N4=(Eb4/Eb1)*N1
+E04=d4*Ea
+Pin=E04*Ia2
+printf('\n\n Duty Ratio Of Chopper=%0.1f\n\n',d1)
+printf('\n\n Motor Speed for duty ratio 0.6=%0.1f rpm\n\n',N3)
+printf('\n\n Maximum Aloowable Speed=%0.1f rpm\n\n',N4)
+printf('\n\n Power Fed to Source=%0.1f KW\n\n',Pin*10^-3)
+//The answers vary due to round off error