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+clc
+//Variable Initialisation
+Ea=200//Input Voltage of motor in volts
+Ia=20//Armature Current in Ampere
+Ra=0.33//Armature resistance in ohm
+La=11e-3//Armature Inductance in ohm
+N1=1200//Rated Speed of Motor in rpm
+N2=800//Rated Speed of Motor in rpm
+f=500//Chopper Frequency in Hz
+//Solution
+T=1/f
+t=Ra/La
+t1=1/t
+Eb1=Ea-(Ia*Ra)
+Eb2=(N2/N1)*Eb1
+E0=Eb2+(Ia*Ra)
+d=E0/Ea
+ton1=d*T
+A=log(1+((Eb2/Ea)*((%e^(T/t1))-1)))//Ia2=0 & A=ton2/t
+ton2=A*t1
+printf('\n\n Duty Cycle=%0.1f\n\n',ton2)
+//The answer provided in the textbook is wrong(answer given in textbook is in invalid range)
+if ton2<ton1 then disp('Current is Continuous')
+end