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+clc
+//Variable Initialisation
+Ea=200//Input Voltage of motor in volts
+Ra=0.12//Armature resistance in ohm
+La=12e-3//Armature Inductance in ohm
+K=2//Motor constant in V-s/rad
+Eb=150//Motor back EMF
+Ia=30//Armature Current in Ampere
+f=300//Chopper Frequency
+//Solution
+T=1/f
+d=(Eb+(Ia*Ra))/Ea
+ton=d*T
+toff=(1-d)*T
+t=Ra/La
+Ea1=Ea
+Imin=poly(0,'Imin');
+Ia1=((Ea1-Eb)/Ra)*(1-%e^(-ton*t))+(Imin*%e^(-ton*t))
+disp (Ia1 ,"Imax is")
+Ea2=0
+Imax=poly(0,'Imax');
+Ia2=((Ea2-Eb)/Ra)*(1-%e^(-toff*t))+(Imax*%e^(-toff*t))
+disp (Ia2 ,"Imin is")
+a=poly(0,'a');
+b=poly(0,'b');
+Imax1=(10.409+(0.975*(-9.96)))/(1-(0.975*0.992))//From above displayed values and rounding off
+Imin1=(-9.960)+(0.992*Imax1)
+Im=Imax1-Imin1//Armature Current Excursion
+printf('\n\n Maximum Armature Current=%0.1f Amp\n\n',Imax1)
+printf('\n\n Minimum Armature Current=%0.1f Amp\n\n',Imin1)
+printf('\n\n Armature Current Excursion=%0.1f Amp\n\n',Im)