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+clc
+//Variable Initialisation
+Ia=180//Armature Current in Ampere
+Ra=0.06//Armature resistance in ohm
+Rb=8
+If=2//Field Current in Ampere
+d=0.5
+K=1.527
+//Solution
+E0=Ia*Rb*(1-d)
+Req=Rb*(1-d)+Ra
+Pb=(Ia^2)*(Rb*(1-d))
+Eb=E0+(Ia*Ra)
+W=Eb/(K*If)
+W1=(W*60)/(2*%pi)
+Ep=Ia*Rb
+printf('\n\n The Average Voltage across chopper=%0.1f Volts\n\n',E0)
+printf('\n\n Equivalent Resistance of motor=%0.1f ohm\n\n',Req)
+printf('\n\n Power dissipated in braking resistor=%0.1f KW\n\n',Pb*10^-3)
+printf('\n\n The Motor Speed=%0.1f rpm\n\n',W1)
+printf('\n\n Peak to Peak Voltage=%0.1f Volts\n\n',Ep)