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+clc
+// Variable Initialization 
+Vm=230//Supply Voltage in Volts
+Ra=0.5//Armature circuit resistance in Ohm
+Irms=25 //Armature current in Amp
+Nr=800 //Motor speed in Rpm
+Kaf=0.172 //Motor Voltage Constant in V/rpm
+a=60//firing angle in Degree
+
+//Solution
+//CASE:A
+//For motoring action
+Ka=(Kaf*60)/(2*%pi)//Constant in V-s/rad
+T=Ka*Irms //Torque of motor in N-m
+Va=(2*Vm*1.414)*cosd(a)*(1/%pi)//Average voltage in Volts
+Eb=Va-(Irms*Ra)//Back Emf in Volts
+N=Eb/Kaf//Speed of motor in Rpm
+//The supply current is square wave if motor current is constant and ripple-free with Amplitude 25A
+P=Vm*Irms //Supply VA in Watt
+//Power from supply is real power if losses in converter are neglected
+Ps=Va*Irms //Power in Watt
+pf=Ps/P //Power factor lag
+
+//CASE:B
+//For polarity reversal (regeneration action)
+Eb1=-Eb //Back emf in Volts
+Va1=Eb1+(Irms*Ra)
+af=acosd((Va1*%pi)/(2*Vm*1.414))//Firing angle in Degree
+//Power fed from DC Machine
+Pdc=Eb*Irms //Power in watt
+//Power  lost in armature resistance
+PL=((Irms)^2)*Ra //Power in Watt
+//Power fed back to ac supply is
+PF=Pdc-PL //Power in watt
+
+//Results
+printf('\n\n The motor Torque=%0.1f N-m \n\n',T)
+printf('\n\n The motor Speed  =%0.1f RPM \n\n',N)
+printf('\n\n The Supply Power Factor=%0.1f Lag\n\n',pf)
+printf('\n\n The Firing Angle=%0.1f Degree\n\n',af)
+printf('\n\n The Power fed back to Supply=%0.1f Watt\n\n',PF)
+//The answers vary due to round off error