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+clc
+// Variable Initialization 
+Vm=230//Supply Voltage in Volts
+Ra=0.05//Armature circuit resistance in Ohm
+L=0.85e-3//Inductance in mH
+N=750//Motor speed in Rpm
+V=220//Rated voltage of motor in Volts
+a=60//firing angle in Degree
+La=0.75e-3//External Inductance in mH
+I=175 //motor current in Amp
+f=50 //source Frequency in Hz
+
+//Solution
+P=atand((2*%pi*f*L)/Ra) // In Degree
+Z=sqrt((Ra^2)+((2*%pi*f*L)^2)) //Impedance In Ohm
+Eb=V-(I*Ra) //back emf in Volts
+w=(2*%pi*N)/60 //Angular Speed in rad/sec
+K=Eb/w
+Cot_P=1/(tand(P))
+A=exp(-%pi*Cot_P)
+AA=(Ra*Vm*1.414)/(K*Z)
+Wmc=AA*sind(a-P)*((1+A)/(A-1))//Critical Speed in rad/Sec
+Wrpm=(Wmc*60)/(2*%pi) //speed in rpm
+//Since the motor Speed is greater than critical Value Wmc ,The drive is operating under discontinuous conduction
+//At N=600 Rpm
+N1=600 //Speed in Rpm
+Eb1=(N1/N)*Eb//Back emf in Volts
+//By trial and error method
+b=201.45//Bita in Degree
+Va=((Vm*1.414)*(cosd(a)-cosd(b))+(%pi+(a-b)*(%pi/180))*(Eb1))*(1/%pi) //Armature voltage in Volts
+Ia=(Va-Eb1)/Ra //Armature current in Amp
+T=K*Ia//Torque in N-m
+
+//La=2.85 mH
+N2=-400
+a1=120
+P1=atand((2*%pi*f*La)/Ra) // In Degree
+Cot_P1=1/(tand(P1))
+Eb2=(N2/N)*Eb//Back emf in Volts
+Z1=sqrt((Ra^2)+((2*%pi*f*La)^2)) //Impedance In Ohm
+AA1=(Ra*Vm*1.414)/(K*Z1)
+A1=exp(-%pi*Cot_P1)
+Wmc1=AA1*sind(a1-P1)*((1+A1)/(A1-1))//Critical Speed in rad/Sec
+Wrpm1=(Wmc1*60)/(2*%pi) //speed in rpm
+//By trial and error method
+b1=297.5//Bita in Degree
+Va1=((Vm*1.414)*(cosd(a1)-cosd(b1))+(%pi+(a1-b1)*(%pi/180))*(Eb2))*(1/%pi) //Armature voltage in Volts
+Ia1=(Va1-Eb2)/Ra //Armature current in Amp
+T1=K*Ia1//Torque in N-m
+
+//Since the motor speed (-600 rpm) is less than critical speed (-409.17 Rpm) the drive's opertion is continuous condition
+N3=-600
+Va2=(2*1.414*Vm)*(cosd(a1))*(1/%pi)
+Eb3=(N3/N)*Eb//Back emf in Volts
+Ia2=(Va2-Eb3)/Ra //Armature current in Amp
+T2=K*Ia2//Torque in N-m
+
+//Results
+printf('\n\n The motor Torque =%0.1f N-m \n\n',T)
+printf('\n\n The motor Torque =%0.1f N-m \n\n',T1)
+printf('\n\n The motor Torque =%0.1f N-m \n\n',T2)
+//The answers vary due to round off error