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+clc
+// Variable Initialization 
+Vm=200 //Supply Voltage in Volts
+N=1000 //speed of motor in RPM
+I=13 //Motor current in Ampere
+Ra=3 //Armature circuit resistance in Ohm
+L=100e-3 //Armature circuit Inductance in mH
+V=230 //Ac Source voltage in Volts
+f=50 //source Frequency in Hz
+a=30
+aa=(30*%pi)/180 //in rad/sec
+N2=400
+//Solution
+P=atand((2*%pi*f*L)/Ra) // In Degree
+Cot_P=1/(tand(P))
+A=exp(-%pi*Cot_P)
+B=exp((-aa)*Cot_P)
+Z=sqrt((Ra^(2))+((2*%pi*f*L)^2)) //Impedance In Ohm
+Eb=Vm-(I*Ra) //back emf in Volts
+w=(2*%pi*N)/60 //Angular Speed in rad/sec
+K=Eb/w
+AA=(Ra*V*1.41)/(K*Z)
+Wmc=AA*((sind(P)*B)-(sind(a-P)*A))*(1/(1-A)) //Critical Speed in rad/Sec
+Wrpm=(Wmc*60)/(2*%pi) //speed in rpm
+//As the motor speed of 400 rpm is less than the critical speed ,the drive is operating under continuous conduction mode
+af=30 //firing angle in Degree
+Va=(V*1.414)*(1+cosd(af))*(1/%pi) //Armature voltage in volts
+//At 400 RPM
+Eb1=Eb*(N2/N) //This Value Is Wrongly Calculated in Textbook
+T=K*(Va-Eb1)*(1/Ra) //Torque in N-m
+//Motor back emf for critical speed equal to 1149.67 rpm
+Ec=(Wrpm*Eb)/N //critical emf in volts
+Tc=K*(Va-Ec)*(1/Ra) //Critical Torque in N-m
+//Since the motor torque of 70 N-m is greater than the critical torque Tc ,the drive is operating in continuous conduction
+Ia=T/K //Armature current in Amp
+Eb2=Va-(Ia*Ra) // Back emf in Volts
+Nm=(Eb2*N)/161 //Answer changed due wrong value is taken in book of Eb1
+
+//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',Nm)
+//The answer provided in the textbook is wrong