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+clc
+//variable initialization
+Vm=600 //Voltage of motor in volt
+Pout=30000 // Output Of Motor In Watt
+F=50 //Supply frequency in hrtz
+P=4 //Number Of Poles
+N1=100 //Speed OF Motor In rpm
+N2=1000 //Speed OF Motor In rpm
+R1=0.05 //Resistance of stator in ohm
+R2=0.07 //Resistance of rotor in ohm
+R0=53 //Resistance of rotor in ohm
+X=0.153 //Reactance of Motor in ohm
+X0=23 //Reactance of Motor in ohm
+n=1.3 //Stator To Rotr Ratio
+N3=300 //Speed OF Motor In rpm
+
+
+//Solution
+Vph=Vm/(sqrt(3))
+a=1/n
+Ns=(120*F)/(P)
+S=(Ns-N2)/Ns
+Wm=(2*%pi)/60
+Tl=(Pout)/(Wm*N3)
+Id=(Tl*Wm*Ns)/(2.339*a*Vph)
+I0=Vph/(X0)
+I2=(sqrt(2/3))*(Id*a)
+Pi=Pout+(R1*((I2)^2))+(R2*((I2)^2))
+e=(Pout/Pi)*100
+theta=-(atand(Vph/(0.779*Id*a*X0)))
+pf=cosd(theta)
+printf('\n\n The Motor Efficiency=%0.1f\n\n',e)
+printf('\n\n The Power Factor=%0.1f lag\n\n',pf)
+//The answers vary due to round off error