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+//Chapter-3, Example 3.5, Page 3.13
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+P=6;//Number of poles
+f=50;//Supply frequency in Hz
+Tm=120;//Shaft torque in N.m
+f1=2;//Rotor current frequency in Hz
+L=5;//Amount of constant losses in N.m
+C=500;//Amount of core losses in W
+
+//CALCULATIONS
+Ns=(120*f)/P;//Synchronous speed in rpm
+s=(f1/f);//Slip of the motor 
+N=(1-s)*Ns;//Actual speed in rpm
+P=(2*3.14*N*Tm)/60;//Shaft power in W
+Pm=(2*3.14*N*(Tm+L))/60000;//Mechanical power output in kW
+R=(s*Pm)/(1-s);//Rotor copper losses in kW
+I=(Pm+R+(L/10));//Motor input in kW
+n=(Pm/I)*100;//Machine efficiency
+
+//OUTPUT
+mprintf('a)Mechanical power output is %3.3f kW\nb)Rotor copper losses is %3.2fkW\nc)Motor input is %3.3f kW\nd)Machine efficiency is %3.1f percent',Pm,R,I,n)
+
+//=================================END OF PROGRAM==============================