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+//Book name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi
+//chapter 10
+//example 10.1
+//edition 1
+//publishing place:Thomson Learning
+clc;
+clear;
+V=208;//source voltage in volts
+p=6;//number of poles
+R1=0.6;//given resistance in ohm
+R2=0.4;//given R'2 in ohm
+Xeq=5;//given Xeq in ohm
+Td=30;//load torque of the motor in ohm
+f=60;//frequency for 3 phase line
+ns=(120*f)/p
+disp('a)To find the regenerative speed:')
+Tl=-Td//reversed load torque
+rpss=ns/60;
+omegas=(2*%pi*rpss);//angular speed
+s=(Tl*omegas*R2)/V^2;
+n=ns*(1-s);
+mprintf("The regenerative speed is %f rpm",n)
+disp('b)To calculate the regenerative speed :')
+rps=n/60;
+omega=(2*%pi*rps);
+Pd=Td*omega;
+I2=sqrt(-Pd/(3*(R2/s)*(1-s)));//to find I'2 which is taken as I2
+Ploss=3*(R1+R2)*I2'^(2)
+Pds=Pd-Ploss;
+mprintf("The power delivered to the electric supply is %f watt",Pds)