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diff --git a/3811/CH10/EX10.1/Ex10_1.jpg b/3811/CH10/EX10.1/Ex10_1.jpg
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diff --git a/3811/CH10/EX10.1/Ex10_1.sce b/3811/CH10/EX10.1/Ex10_1.sce
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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)

diff --git a/3811/CH10/EX10.2/Ex10_2.jpg b/3811/CH10/EX10.2/Ex10_2.jpg
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diff --git a/3811/CH10/EX10.2/Ex10_2.sce b/3811/CH10/EX10.2/Ex10_2.sce
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+//Book name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi

+//chapter 10

+//example 10.2

+//edition 1

+//publishing place:Thomson Learning

+clc;

+clear;

+Vdc=200;//voltage at the dc link in volt

+I=25;//motor current in A

+R1=0.5;//stator resistance in ohm

+Ib=3*I;

+Vb=Ib*1.5*R1;//braking voltage in volt

+d=1.5*(Vb/Vdc)^2;

+mprintf("\nThe duty ratio of the FWM is %f",d)

diff --git a/3811/CH10/EX10.3/Ex10_3.jpg b/3811/CH10/EX10.3/Ex10_3.jpg
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diff --git a/3811/CH10/EX10.3/Ex10_3.sce b/3811/CH10/EX10.3/Ex10_3.sce
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+//Book Name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi

+//chapter 10

+//example 10.3

+//edition 1

+//publishing place:Thomson Learning

+clc;

+clear;

+n1=1150;//full load speed in rpm

+V=300;//terminal voltage in volt

+f=80;//frequency in Hz

+Rr=0.5;//rotor resistance of the motor in ohm

+Xeq=3;//equivalent inductive reactance in ohm

+ns=1200;//nearest synchronous speed in rpm

+rpss=ns/60;

+omegas=(2*%pi*rpss);

+s1=(ns-n1)/ns;

+T6=(V^(2)*s1)/(omegas*Rr);//torque at the point 6

+T6=ceil(T6);

+mprintf("\nThe torque developed is %d Nm",T6)//approximated value

+s6=(T6*Rr*(-omegas))/V^(2);

+mprintf("\nThe slip is %f",s6)

+n6=(-ns)*(1-s6);

+mprintf("\nThe current of the induction motor does not surge to high value when the concurrent braking is implemented")

+