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-rw-r--r--3731/CH5/EX5.1/Ex5_1.sce25
-rw-r--r--3731/CH5/EX5.10/Ex5_10.sce85
-rw-r--r--3731/CH5/EX5.11/Ex5_11.sce53
-rw-r--r--3731/CH5/EX5.12/Ex5_12.sce47
-rw-r--r--3731/CH5/EX5.13/Ex5_13.sce45
-rw-r--r--3731/CH5/EX5.14/Ex5_14.sce89
-rw-r--r--3731/CH5/EX5.15/Ex5_15.sce62
-rw-r--r--3731/CH5/EX5.16/Ex5_16.sce62
-rw-r--r--3731/CH5/EX5.17/Ex5_17.sce45
-rw-r--r--3731/CH5/EX5.18/Ex5_18.sce57
-rw-r--r--3731/CH5/EX5.19/Ex5_19.sce46
-rw-r--r--3731/CH5/EX5.2/Ex5_2.sce27
-rw-r--r--3731/CH5/EX5.20/Ex5_20.sce34
-rw-r--r--3731/CH5/EX5.21/Ex5_21.sce38
-rw-r--r--3731/CH5/EX5.22/Ex5_22.sce51
-rw-r--r--3731/CH5/EX5.23/Ex5_23.sce38
-rw-r--r--3731/CH5/EX5.3/Ex5_3.sce22
-rw-r--r--3731/CH5/EX5.4/Ex5_4.sce25
-rw-r--r--3731/CH5/EX5.5/Ex5_5.sce61
-rw-r--r--3731/CH5/EX5.6/Ex5_6.sce68
-rw-r--r--3731/CH5/EX5.7/Ex5_7.sce53
-rw-r--r--3731/CH5/EX5.8/Ex5_8.sce37
-rw-r--r--3731/CH5/EX5.9/Ex5_9.sce79
23 files changed, 1149 insertions, 0 deletions
diff --git a/3731/CH5/EX5.1/Ex5_1.sce b/3731/CH5/EX5.1/Ex5_1.sce
new file mode 100644
index 000000000..ebc684a99
--- /dev/null
+++ b/3731/CH5/EX5.1/Ex5_1.sce
@@ -0,0 +1,25 @@
+//Chapter 5:DC Motor Drives
+//Example 1
+clc;
+
+//Variable Initialization
+
+//Motor ratings
+V1=200 //rated voltage in V
+Ia1=10.5 //rated current in A
+N1=2000 //speed in rpm
+Ra=0.5 //armature resistance in ohms
+Rs=400 //field resistance in ohms
+V2=175 //drop in source voltage in V
+
+//Solution
+
+flux2=V2/V1
+Ia2=1/flux2*Ia1 //since load torque
+E1=V1-Ia1*Ra
+E2=V2-Ia2*Ra
+N2=(E2/E1)*(1/flux2)*N1
+
+//Result
+mprintf("\nMotor speed is:%.1f rpm",N2)
+//Answer provided in the book is incorrect
diff --git a/3731/CH5/EX5.10/Ex5_10.sce b/3731/CH5/EX5.10/Ex5_10.sce
new file mode 100644
index 000000000..ebe71b3aa
--- /dev/null
+++ b/3731/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,85 @@
+//Chapter 5:Dc Motor Drives
+//Example 10
+clc;
+
+//Variable Initialization
+
+
+V=220 // rated voltage in v
+N=1000 // rated speed in rpm
+Ia=175 // rated current in A
+Ra=0.08 // armature resistance in ohms
+N1=1050 // initial speed of the motor in rpm
+J=8 // moment of inertia of the motor load system kg-m2
+La=0.12 // armature curcuit inductance in H
+
+//Solution
+E=V-Ia*Ra
+Wm=N*2*%pi/60 //rated speed in rad/s
+//(a)When the braking current is twice the rated current
+Ia1=2*Ia
+E1=N1/N*E
+x=(V+E1)/Ia1 //x=Rb+Ra
+Rb=x-Ra //required braking resistance
+
+//(b)To obtain the expression for the transient value of speed and current including the effect of armature inductance
+//The values given below are taken from Ex-5.9
+ta=0.194 //time constant in sec
+B=0
+tm1= %inf //tm1=J/B and B=0 which is equal to infinity
+tm2=1.274
+K=1.967
+Trated=E*Ia/Wm //rated torque
+Tl=0.5*Trated //load torque is 50% of the rated torque
+Ra=Rb
+K1=N1*2*%pi/60 //initial speed in rad/s
+//Values of the coefficient of the quadratic equation for Wm
+x1=(1+ta/tm1)/ta
+x2=1/tm2/ta
+x3=-(K*V+Ra*Tl)/J/Ra/ta
+//Values of the coefficient of the quadratic equation ia
+y1=(1+ta/tm1)/ta
+y2=1/tm2/ta
+y3=-B*V/J/Ra/ta+K*Tl/J/Ra/ta
+
+//solving the quadratic equation
+a = 1
+b = x1
+c = x2
+//Discriminant
+d = (b**2) - (4*a*c)
+
+alpha1 = (-b+sqrt(d))/(2*a)
+alpha2 = (-b-sqrt(d))/(2*a)
+
+K3=x3/x2
+K4=y3/y2
+
+Wm_0=K1 ;ia_0=0
+d_Wm_dt_0=(K*ia_0-B*Wm-Tl)/J ;d_ia_dt_0=(-V-Ra*ia_0-K*K1)/La //Wm=K1 at t=0 and during braking rated voltage V is equal to -V
+
+a = [1,1;real(alpha1),real(alpha2)]
+b = [Wm_0;d_Wm_dt_0]
+x = inv(a)*b
+c = [1,1;real(alpha1),real(alpha2)]
+d = [-K4;d_ia_dt_0]
+y = inv(c)*d
+
+//(c)To calculate the time taken for the speed to fall to zero value
+a=-K3/x(1) //a=exp(-0.966*t1)
+t1=alpha1*log(a) //take log base e on both sides
+
+
+//Results
+mprintf("(a)Hence the braking resistance is :%.3f ohm",Rb)
+mprintf("\n(b)The solutions for alpha are %.3f and %.3f",real(alpha1),real(alpha2))
+mprintf("\nWm=%.2f + A*exp(%.3f*t) + B*exp(%.2f*t)",K3,real(alpha1),real(alpha2))
+mprintf("\nia=%.2f+ C*exp(%.3f*t) + D*exp(%.2f*t)",K4,real(alpha1),real(alpha2))
+mprintf("\nWe have to find the value of A,B,C and D in the linear equation using the initial condition")
+mprintf("\nA=%.2f B=%.2f C=%.2f D=%.2f",x(1),x(2),y(1),y(2))
+mprintf("\nHence the expression for the transient value for the speed is")
+mprintf("\nWm=%.2f+%.2f*exp(%.3f*t)%.2f*exp(%.2f*t)",K3,x(1),real(alpha1),x(2),real(alpha2))
+mprintf("\nHence the expression for the transient value for the current is")
+mprintf("\nia=%.2f %.1f *exp(%.3f*t) +%.2f*exp(%.2f*t)",K4,y(1),real(alpha1),y(2),real(alpha2))
+mprintf("\n(c)Hence the time taken is :%.2f sec",real(t1))
+//There is slight difference in the answers due to accuracy
diff --git a/3731/CH5/EX5.11/Ex5_11.sce b/3731/CH5/EX5.11/Ex5_11.sce
new file mode 100644
index 000000000..b63945d9f
--- /dev/null
+++ b/3731/CH5/EX5.11/Ex5_11.sce
@@ -0,0 +1,53 @@
+
+//Chapter 5:Dc Motor Drives
+//Example 11
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor
+V=220 // rated voltage in V
+N=600 // rated speed in rpm
+Ia=500 // rated current in A
+Ra=0.02 // armature resistance in ohms
+Rf=10 // field resistance in ohms
+
+//Solution
+Ia1=2*Ia
+E1=V-Ia*Ra //rated back emf at rated operation
+Wm1=2*%pi*N/60 //angular speed
+Trated=E1*Ia1/Wm1 //rated torque
+
+//(i) When the speed of the motor is 450rpm
+N1=450 //given speed in rpm
+Tl=2000-2*N1 //load torque is a function of the speed as given
+Ia2=Tl/Trated*Ia1 //for a torque of Tl as a function of current
+E2=N1/N*E1 //for a given speed of 450rpm
+V2=E2+Ia2*Ra //terminal voltage for a given speed of 450 rpm
+
+//(ii) when the speed of the motor is 750rpm
+N1=750 //given speed in rpm
+Tl=2000-2*N1 //load torque is a function of the speed as given
+Wm_=2*%pi*N1/60
+Ke_phi1=E1/Wm1
+
+//Since we know that V=Ke*phi*Wm+Ia*Ra by solving we get that 0.02*(Ia_)**2 -220*Ia_ + 39270 = 0"
+a = 0.02
+b = -220
+c = 39270
+
+//Discriminant
+d = (b**2) - (4*a*c)
+
+Ia_1 = (-b-sqrt(d))/(2*a)
+Ia_2 = (-b+sqrt(d))/(2*a)
+
+Ke_phi=Tl/abs(Ia_1)
+V1=V*Ke_phi/Ke_phi1 //required field voltage
+
+//Results
+mprintf("(i)Hence motor terminal voltage is :%.1f V",V2)
+mprintf("\nAnd the armature current is :%.1f A",Ia2)
+mprintf("\n(ii)The solutions for Ia_ are %.1f A and %.1f A",abs(Ia_1),abs(Ia_2))
+mprintf("\nWe ignore %d A since it is infeasible,\n Hence armature current is :%.1f A",abs(Ia_2),abs(Ia_1))
+mprintf("\nHence the required field voltage is :%.1f V",V1)
diff --git a/3731/CH5/EX5.12/Ex5_12.sce b/3731/CH5/EX5.12/Ex5_12.sce
new file mode 100644
index 000000000..5e410cde3
--- /dev/null
+++ b/3731/CH5/EX5.12/Ex5_12.sce
@@ -0,0 +1,47 @@
+
+//Chapter 5:Dc Motor Drives
+//Example 12
+clc;
+
+//Variable Initialization
+
+//Ratings of the 2-pole separately excited DC motor with the fields coils connected in parallel
+V=220 // rated voltage in V
+N=750 // rated speed in rpm
+Ia1=100 // rated current in A
+Ra=0.1 // armature resistance in ohms
+
+//Solution
+E1=V-Ia1*Ra //rated back emf at rated operation
+Wm1=2*%pi*N/60 //angular speed
+Trated=E1*Ia1/Wm1 //rated torque
+Ke_phi1=E1/Wm1
+
+//(i) When the armature voltage is reduced to 110V
+Wm2=2*%pi*N/60 //angular speed
+E2=Ke_phi1*Wm2
+//Now there are two linear equations...that we have to solve
+//They are given by 0.3*N2+2.674*Ia2=500 and 0.28*N2+0.1*Ia2=110
+a = [0.3,2.674;0.28,0.1]
+b = [500;110]
+x = inv(a)*b
+N2=x(1) //let the motor speed be N2
+Ia2=x(2) //let the motor current be Ia2
+
+//(ii)When the field coils are connected in series
+K=Ke_phi1/2
+Wm3=2*%pi*N/60 //angular speed
+E3=K*Wm3
+//Now there are two linear equations...that we have to solve"
+//They are given by 0.3*N3+1.337*Ia3=500 and 0.14*N3+0.1*Ia3=220"
+a = [0.3,1.337;0.14,0.1]
+b = [500;220]
+x = inv(a)*b
+N3=x(1) //let the motor speed be N3
+Ia3=x(2) //let the motor current be Ia3
+
+//Results
+mprintf("(i)Hence the motor armature current is Ia2 :%.1f A",Ia2)
+mprintf("\nAnd the required speed is N2 :%.1f rpm",N2)
+mprintf("\n(ii)Hence the motor armature current is Ia3 :%.1f A",Ia3)
+mprintf("\nAnd the required speed is N3 :%.1f rpm",N3)
diff --git a/3731/CH5/EX5.13/Ex5_13.sce b/3731/CH5/EX5.13/Ex5_13.sce
new file mode 100644
index 000000000..aa644aa15
--- /dev/null
+++ b/3731/CH5/EX5.13/Ex5_13.sce
@@ -0,0 +1,45 @@
+//Chapter 5:Dc Motor Drives
+//Example 13
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor
+V=200 // rated voltage in V
+N=875 // rated speed in rpm
+Ia=150 // rated current in A
+Ra=0.06 // armature resistance in ohms
+Vs=220 // source voltage in V
+f=50 // frequency of the source voltage in Hz
+
+//Solution
+E=V-Ia*Ra //back emf
+Vm=sqrt(2)*Vs //peak voltage
+
+//(i)When the speed is 750 rpm and at rated torque
+N1=750 //given speed in rpm
+E1=N1/N*E //back emf at the given speed N1
+Va=E1+Ia*Ra //terminal voltage
+cos_alpha=Va*%pi/2/Vm
+alpha=acos(cos_alpha) //required firing angle in radian
+alpha1=alpha*180/%pi //required firing angle in degrees
+
+//(ii)When the speed is -500rpm and at rated torque
+N1=-500 //given speed in rpm
+E1=N1/N*E //back emf at the given speed N1
+Va=E1+Ia*Ra //terminal voltage
+cos_alpha=Va*%pi/2/Vm
+alpha=acos(cos_alpha) //required firing angle in radian
+alpha2=alpha*180/%pi //required firing angle in degrees
+
+//(iii)When the firing angle is 160 degrees
+alpha=160 //firing angle in degrees
+alpha=alpha*%pi/180
+Va=2*Vm/%pi*cos(alpha)
+E1=Va-Ia*Ra //since Va=E1+Ia*Ra
+N1=E1/E*N //the required speed at the given firing angle
+
+//Results
+mprintf("(i)Hence the required firing angle is :%.1f °",alpha1)
+mprintf("\n(ii)Hence the required firing angle is :%.1f °",alpha2)
+mprintf("\n(iii)Hence the required speed is :%.1f rpm",N1)
diff --git a/3731/CH5/EX5.14/Ex5_14.sce b/3731/CH5/EX5.14/Ex5_14.sce
new file mode 100644
index 000000000..4b499ef77
--- /dev/null
+++ b/3731/CH5/EX5.14/Ex5_14.sce
@@ -0,0 +1,89 @@
+//Chapter 5:Dc Motor Drives
+//Example 14
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor is same as that of Ex-5.13
+V=200 // rated voltage in v
+N=875 // rated speed in rpm
+Ia=150 // rated current in A
+Ra=0.06 // armature resistance in ohms
+Vs=220 // source voltage in v
+f=50 //frequency of the source voltage in hz
+La=0.85e-3 // armature curcuit inductance in H
+
+//Solution
+E=V-Ia*Ra //back emf
+Vm=sqrt(2)*Vs //peak voltage
+Wm=2*%pi*N/60 //synchronous angular speed
+
+//(i)When the speed is 400 rpm and firing angle is 60 degrees
+N1=400 //given speed in rpm
+alpha=60 //firing angle in degrees
+W=2*%pi*f
+x=W*La/Ra
+phi=atan(x)
+cot_phi=1/tan(phi)
+Z=sqrt(Ra**2+(W*La)**2)
+K=E/Wm
+
+y=Ra*Vm/Z/K
+a=(1+exp(-(%pi*cot_phi)))/(exp(-(%pi*cot_phi))-1)
+alpha=alpha*%pi/180
+Wmc=y*sin(alpha-phi)*a //required angular speed in rps
+Nmc=Wmc*60/2/%pi //required angular speed in rpm
+
+E1=N1/N*E
+
+//The equation Vm/Z*sin(beta-phi)-E/Ra+(E/Ra-Vm/Z*sin(alpha-phi))*exp(-(beta-alpha)*cot_phi)=0
+//can be solved using trial method such that beta=230 degrees
+beta=230 //in degrees
+beta=beta*%pi/180//in radians
+
+Va=(Vm*(cos(alpha)-cos(beta))+(%pi+alpha-beta)*E1)/%pi
+Ia=(Va-E1)/Ra
+T1=K*Ia
+
+//(ii)When the speed is -400 rpm and firing angle is 120 degrees
+Le=2e-3 //external inductance added to the armature
+L=La+Le
+N2=-400 //given speed in rpm
+alpha=120 //firing angle in degrees
+x=W*L/Ra
+phi=atan(x)
+cot_phi=1/tan(phi)
+Z=sqrt(Ra**2+(W*L)**2)
+K=E/Wm
+
+y=Ra*Vm/Z/K
+alpha=alpha*%pi/180
+a=(1+exp(-(%pi*cot_phi)))/(exp(-(%pi*cot_phi))-1)
+Wmc=y*sin(alpha-phi)*a //required angular speed in rps
+Nmc1=Wmc*60/2/%pi //required angular speed in rpm
+//The motor is operating under discontinous condition"
+E2=N2/N*E
+
+//The equation Vm/Z*sin(beta-phi)-E/Ra+(E/Ra-Vm/Z*sin(alpha-phi))*exp(-(beta-alpha)*cot_phi)=0
+//can be solved using trial method such that beta=281 degrees
+beta=281 //in degrees
+beta=beta*%pi/180//in radians
+
+Va=(Vm*(cos(alpha)-cos(beta))+(%pi+alpha-beta)*E2)/%pi
+Ia=(Va-E2)/Ra
+T2=K*Ia
+
+//(iii)When the speed is -600 rpm and firing angle is 120 degrees
+N3=-600 //speed in rpm
+alpha=120 //firing angle in degrees
+alpha=alpha*%pi/180
+Va=2*Vm/%pi*cos(alpha)
+E3=N3/N*E //since Va=E1+Ia*Ra
+Ia=(Va-E3)/Ra
+T3=K*Ia
+
+//Results
+mprintf("(i)Hence the required torque is :%.2f N-m",T1)
+mprintf("\n(ii)Hence the required torque is :%.1f N-m",T2)
+mprintf("\n(iii)Hence the required torque is :%.1f N-m",T3)
+//There is a minor difference in the answers because of accuracy
diff --git a/3731/CH5/EX5.15/Ex5_15.sce b/3731/CH5/EX5.15/Ex5_15.sce
new file mode 100644
index 000000000..9e719a1ed
--- /dev/null
+++ b/3731/CH5/EX5.15/Ex5_15.sce
@@ -0,0 +1,62 @@
+//Chapter 5:Dc Motor Drives
+//Example 15
+clc;
+
+//Variable Initialization
+
+V=200 // rated voltage in v
+N=875 // rated speed in rpm
+Ia=150 // rated current in A
+Ra=0.06 // armature resistance in ohms
+Vs=220 // source voltage in v
+f=50 //frequency of the source voltage in hz
+La=2.85e-3 // armature curcuit inductance in H
+
+//Solution
+E=V-Ia*Ra //back emf
+Vm=sqrt(2)*Vs //peak voltage
+Wm=2*%pi*N/60 //angular speed
+W=2*%pi*f
+
+alpha=120 //firing angle in degrees
+x=W*La/Ra
+phi=atan(x)
+cot_phi=1/tan(phi)
+Z=sqrt(Ra**2+(W*La)**2)
+K=E/Wm
+
+y=Ra*Vm/Z/K
+a=(1+exp(-(%pi*cot_phi)))/(exp(-(%pi*cot_phi))-1)
+alpha=alpha*%pi/180
+Wmc=y*sin(alpha-phi)*a //required angular speed in rps
+Nmc=Wmc*60/2/%pi //required angular speed in rpm
+
+Va=2*Vm/%pi*cos(alpha)
+E1=Nmc/N*E //value of back emf at the critical speed of Nmc
+Ia=(Va-E1)/Ra
+Tc=K*Ia
+
+//(i)When the torque is 1200 N-m and firing angle is 120 degrees
+T2=1200 //given torque in N-m
+Ia2=T2/K //given terminal current for the given torque and the answer in the book is wrong
+E2=Va-Ia*Ra
+N2=E2/E*N
+
+//(ii)When the torque is 300 N-m and firing angle is 120 degrees
+T=300 //required torque in N-m
+beta=233.492 //required angle in degrees
+beta=beta*%pi/180 //in radians
+x=beta-alpha
+E1=(Vm*(cos(alpha)-cos(beta)))/x-(%pi*Ra*T)/(K*x)
+N1=E1/E*N //required speed
+
+
+//Results
+mprintf("\nThe motor is operating under continuous condition")
+mprintf("\nThe torque Tc is :%.2f N-m",Tc)
+//The answer for torque Tc in the book is wrong due to accuracy which leads to other incorrect answers
+mprintf("\n(i)Hence the required speed is :%.1f rpm",N2)
+mprintf("\n(ii)The equation Vm/Z*sin(beta-phi)-sin(alpha-phi))*exp(-(beta-alpha)*cot_phi)=")
+mprintf("\n(Vm*(cos(alpha)-cos(beta))/Ra/(beta-alpha)-pi*T/K/(beta-alpha) )*(1-exp(-(beta-alpha)*cot_phi)")
+mprintf("\ncan be solved using trial method such that beta=233.492 degrees")
+mprintf("\n Hence the required speed is :%.1f rpm",N1)
diff --git a/3731/CH5/EX5.16/Ex5_16.sce b/3731/CH5/EX5.16/Ex5_16.sce
new file mode 100644
index 000000000..3e7e9948f
--- /dev/null
+++ b/3731/CH5/EX5.16/Ex5_16.sce
@@ -0,0 +1,62 @@
+//Chapter 5:Dc Motor Drives
+//Example 16
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor
+V=220 // rated voltage in v
+N=960 // rated speed in rpm
+Ia=12.8 // rated current in A
+Ra=2 // armature resistance in ohms
+Vs=230 // source voltage in v
+f=50 //frequency of the source voltage in hz
+La=150e-3 // armature curcuit inductance in H
+
+//Solution
+E=V-Ia*Ra //back emf
+Vm=sqrt(2)*Vs //peak voltage
+Wm=2*%pi*N/60 //angular speed
+W=2*%pi*f
+
+//(i)When speed is 600rpm and the firing angle is 60 degrees
+alpha=60 //firing angle in degrees
+N1=600 //motor speed in rpm
+x=W*La/Ra
+phi=atan(x)
+cot_phi=1/tan(phi)
+Z=sqrt(Ra**2+(W*La)**2)
+K=E/Wm
+
+y=Ra*Vm/Z/K
+alpha=alpha*%pi/180
+b=sin(phi)*exp(-(alpha*cot_phi))
+c=sin(alpha-phi)*exp(-(%pi*cot_phi))
+a=1-exp(-(%pi*cot_phi))
+Wmc=y*(b-c)/a //required angular speed in rps
+Nmc=Wmc*60/2/%pi //required angular speed in rpm
+
+Va=Vm/%pi*(1+cos(alpha))
+E1=N1/N*E //value of back emf at the speed of N1
+Ia=(Va-E1)/Ra
+T=K*Ia
+
+//(ii)When the torque is 20 N-m and firing angle is 60 degrees
+T1=20 //required torque in N-m
+alpha=60 //required firing angle in degrees
+Ec=Nmc/N*E //motor back emf at critical speed of Nmc
+Tc=K*(Va-Ec)/Ra //torque at the critical speed
+
+Ia=T1/K
+E1=Va-Ia*Ra
+N1=E1/E*N //required speed
+
+//Results
+//if N1<Nmc then
+mprintf("(i)The motor is operating under continuous condition")
+mprintf("\nHence the required torque is :%.2f N-m",T)
+//end
+//if Tc<T1 then
+mprintf("\n(ii)The motor is operating under continuous condition")
+mprintf("\nHence the required speed is :%.1f rpm",N1)
+//end
diff --git a/3731/CH5/EX5.17/Ex5_17.sce b/3731/CH5/EX5.17/Ex5_17.sce
new file mode 100644
index 000000000..c07442049
--- /dev/null
+++ b/3731/CH5/EX5.17/Ex5_17.sce
@@ -0,0 +1,45 @@
+//Chapter 5:Dc Motor Drives
+//Example 17
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor
+V=220 // rated voltage in V
+N=1500 // rated speed in rpm
+Ia=50 // rated current in A
+Ra=0.5 // armature resistance in ohms
+Vl=440 // line voltage inV with 3-phase ac supply
+f=50 //frequency of the source voltage in Hz
+
+//Solution
+//(i) Tranformer ratio
+alpha=0 //firing angle in degrees
+Va=V //motor terminal voltage is equal to the rated voltage when the firing angle is 0 degrees
+Vm=%pi/3*Va/cos(alpha)
+Vrms=Vm/sqrt(2) //rms value of the converter input voltage
+a=(Vl/sqrt(3))/Vrms //required transformer ratio
+
+//(ii)Value of the firing angle
+E=V-Ia*Ra //back emf at the rated speed
+
+//(a)When the speed of the motor is 1200 rpm and rated torque
+N1=1200 //speed of the motor i rpm
+E1=N1/N*E //back emf at the given speed N1
+Va=E1+Ia*Ra //terminal voltage at the given speed N1
+alpha=acos(%pi/3*Va/Vm) //required firing angle in radians
+alpha1=alpha*180/%pi //required firing angle in degrees
+
+//(b)When the speed of the motor is -800 rpm and twice the rated torque
+N1=-800 //speed of the motor in rpm
+E1=N1/N*E //back emf at the given speed N1
+Ia=2*Ia //torque is directly proportional to the current hence twice the rated current
+Va=E1+Ia*Ra //terminal voltage at the given speed N1
+alpha=acos(%pi/3*Va/Vm) //required firing angle in radians
+alpha2=alpha*180/%pi //required firing angle in degrees
+
+
+//Results
+mprintf("(i)Hence the required transformer ratio is :%.3f",a)
+mprintf("\n(ii)(a)Hence the required firing angle is :%.2f °",alpha1)
+mprintf("\n(b)Hence the required firing angle is :%.2f °",alpha2)
diff --git a/3731/CH5/EX5.18/Ex5_18.sce b/3731/CH5/EX5.18/Ex5_18.sce
new file mode 100644
index 000000000..342b13451
--- /dev/null
+++ b/3731/CH5/EX5.18/Ex5_18.sce
@@ -0,0 +1,57 @@
+//Chapter 5:Dc Motor Drives
+//Example 18
+clc;
+
+//Variable Initialization
+
+//The separately excited motor is fed from a circulating dual converter
+V=220 // rated voltage in V
+N=1500 // rated speed in rpm
+Ia=50 // rated current in A
+Ra=0.5 // armature resistance in ohms
+Vl=165 // line voltage in V
+f=50 // frequency of the source voltage in Hz
+
+//Solution
+E=V-Ia*Ra //back emf at the rated speed
+Vm=Vl*sqrt(2) //peak voltage
+
+//(i)During motoring operation when the speed is 1000 rpm and at rated torque
+N1=1000 //speed of the motor in rpm
+E1=N1/N*E //back emf at the given speed N1
+Va=E1+Ia*Ra //terminal voltage at the given speed N1
+alpha_A=acos(%pi/3*Va/Vm)
+alpha_A=alpha_A*180/%pi//required converter firing angle in degrees
+alpha_B=180-alpha_A
+
+//(ii)During braking operation when the speed is 1000 rpm and at rated torque
+N1=1000 //speed of the motor in the book is given as 100 rpm which is wrong
+E1=N1/N*E //back emf at the given speed N1
+Va=E1-Ia*Ra //terminal voltage at the given speed N1
+alpha_A1=acos(%pi/3*Va/Vm)
+alpha_A1=alpha_A1*180/%pi//required converter firing angle in degrees
+alpha_B1=180-alpha_A1
+
+//(iii)During motoring operation when the speed is -1000 rpm and at rated torque
+N1=-1000 //speed of the motor in rpm
+E1=N1/N*E //back emf at the given speed N1
+Va=E1-Ia*Ra //terminal voltage at the given speed N1
+alpha_A2=acos(%pi/3*Va/Vm)
+alpha_A2=alpha_A2*180/%pi//required converter firing angle in degrees
+alpha_B2=180-alpha_A2
+
+//(iv)During braking operation when the speed is -1000 rpm and at rated torque
+N1=-1000 //speed of the motor in the book it is given as 100 rpm which is wrong
+E1=N1/N*E //back emf at the given speed N1
+Va=E1+Ia*Ra //terminal voltage at the given speed N1
+alpha_A3=acos(%pi/3*Va/Vm)
+alpha_A3=alpha_A3*180/%pi//required converter firing angle in degrees
+alpha_B3=180-alpha_A3
+
+//Results
+mprintf("\n(i)Hence the required firing angle is :%.1f °",alpha_B)
+mprintf("\n(ii)Hence the required firing angle is :%.1f °",alpha_B1)
+mprintf("\n(iii)Hence for negative speed during motoring operation the required firing angles are :")
+mprintf("\nalpha_A :%.1f ° and alpha_B :%.1f °",alpha_A2,alpha_B2)
+mprintf("\n(iv)Hence for negative speed during braking operation the required firing angles are :")
+mprintf("\nalpha_A :%.1f ° and alpha_B :%.1f °",alpha_A3,alpha_B3)
diff --git a/3731/CH5/EX5.19/Ex5_19.sce b/3731/CH5/EX5.19/Ex5_19.sce
new file mode 100644
index 000000000..aff7bfc47
--- /dev/null
+++ b/3731/CH5/EX5.19/Ex5_19.sce
@@ -0,0 +1,46 @@
+//Chapter 5:Dc Motor Drives
+//Example 19
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor
+V=230 // rated voltage in V
+N=960 // rated speed in rpm
+Ia=200 // rated current in A
+Ra=0.02 // armature resistance in ohms
+Vs=230 // source voltage in V
+
+//Solution
+E=V-Ia*Ra //back emf
+
+//(i) When the speed of motor is 350 rpm with the rated torque during motoring operation
+N1=350 //given speed in rpm
+E1=N1/N*E //given back emf at N1
+Va=E1+Ia*Ra //motor terminal voltage
+delta=Va/V //duty ratio
+
+//(ii) When the speed of motor is 350 rpm with the rated torque during braking operation
+Va=E1-Ia*Ra //motor terminal voltage
+delta1=Va/V //duty ratio
+
+//(iii)Maximum duty ratio is 0.95
+delta2=0.95 //maximum duty ratio
+Va=delta2*V //terminal voltage
+Ia1=2*Ia //maximum permissable current
+E1=Va+Ia1*Ra //back emf
+N1=E1/E*N //maximum permissible speed
+Pa=Va*Ia1 //power fed to the source
+
+//(iv) If the speed of the motor is 1200 rpm and the field of the motor is also controlled
+N2=1200 //given speed in rpm
+//Now the field current is directly proportional to the speed of the motor
+If=N/N2 //field current as a ratio of the rated current
+
+
+//Results
+mprintf("(i) Duty ratio is :%.3f",delta)
+mprintf("\n(ii)Duty ratio is :%.2f",delta1)
+mprintf("\n(iii)Maximum permissible speed is :%d rpm",N1)
+mprintf("\nPower fed to the source is :%.1f kW",Pa/1000)
+mprintf("\n(iv)Field current as a ratio of the rated current is :%.1f",If)
diff --git a/3731/CH5/EX5.2/Ex5_2.sce b/3731/CH5/EX5.2/Ex5_2.sce
new file mode 100644
index 000000000..6bb6466c9
--- /dev/null
+++ b/3731/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,27 @@
+//Chapter 5:DC Motor Drives
+//Example 2
+clc;
+
+//Variable Initialization
+
+V1=220 //rated voltage in V
+Ia1=100 //rated current in A
+N1=1000 //rated speed in rpm clockwise
+Ra=0.05 //armature resistance in ohms
+Rs=0.05 //field resistance in ohms
+
+//Solution
+//Turns is reduced to 80% then flux is also reduced by the same value and hence current is also reduced
+T1=Ia1**2 //flux is directly proportional to current Ia
+T2=0.8*1**2 //flux is directly proportional to current Ia
+Ia2=-Ia1/sqrt(0.8) //since T1=T2 and the direction is opposite
+
+E1=V1-Ia1*(Ra+Rs)
+
+Rs=.8*Rs //Rs=80% of the field resistance 0.05ohm since the flux is reduced to 80%
+E2=-(V1+Ia2*(Ra+Rs))
+
+N2=(E2/E1)*(Ia1/Ia2)*(N1/0.8) //since E=Kn*flux*N
+
+//Result
+mprintf("\nMotor speed is:N2=%.1f rpm",N2)
diff --git a/3731/CH5/EX5.20/Ex5_20.sce b/3731/CH5/EX5.20/Ex5_20.sce
new file mode 100644
index 000000000..f3e6054c5
--- /dev/null
+++ b/3731/CH5/EX5.20/Ex5_20.sce
@@ -0,0 +1,34 @@
+//Chapter 5:Dc Motor Drives
+//Example 20
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor when it is operated in dynamic breaking
+V=230 // rated voltage in V
+N=960 // rated speed in rpm
+Ia=200 // rated current in A
+Ra=0.02 // armature resistance in ohms
+Vs=230 // source voltage in V
+Rb=2 // braking resistance in ohm
+
+//Solution
+//When the motor speed is 600 rpm and the braking torque is twice the rated value
+Ia1=2*Ia //torque is directly proportional to current
+N1=600 //speed of the motor in rpm
+E=V-Ia*Ra //back emf
+E1=N1/N*E
+x=E1/Ia1-Ra //x=(1-delta)*Rb
+y=x/Rb //y=1-delta
+delta=1-y //duty ratio
+
+//(ii)If the duty ratio is 0.6 and and the braking torque is twice the rated value
+delta1=0.6 //duty ratio
+Ia1=2*Ia //torque is directly proportional to current
+E1=Ia1*((1-delta1)*Rb+Ra) //back emf
+N1=E1/E*N
+
+
+//Results
+mprintf("(i)Duty ratio is :%.2f",delta)
+mprintf("\n(ii)Hence the motor speed is :%.1f rpm",N1)
diff --git a/3731/CH5/EX5.21/Ex5_21.sce b/3731/CH5/EX5.21/Ex5_21.sce
new file mode 100644
index 000000000..eede688eb
--- /dev/null
+++ b/3731/CH5/EX5.21/Ex5_21.sce
@@ -0,0 +1,38 @@
+//Chapter 5:Dc Motor Drives
+//Example 21
+clc;
+
+//Variable Initialization
+
+//Ratings of the series motor
+N=600 //speed in rpm
+Vs=220 //source voltage in V
+Ra_Rf=0.12 //combine armature resistance field resistance
+
+//Magnetisation curve at N
+If=[10, 20,30, 40, 50, 60, 70, 80] //field current in A
+E =[64,118,150,170,184,194,202,210] //terminal voltage in V
+
+//Solution
+//(i)When the duty ratio is 0.6 and motor current is 60 A
+delta=0.6 //duty ratio
+Ia1=60 //motor current in A
+Va1=delta*Vs //terminal voltage for the given duty ratio
+E1=Va1-Ia1*Ra_Rf //back emf for the given duty ratio
+
+//For Ia1=60 A the terminal voltage is 194 V as given in the magnetization curve
+N1=E1/E(6)*N //motor speed for the given duty ratio
+
+//(ii)When the speed is 400rpm and the duty ratio is 0.65
+delta=0.65 //duty ratio
+N2=400 //speed in rpm
+Va1=delta*Vs //terminal voltage for the given duty ratio
+
+//From the magnetization characteristic for the speed of 400rpm the current Ia=70 A
+E1=Va1-If(7)*Ra_Rf //back emf for the given duty ratio
+T=(E1*If(7))/N2/(2*%pi/60) //required torque
+
+
+//Results
+mprintf("(i)Hence the motor speed is :%.1f rpm",N1)
+mprintf("\n(ii)Hence the required torque is :%.1f N-m",T)
diff --git a/3731/CH5/EX5.22/Ex5_22.sce b/3731/CH5/EX5.22/Ex5_22.sce
new file mode 100644
index 000000000..45a1f3d02
--- /dev/null
+++ b/3731/CH5/EX5.22/Ex5_22.sce
@@ -0,0 +1,51 @@
+//Chapter 5:Dc Motor Drives
+//Example 22
+clc;
+
+//Variable Initialization
+
+//The motor is operated using regenarative braking method
+N=600 //speed in rpm
+Vs=220 //source voltage in V
+Ra_Rf=0.12 //combine armature resistance field resistance
+
+//Magnetisation curve at N
+If=[10, 20,30, 40, 50, 60, 70, 80] //field current in A
+E =[64,118,150,170,184,194,202,210] //terminal voltage in V
+
+//Solution
+//(i)When the duty ratio is 0.5 and the braking torque is equal to the motor torque
+delta=0.5 //duty ratio
+Va1=delta*Vs //terminal voltage
+Ia1=If(7) //current at rated motor torque
+E1=Va1+Ia1*Ra_Rf //back emf for the given duty ratio
+N1=E1/E(7)*N //for a current of 70 A E=202 V from the magnetization curve
+
+//(ii)When maximum permisssible duty ratio is 0.95 and current is 70A
+delta_max=0.95 //maximum duty ratio
+Va1=delta_max*Vs //terminal voltage
+Ia1=70 //maximum permissible current in A
+E2=Va1+Ia1*Ra_Rf //back emf for the given duty ratio
+N2=E2/E(7)*N //for a current of 70 A E=202 V
+
+//(iii)When the motor speed is 1000rpm and maximum current is 70A with duty ratio in the range of 0.05 to 0.95
+Ia1=70 //maximum permissible current in A
+N3=1000 //given speed in rpm
+delta_max=0.95 //maximum duty ratio
+E3=N3/N*E(7) //terminal voltage
+x=(E3-delta_max*Vs)/Ia1 //x=R+Ra_Rf where R is the required external resistance
+R=x-Ra_Rf //external resistance
+
+//(iv)when the motor is running at 1000rpm with current at 70
+Ia1=70 //maximum permissible current in A
+N4=1000 //given speed in rpm
+Ra=Ra_Rf //total value of armature resistance is assumed to be the same
+E4=Va1+Ia1*Ra //back emf for the given speed N4
+E_=N/N4*E4
+ratio=E_/E(7) //fraction of the requuired number of turns to be reduced
+
+//Results
+mprintf("(i)Hence the motor speed is :%.1f rpm",N1)
+mprintf("\n(ii)Hence the motor speed is :%.1f rpm",N2)
+mprintf("\n(iii)Hence the required external resistance is :%.1f ohm",R)
+mprintf("\n(iv)Hence fraction of the number of turns to be reduced is :%.3f",ratio)
diff --git a/3731/CH5/EX5.23/Ex5_23.sce b/3731/CH5/EX5.23/Ex5_23.sce
new file mode 100644
index 000000000..b6ec994f6
--- /dev/null
+++ b/3731/CH5/EX5.23/Ex5_23.sce
@@ -0,0 +1,38 @@
+//Chapter 5:Dc Motor Drives
+//Example 23
+clc;
+
+//Variable Initialization
+
+//The motor is operated using dynamic braking method
+N=600 //speed in rpm
+Vs=220 //source voltage in v
+Ra=0.12 // armature resistance in ohms
+delta_min=0.1 //manimum value of duty ratio
+delta_max=0.9 //maximum value of duty ratio
+
+//Magnetisation curve at N
+If=[10, 20,30, 40, 50, 60, 70, 80] //field current in A
+E =[64,118,150,170,184,194,202,210] //terminal voltage in V
+
+//Solution
+//(i) Maximum braking speed is 800rpm with armature current of 70 A
+N1=800 //maximum braking speed in rpm
+Ia=70 //armature current in A
+E1=N1/N*E(7) //at 70A motor back emf is 202V
+Rbe=E1/Ia-Ra //effective value of braking resistance
+Rb=Rbe/(1-delta_min) //required braking resistance
+
+//(ii)When the speed of the motor is 87 rpm
+//now torque is maximum when the duty ratio is maximum
+N1=87 //speed in rpm
+R=Rb*(1-delta_max)+Ra
+
+Ia=If(5) //value of armature current for the given value of E=184V
+Ke_phi=E(5)/(2*%pi*N)*60
+T=Ke_phi*Ia //required torque
+
+
+//Results
+mprintf("(i)Hence braking resistance is:%.2f ohm",Rb)
+mprintf("\n(ii)Hence the required torque is :%.1f N-m",T)
diff --git a/3731/CH5/EX5.3/Ex5_3.sce b/3731/CH5/EX5.3/Ex5_3.sce
new file mode 100644
index 000000000..79a209292
--- /dev/null
+++ b/3731/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,22 @@
+//Chapter 5:Dc Motor Drives
+//Example 3
+clc;
+
+//Variable Initialization
+
+//Motor ratings
+V1=220 //rated voltage in V
+Ia1=200 //rated current in A
+Ra=0.06 //armature resistance in ohms
+Rb=0.04 //internal resistance of the variable source in ohms
+N1=800 //speed in rpm
+N2=600 //speed when motor is operatingin regenerative braking in rpm
+
+//Solution
+Ia2=0.8*Ia1 //motor is opereting in regenerative braking at 80% of Ia1
+E1=V1-Ia1*Ra //back emf at rated operation
+E2=(N2/N1)*E1 //back emf at the given speed N2
+V2=E2-Ia2*(Ra+Rb) //internal voltage of thevariable source
+
+//Results
+mprintf("\n Internal voltage of the variable source:%.1f V",V2)
diff --git a/3731/CH5/EX5.4/Ex5_4.sce b/3731/CH5/EX5.4/Ex5_4.sce
new file mode 100644
index 000000000..0774feecb
--- /dev/null
+++ b/3731/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,25 @@
+//Chapter 5:DC Motor Drives
+//Example 4
+clc;
+
+//Variable Initialization
+
+//The ratings of the motor are same as that of Ex-5.2
+V1=220 //rated voltage in V
+Ia1=100 //rated current in A
+N1=1000 //speed in rpm clockwise
+N2=800 //given speed during the dynamic braking in rpm
+Ra=0.05 //armature resistance in ohms
+Rs=0.05 //field resistance in ohms
+
+//Solution
+
+T2 = 2 //dynamic torque is twice the rated torque
+Ia2=Ia1*sqrt(T2) //since T=Kf*Ia**2
+E1=V1-Ia1*(Ra+Rs)
+E2=(Ia2/Ia1)*(N2/N1)*E1 //since E=Ke*Ia*N
+Rb=E2/Ia2-(Ra+Rs) //since E2=Ia2(Rb+Ra+Rs) during braking
+
+//Results
+mprintf("\n Braking current Ia2: %.1f A",Ia2)
+mprintf("\n Required braking resistance Rb: %.2f ohm",Rb)
diff --git a/3731/CH5/EX5.5/Ex5_5.sce b/3731/CH5/EX5.5/Ex5_5.sce
new file mode 100644
index 000000000..168a12d7e
--- /dev/null
+++ b/3731/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,61 @@
+//Chapter 5:Dc Motor Drives
+//Example 5
+clc;
+
+//Variable Initialization
+
+//Ratings of the DC shunt motor which operated under dynamic braking
+Rb=1 //braking resisance in ohms
+Ra=0.04 //armature resistance in ohms
+Rf=10 //field resistance in ohms
+T=400 //load torque in N-m
+
+//Magnetisation curve at N1
+N1=600 //speed in rpm
+If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25] //field current in A
+E =[25,50,73.5,90,102.5,110,116,121,125,129] //back emf in V
+
+//Solution
+disp(If,"Field current If:in A")
+x=(Rb+Rf)/Rb
+Ia = If * x //armature current
+Wm=2*%pi*N1/60
+Ke_flux=E / Wm //Ke*flux=constant
+T=[]
+for i=1:10
+T($+1)=(Ke_flux(i))*(Ia(i)) //torque
+end
+disp(Ke_flux,"Ke_flux :")
+disp(T,"Torque :in N-m")
+
+
+//Results
+
+//Plotting the values of Ke*flux vs If
+If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25] //field current in A
+subplot(2,1,1)
+plot(If,Ke_flux,'y')
+xlabel('field current I_f')
+ylabel('Ke*flux')
+title('If vs Ke*flux')
+xgrid(2)
+
+//Plotting the values of T vs If
+If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25] //field current in A
+subplot(2,1,2)
+plot(T,If)
+xlabel('Torque T')
+ylabel('field current I_f')
+title('T vs If')
+xgrid(2)
+
+
+mprintf("\nFrom the plot we can see that when the torque is 400 N-m, ")
+mprintf("\nthe field current is If=19.3 A, and Ke*flux=1.898 when If=19.3 A")
+T=400 // braking torque in N-m
+If=19.13 // field current in A
+Ke_flux=1.898 // Ke*flux
+Ia=x*If
+E=If*Rf+Ia*Ra //since E=V+Ia*Ra
+N2=(E/Ke_flux)*(60/(2*%pi)) //required speed
+mprintf("\nHence the required speed in is :%.1f rpm",N2)
diff --git a/3731/CH5/EX5.6/Ex5_6.sce b/3731/CH5/EX5.6/Ex5_6.sce
new file mode 100644
index 000000000..7240174df
--- /dev/null
+++ b/3731/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,68 @@
+//Chapter 5:Dc Motor Drives
+//Example 6
+clc;
+
+//Variable Initialization
+
+//The motor rating is same as that of Ex-5.5
+N=600 //value of the speed given from the magnetization curve in Ex-5.5
+
+Ra=0.04 //armature resistance in ohms
+Rf=10 //field resistance in ohms
+T=400 //load torque in N-m
+N1=1200 //given speed in rpm to hold the overhauling torque
+
+//Solution
+Wm=2*%pi*N1/60 //angular speed at the given speed N1
+
+//Magnetisation curve at N=600rpm
+If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25] //field current in A
+E =[25,50,73.5,90,102.5,110,116,121,125,129] //value of the back emf as given in Ex-5.5 for the speed N in V
+
+//Magnetisation curve at N=1200rpm
+If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25] //field current in A
+E1=N1/N*E //back emf at the speed N1
+mprintf("Hence the magnetization curve at 1200rpm is")
+disp(If,"Field current If:in A")
+disp(E1,"Back emf is E1 in V:")
+
+Pd=T*Wm //power developed
+x=Pd*Ra
+V=[]
+for i=1:10
+V($+1)=E1(i)-x/E1(i)
+end
+disp(V,"Terminal voltage V:in V")
+
+//Results
+//Plotting the values of V vs If
+subplot(2,1,1)
+plot(V,If)
+xlabel('Terminal voltage V')
+ylabel('Field current I_f')
+title('V vs If')
+xgrid(2)
+
+//Plotting the values of E vs If
+If=[2.5,5,7.5,10,12.5,15,17.5,20,22.5,25] //field current in A
+E =[25,50,73.5,90,102.5,110,116,121,125,129] //value of the back emf as given in Ex-5.5 for the speed N in V
+E1=N1/N*E //back emf at the speed N1
+
+subplot(2,1,2)
+plot(E1,If,'y')
+xlabel('E')
+ylabel('Field current I_f')
+title('E vs If')
+xgrid(2)
+
+mprintf("\nFrom the plot we can see that when the current If=25 A the terminal voltage is V=250 V with the back emf E=258V")
+
+E=258 //value of the back emf in V at from the plot
+V=250 //value of terminal voltage in V from the plot at E=258 V
+If=25 //value of If in A from the plot at E=258 V
+Ia=(E-V)/Ra //armature current
+If=V/Rf //field current
+Ir=Ia-If
+Rb=V/Ir //braking resistance
+
+mprintf("\nHence the rquired braking resistance is %.3f ohm",Rb)
diff --git a/3731/CH5/EX5.7/Ex5_7.sce b/3731/CH5/EX5.7/Ex5_7.sce
new file mode 100644
index 000000000..c4238f101
--- /dev/null
+++ b/3731/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,53 @@
+//Chapter 5:Dc Motor Drives
+//Example 7
+clc;
+
+//Variable Initialization
+
+//Ratings of the DC series motor which operated under dynamic braking
+Ra=0.5 //total resistance of armature and field windings in ohms
+Rf=10 //field resistance in ohms
+T=500 //overhauling load torque in N-m
+N=600 //speed at the overhauling torque T in rpm
+
+//Nagnetisation curve at a speed of 500 rpm
+N1=500 //speed in rpm
+Ia=[20, 30, 40, 50, 60, 70, 80] //armature current in A
+E =[215,310,381,437,482,519,550] //back emf in V
+
+//Solution
+Wm1=2*%pi*N1/60
+disp(Ia,"Armature current : in A")
+Ke_flux=E / Wm1 //Ke*flux=constant
+disp(Ke_flux,"Ke_flux :")
+T=[]
+for i=1:7
+T($+1)=(Ke_flux(i))*(Ia(i)) //torque
+end
+disp(T,"Torque :in N-m")
+
+
+//Results
+//Plotting the values of Ke*flux vs Ia and T vs Ia
+subplot(2,1,1)
+plot(Ia,Ke_flux,'y')
+xlabel('Armature current I_a')
+ylabel('Ke*flux')
+title('Ke*flux vs Ia')
+xgrid(2)
+
+subplot(2,1,2)
+plot(T,Ia)
+xlabel('Torque T')
+ylabel('Armature current I_a')
+title('T vs Ia')
+xgrid(2)
+
+mprintf("\nFrom the plot we can see that at the given torque T=500 N-m the current Ia is 56 A, and Ke*flux is 8.9 at Ia=56 A")
+Ke_flux=8.9 //value of Ke*flux at T=500 N-m from the plot
+Ia=56 //value of Ia at at T=500 N-m from the plot
+Wm=2*%pi*N/60
+E=Ke_flux*Wm //required emf
+x=E/Ia //x=Ra+Rb
+Rb=x-Ra //required braking resistance
+mprintf("\nHence the rquired braking resistance is %.3f ohm",Rb)
diff --git a/3731/CH5/EX5.8/Ex5_8.sce b/3731/CH5/EX5.8/Ex5_8.sce
new file mode 100644
index 000000000..02f4e22e9
--- /dev/null
+++ b/3731/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,37 @@
+//Chapter 5:Dc Motor Drives
+//Example 8
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor
+V=220 // rated voltage in V
+N=970 // rated speed in rpm
+Ia=100 // rated current in A
+Ra=0.05 // armature resistance in ohms
+N1=1000 // initial speed of the motor in rpm
+
+//Solution
+E=V-Ia*Ra
+E1=N1/N*E //value of back emf at the speed N1
+
+//(a)The resistance to be placed
+Ia1=2*Ia //value of the braking current is twice the rated current
+Rb=(E1+V)/Ia1-Ra //required resistance
+
+//(b)The braking torque
+Wm=(2*%pi*N1)/60
+T=E1*Ia1/Wm
+
+//(c)When the speed has fallen to zero the back emf is zero
+E2=0
+Ia2=V/(Ra+Rb)
+T2=Ia2/Ia1*T //since the torque is directly proportional to the current
+
+
+//Results
+mprintf("(a)Hence required resistance is :%.2f ohm",Rb)
+//Answer given for the resistance in the book is wrong
+
+mprintf("\n(b)Hence the required braking torque is :%.1f N-m",T)
+mprintf("\n(c)Hence the required torque is :%.1f N-m",T2)
diff --git a/3731/CH5/EX5.9/Ex5_9.sce b/3731/CH5/EX5.9/Ex5_9.sce
new file mode 100644
index 000000000..d98a78d64
--- /dev/null
+++ b/3731/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,79 @@
+//Chapter 5:Dc Motor Drives
+//Example 9
+clc;
+
+//Variable Initialization
+
+//Ratings of the separately excited motor which operates under rheostatic braking
+V=220 // rated voltage in V
+N=1000 // rated speed in rpm
+Ia=175 // rated current in A
+Ra=0.08 // armature resistance in ohms
+N1=1050 // initial speed of the motor in rpm
+J=8 // moment of inertia of the motor load system kg-m2
+La=0.12 // armature curcuit inductance in H
+
+//Solution
+E=V-Ia*Ra
+Wm=N*2*%pi/60 //rated speed in rad/s
+
+//(a)When the braking current is twice the rated current
+Ia1=2*Ia
+E1=N1/N*E
+x=E1/Ia1 //x=Rb+Ra
+Rb=x-Ra //required braking resistance
+
+//(b)To obtain the expression for the transient value of speed and current including the effect of armature inductance
+Ra=x //total armature current
+K1=N1*2*%pi/60 //initial speed in rad/s
+K=E/Wm
+B=0
+ta=La/Ra //time constant in sec
+Trated=E*Ia/Wm //rated torque
+Tl=0.15*Trated //load torque is 15% of the rated torque
+tm1= %inf //tm1=J/B and B=0 which is equal to infinity
+tm2=J*Ra/(B*Ra+K**2)
+
+a = ta
+b = -(1+ta/tm1)
+c = 1/tm2
+
+//Discriminant
+d = (b**2) - (4*a*c)
+
+alpha1 = (-b-sqrt(d))/(2*a)
+alpha2 = (-b+sqrt(d))/(2*a)
+
+K3=tm2*Tl/J
+K4=tm2*K*Tl/J/Ra
+
+//Transient value for speed
+x1=((J*alpha2-B)*K1-(Tl-J*alpha2*K3))/(J*(alpha2-alpha1))
+y1=((J*alpha1-B)*K1-(Tl-J*alpha1*K3))/(J*(alpha1-alpha2))
+
+//Transient value for the current
+x2=(K*K1+alpha2*La*K4)/(La*(alpha2-alpha1))
+y2=(K*K1+alpha1*La*K4)/(La*(alpha1-alpha2))
+
+
+//(c) To calculate the time taken by braking operation and the maximum value of the armature current
+//now Wm=0 for the braking operation and hence 151.5 exp(-0.963*t1)- 8.247 = 0 from the previous answer in (b)
+a=K3/x1 //a=exp(-0.963*t1)
+t1=-alpha1*log(real(a)) //take log base e on both sides
+//now d/dt(ia)=0 for themaximum current and hence d/dt(26.25-593.1exp(-0.963*t)+566.8exp(-4.19*t) = 0 from the previous answer in (b)
+b=abs(alpha2*y2)/abs(alpha1*x2) //b=exp(-0.963*t)/exp(-4.19*t)
+t2=log(b)/(-alpha1+alpha2) //take log base e on both sides
+t2=abs(t2)
+ia=K4-real(x2)*exp(real(-alpha1)*t2)-real(y2)*exp(real(-alpha2)*t2)
+
+
+//Results
+mprintf("(a)Hence the braking resistance is :%.3f ohm",Rb)
+mprintf("\nb)The value of alpha1 :%.3f and alpha2 :%.3f ",real(alpha1),real(alpha2))
+mprintf("\nHence the expression for the transient value for the speed is")
+mprintf("\nWm=%.1f exp( -%.3f *t)%.1f exp( -%.2f *t) - %.3f",real(x1),real(alpha1),real(y1),real(alpha2),K3)
+mprintf("\nHence the expression for the transient value for the current is")
+mprintf("\nia=%.2f -%.1f exp(-%.3f*t) +%.1f exp(%.2f*t)",K4,real(x2),real(alpha1),-real(y2),-real(alpha2))
+mprintf("\n(c)Hence the time taken is :%.2f sec",t2)
+mprintf("\nHence the maximum current is:%.2f A",ia)
+//There is a slight difference in the answers because of rounding
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