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+//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