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+//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