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+//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)