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+
+//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)