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+//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)