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+//CHAPTER 8- DIRECT CURRENT MACHINES
+//Example 32
+
+disp("CHAPTER 8");
+disp("EXAMPLE 32");
+
+//250 V series motor at 1000 rpm
+//VARIABLE INITIALIZATION
+v_t=250;                     //in Volts
+I=20;                        //in Amperes
+N1=1000;                     //in rpm
+P=4;                         //number of poles
+r_p=0.05;                    //resistance of field coil on each pole in Ohms
+r_a=0.2;                     //in Ohms
+
+//SOLUTION
+
+r_se=P*r_p;                  // series field resistance
+r_m=r_a+r_se;                //resistance of motor
+E_b1=v_t-(I*r_m);            // back emf
+//Torque t1 dir prop phi1.Ia
+//=> since phi dir prop Ia
+//=> torque dir prop Ia^2
+T1=I^2;                      // torque
+
+//solution (a)
+//10 ohm resistance in parallel with armature
+//let I be input currnet then, drop in series field = r_a.I
+//Voltage across the terminals = V = Vt-r_a.I
+//=> current in 10 ohm resistance (=r) = (Vt-r_a.I)/r            (eq 1)
+// now, Armature current Ia
+// Ia= I - (Vt-r_a.I)/r                                          (eq 2)
+//Torque developed t2 dir prop phi2.Ia
+//=> since phi dir prop I
+//=> torque dir prop I.Ia
+//However, T2=T1, as torque developed in two cases is equal
+//=> I.Ia = T1
+//substituting value of Ia from eq 2, we get
+//I.(I - (Vt-r_a.I)/r) =T1
+//=>I. (I.r+r_a.I -Vt)/r = T1
+//=> (r+r_a).I^2 -Vt.I =T1.r
+//=>  (r+r_a).I^2 -Vt.I - T1.r =0
+//solving the quadratic equation directly,
+r=10;                        //in Ohms
+a=10.2; //(r+r_a). value 1.02 in text book, as it was divided by r=10
+b=-250; //Vt ; -25 in text book, as it was divided by r=10
+c=-4000; // T1.r; 400 in text book, as it was not multiplied by r=10
+D=b^2-(4*a*c);
+x1=(-b+sqrt(D))/(2*a);
+x2=(-b-sqrt(D))/(2*a);
+//to extract the positive root out of the two 
+if (x1>0 & x2<0)
+I1=x1;
+else (x1<0 & x2>0)
+I1=x2;
+end;
+I_a=((10.2*I1)-v_t)/r;      // armature current
+E_b2=v_t-(I_a*r_a);         // back emf
+N2=((E_b2/E_b1)*I*N1)/I1;
+N2=round(N2);               //to round off the value
+disp(sprintf("(a) The speed with 10 Ω resistance in parallel with the armature is %d rpm",N2));
+
+//solution (b)
+//0.5 ohmic diverter resistance
+//resistance in the field winding = 0.5/(0.5+r_a)
+// since r_a=0.2,the value becomes 0.5/0.7 = 5/7
+//Torque T3 dir prop phi3.Ia
+// => dir prop 5/7 . I. I.
+//=> dir prop 5/7 I^2
+//since T3=T1
+//=> 5/7 I^2= T1
+//=> 5/7. I^2 - T1=0
+//solving the quadratic equation directly,with new values
+a=5/7;
+b=0;
+c=-400;
+D=b^2-(4*a*c); 
+y1=(-b+sqrt(D))/(2*a);
+y2=(-b-sqrt(D))/(2*a);
+//to extract the positive root out of the two 
+if (y1>0 & y2<0)
+I2=y1;
+else (y1<0 & y2>0)
+I2=y2;
+end;
+E_b3=v_t-(I2*r_a);          // back emf
+N3=((E_b3/E_b1)*I*N1)/(I2*a);
+N3=round(N3);               //to round off the value
+disp(sprintf("(b) The speed with 0.5 Ω resistance in parallel with series field is %d rpm",N3));
+
+//The answers are slightly different due to the precision of floating point numbers
+
+//END
+
+