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+//CHAPTER 8- DIRECT CURRENT MACHINES
+//Example 7
+
+disp("CHAPTER 8");
+disp("EXAMPLE 7");
+
+//VARIABLE INITIALIZATION
+p_o=20*746;                  //output power from H.P. to Watts (1 H.P.=745.699 or 746 W) 
+v_t=230;                     //in Volts
+N1=1150;                     //speed in rpm
+P=4;                         //number of poles
+Z=882;                       //number of armature conductors
+r_a=0.188;                   //armature resistance in Ohms
+I_a1=73;                     //armature current in Amperes
+I_f=1.6;                     //field current in Amperes
+ratio=0.8;                   //phi2:phi1=0.8  (here phi=flux)
+
+//SOLUTION
+
+E_b1=v_t-(I_a1*r_a);  
+I_a2=I_a1/ratio;             //(phi2*I_a2)=(phi1*I_a1)
+E_b2=v_t-(I_a2*r_a);
+N2=(E_b2/E_b1)*(1/ratio)*N1; //N2:N1=(E_b2/E_b1)*(phi1/phi2)
+N2=round(N2);                //to round off the value of N2 (before rounding off N2=1414.695516 rpm)
+disp(sprintf("The new operating speed is %d rpm",N2));
+
+//The answer is slightly different due to the precision of floating point numbers
+
+//END
+
+
+