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+//CHAPTER 8- DIRECT CURRENT MACHINES
+//Example 6
+
+disp("CHAPTER 8");
+disp("EXAMPLE 6");
+
+//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
+N=1150;            //speed in rpm
+P=4;               //number of poles
+Z=882;             //number of armature conductors
+r_a=0.188;         //armature resistance in Ohms
+I_a=73;            //armature current in Amperes
+I_f=1.6;           //field current in Amperes
+
+//SOLUTION
+
+//solution (i)
+E_b=v_t-(I_a*r_a); 
+w=(2*%pi*N)/60;    //in radian/sec
+T_e=(E_b*I_a)/w;
+disp(sprintf("(i) The electromagnetic torque is %f N-m",T_e));
+
+//solution (ii)
+A=P;               //since it is lap winding, so A=P and A=number of parallel paths
+phi=(E_b*60*A)/(P*N*Z);
+disp(sprintf("(ii) The flux per pole is %f Wb",phi));
+
+//solution (iii)
+p_rotor=E_b*I_a;   //power developed on rotor
+p_rot=p_rotor-p_o; //p_shaft=p_out
+disp(sprintf("(iii) The rotational power is %f W",p_rot)); 
+
+//solution (iv)
+tot_loss=p_rot+((I_a^2)*r_a)+(v_t*I_f);
+p_i=p_o+tot_loss;
+eff=(p_o/p_i)*100;
+disp(sprintf("(iv) The efficiency is %f %%",eff));
+
+//solution (v)
+T=p_o/w;
+disp(sprintf("(v) The shaft torque is %f N-m",T));
+
+//The answers are slightly different due to the precision of floating point numbers
+
+//END
+
+
+
+
+