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+//CHAPTER 8- DIRECT CURRENT MACHINES
+//Example 8
+
+disp("CHAPTER 8");
+disp("EXAMPLE 8");
+
+//VARIABLE INITIALIZATION
+v_t=250;                     //in Volts
+r_a=0.1;                     //armature resistance in Ohms
+r_f=125;                     //field resistance in Ohms
+p_o=20*1000;                 //output power in Watts
+N_g=1000;                    //speed as generator in rpm
+
+//SOLUTION
+
+//machine as a generator
+I_l=p_o/v_t;
+I_f=v_t/r_f;                 //I_f is same as I_sh 
+I_ag=I_l+I_f;
+E_a=v_t+(I_ag*r_a);          //induced emf = E_a = E_g
+
+//machine as a motor
+I_l=p_o/v_t;
+I_f=v_t/r_f;
+I_am=I_l-I_f;
+E_b=v_t-(I_am*r_a);         //back emf = E_b = E_m
+
+//solution (a)
+N_m=(N_g*E_b)/E_a;
+N_m=round(N_m);             //to round off the value of N_m
+disp(sprintf("(a) The speed of the same machine as a motor is %d rpm",N_m));
+
+//solution (b)
+
+//(i)
+p1=(E_a*I_ag)/1000;          //to express the answer in kW
+disp(sprintf("(b) (i) The internal power developed as generator is %f kW",p1));
+
+//(ii)
+p2=(E_b*I_am)/1000;        
+disp(sprintf("(b) (ii) The internal power developed as motor is %f kW",p2));
+
+//END
+