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+//CHAPTER 8- DIRECT CURRENT MACHINES
+//Example 8
+
+disp("CHAPTER 8");
+disp("EXAMPLE 8");
+
+//250 V DC shunt machine
+//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; //load current
+I_f=v_t/r_f; //field current, I_f is same as I_sh
+I_ag=I_l+I_f; //Output current as generator
+E_a=v_t+(I_ag*r_a); //induced emf = E_a = E_g
+
+//machine as a motor
+I_l=p_o/v_t; //full load current
+I_f=v_t/r_f;
+I_am=I_l-I_f; //output current as motor
+E_b=v_t-(I_am*r_a); //back emf = E_b = E_m
+
+//solution (a)
+N_m=(N_g*E_b)/E_a; //Speed of motor in RPM
+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)
+//internal power developed as generator
+//(i)
+//total power developed in the armature
+//=Eg.Iag
+p_g=(E_a*I_ag)/1000; //to express the answer in kW divide by 1000
+disp(sprintf("(b) (i) The internal power developed as generator is %.1f kW",p_g));
+
+//(ii)
+//internal power developed as motor
+// is total power developed in armature
+//=Em.Iam
+p_m=(E_b*I_am)/1000;
+disp(sprintf("(b) (ii) The internal power developed as motor is %.1f kW",p_m));
+
+//END
+