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+//CHAPTER 8- DIRECT CURRENT MACHINES
+//Example 18
+
+disp("CHAPTER 8");
+disp("EXAMPLE 18");
+
+//VARIABLE INITIALIZATION
+v_t=460;                     //in Volts
+p_o=10*736;                  //in Watts (1 metric H.P=735.5 W)
+ratio=85/100;                //as given in the question 
+eff=84/100; 
+I_f=1.1;                     //in Amperes
+r_a=0.2;                     //in Ohms
+
+//SOLUTION
+p_i=p_o/eff;
+I_l=p_i/v_t;
+I_a=I_l-I_f;
+E1=v_t-(I_a*r_a);
+E2=E1*ratio;                 //E2:E1=N2:N1=ratio
+v=v_t-E2;                    //voltage drop across r_a and r_s (r_s is the series resistance to be inserted)
+r_s=(v/I_a)-r_a;
+disp(sprintf("The resistance required is %f Ω",r_s));
+
+//The answer is different because ratio equals 85/100 and not 75/100
+
+//END
+