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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
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