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//CHAPTER 8- DIRECT CURRENT MACHINES
//Example 7
disp("CHAPTER 8");
disp("EXAMPLE 7");
//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
N1=1150; //speed in rpm
P=4; //number of poles
Z=882; //number of armature conductors
r_a=0.188; //armature resistance in Ohms
I_a1=73; //armature current in Amperes
I_f=1.6; //field current in Amperes
ratio=0.8; //phi2:phi1=0.8 (here phi=flux)
//SOLUTION
E_b1=v_t-(I_a1*r_a);
I_a2=I_a1/ratio; //(phi2*I_a2)=(phi1*I_a1)
E_b2=v_t-(I_a2*r_a);
N2=(E_b2/E_b1)*(1/ratio)*N1; //N2:N1=(E_b2/E_b1)*(phi1/phi2)
N2=round(N2); //to round off the value of N2 (before rounding off N2=1414.695516 rpm)
disp(sprintf("The new operating speed is %d rpm",N2));
//The answer is slightly different due to the precision of floating point numbers
//END
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