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//CHAPTER 8- DIRECT CURRENT MACHINES
//Example 32
disp("CHAPTER 8");
disp("EXAMPLE 32");
//VARIABLE INITIALIZATION
v_t=250; //in Volts
I=20; //in Amperes
N1=1000; //in rpm
P=4; //number of poles
r_p=0.05; //resistance of field coil on each pole in Ohms
r_a=0.2; //in Ohms
//SOLUTION
r_se=P*r_p;
r_m=r_a+r_se; //resistance of motor
E_b1=v_t-(I*r_m);
T1=I^2;
//solution (a)
//solving the quadratic equation directly,
r=10; //in Ohms
a=1.02;
b=-25;
c=-400;
D=b^2-(4*a*c);
x1=(-b+sqrt(D))/(2*a);
x2=(-b-sqrt(D))/(2*a);
//to extract the positive root out of the two
if (x1>0 & x2<0)
I1=x1;
else (x1<0 & x2>0)
I1=x2;
end;
I_a=((10.2*I1)-v_t)/r;
E_b2=v_t-(I_a*r_a);
N2=((E_b2/E_b1)*I*N1)/I1;
N2=round(N2); //to round off the value
disp(sprintf("(a) The speed with 10 Ω resistance in parallel with the armature is %d rpm",N2));
//solution (b)
//solving the quadratic equation directly,
a=5/7;
b=0;
c=-400;
D=b^2-(4*a*c);
y1=(-b+sqrt(D))/(2*a);
y2=(-b-sqrt(D))/(2*a);
//to extract the positive root out of the two
if (y1>0 & y2<0)
I2=y1;
else (y1<0 & y2>0)
I2=y2;
end;
E_b3=v_t-(I2*r_a);
N3=((E_b3/E_b1)*I*N1)/(I2*a);
N3=round(N3); //to round off the value
disp(sprintf("(b) The speed with 0.5 Ω resistance in parallel with series field is %d rpm",N3));
//The answers are slightly different due to the precision of floating point numbers
//END
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