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//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
//Example 15
clc;
disp("CHAPTER 2");
disp("EXAMPLE 15");
//VARIABLE INITIALIZATION
I=2; //in Amperes
angle_I=60; //in degrees
v1=200; //in Volts
f1=50; //in Hertz
v2=100; //in Volts
f2=25; //in Hertz
//SOLUTION
//solution (i): when supply is 200V and frequency is 50 Hz
z1=v1/I;
disp(sprintf("(i) When the supply is 200V and frequency is 50 Hz:"));
disp(sprintf("The impedance is %d Ω, %d degrees",z1,angle_I));
function [x,y]=pol2rect(mag,angle); //function 'pol2rect()' converts impedance in polar form to rectangular form
x=mag*cos(angle*(%pi/180)); //to convert the angle from degrees to radians
y=mag*sin(angle*(%pi/180));
endfunction;
[r,x1]=pol2rect(z1,angle_I);
disp(sprintf("The resistance is %d Ω",r));
L=x1/(2*%pi*f1);
disp(sprintf("The inductance is %f H",L));
//solution (ii): when supply is 100V and frequency is 25 Hz
x2=2*%pi*f2*L;
z2=sqrt((r^2)+(x2^2));
angle=atan(x2/r);
I1=v2/z2;
p=v2*I1*cos(-angle);
disp(sprintf("(ii) When supply is 100V and frequency is 25 Hz:"));
disp(sprintf("The power consumed is %f W",p));
//Answer may be slightly different due to precision of floating point numbers
//END
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