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//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
//Example 15
disp("CHAPTER 2");
disp("EXAMPLE 15");
//GIVEN
//choke coil takes current of 2 Amp 60 deg lagging
//Applied voltage 200 V 50Hz
//VARIABLE INITIALIZATION
I=2; //in Amperes
angle_I=60; //in degrees
v1=200; //in Volts
f=50; //in Hertz
//SOLUTION (i)
z1=v1/I;
disp(sprintf("The impedance is %d Ω, %d degrees",z1,angle_I));
//function to convert from polar form to rectangular form
function [x,y]=pol2rect(mag,angle);
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*f);
disp(sprintf("The inductance is %5.3f H",L));
//SOLUTION (ii)
//Choke is now connected to 100 V 25 hz power supply
//Howevetr, R and L of the choke will remain the same
//Reactance will change
v2=100; // in volts
f2=25; // in Hz
x2=2*%pi*f2*L; // inductive reactance in the new system
z2=sqrt((r^2)+(x2^2)); // impedance in the new system
angle=atan(x2/r);
I1=v2/z2; // current in the new system
p=v2*I1*cos(-angle); //power consumed
//
//disp(sprintf("The angle is %5.4f ",angle));// text book value is assumed 0.75
disp(sprintf("The power consumed is %5.1f W",p));
//END
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