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//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
//Example 13
disp("CHAPTER 2");
disp("EXAMPLE 13");
//given
//load of impedance 1+j.1 ohm connected AC Voltage
//AC Voltage represented by V=20.sqrt(2).cos(wt+10) volt
//to find
//current in form of i=Im.sin(wt+phi) A
// real power
//Equations to be used
//real Power pr=Vrms.Irms.cos (phi)
// =(Vm/sqrt(2)).(Im/sqrt(2)).cos(phi)
// apparent power pa=Vrms.Irms
// =(Vm/sqrt(2)).(Im/sqrt(2))
//
//VARIABLE INITIALIZATION
z1=1+(%i*1); //impedance in rectangular form in Ohms
v=20*sqrt(2); //amplitude of rms value of voltage in Volts
//SOLUTION
function [z,angle]=rect2pol(x,y);
z=sqrt((x^2)+(y^2)); //z is impedance & the resultant of x and y
angle=atan(y/x)*(180/%pi); //to convert the angle from radians to degrees
endfunction;
//solution (i)
[z,angle]=rect2pol(1,1);
v=v/sqrt(2);
angle_v=100; //v=(20/sqrt(2))*sin(ωt+100)
I=v/z; //RMS value of current
angle_I=angle_v-angle;
Im=I*sqrt(2);
disp(sprintf("(i) The current in load is i = %d sin(ωt+%d) A",Im,angle_I));
//solution (ii)
pr=(v/sqrt(2))*(I*sqrt(2))*cos(angle*(%pi/180));
disp(sprintf("(ii) The real power is %4.0f W",pr));
//solution (iii)
pa=(v/sqrt(2))*(I*sqrt(2));
disp(sprintf("(ii) The apparent power is %6.2f VAR",pa));
//END
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