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//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
//Example 35 // read it as example 34 in the book on page 2.88
disp("CHAPTER 2");
disp("EXAMPLE 35");
//VARIABLE INITIALIZATION
R=100; //in Ω
L=0.2; //in Henry
C=20*10^(-6); //farads
V=240; // volts
f=50; //Hz
//
//SOLUTION
//Solution (a)
XL=2*%pi*f*L;
XC=1/(2*%pi*f*C);
//impedence Z=sqrt(R^2 +XL^2)
X=XL-XC;
Z=sqrt(R^2 +X^2);
disp("SOLUTION (a)");
disp(sprintf("The total impedence is %d Ω", Z));
I=V/Z;
disp("SOLUTION (b)");
disp(sprintf("The total current is %.3f Amp", I));
Vr=I*R;
Vi=I*XL;
Vc=I*XC;
disp("SOLUTION (c)");
disp(sprintf("The voltage across resistance is %.1f V",Vr));
disp(sprintf("The voltage across inductance is %.1f V",Vi));
disp(sprintf("The voltage across capacitance is %.1f V",Vc));
pf=R/Z;
pc=V*I*pf;
disp("SOLUTION (d)");
disp(sprintf("The Power Factor is %.2f leading", pf));
disp("SOLUTION (e)");
disp(sprintf("The Power consumed in the circuit is %.0f W",pc));
//XL=XC
f0=1/(2*%pi*sqrt(L*C));
disp("SOLUTION (f)");
disp(sprintf("Resonance will occur at %.1f Hz",f0));//The text book answer is 39.8 which is apprently wrong
disp(" ");
//
//END
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