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//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
//Example 30 // read it as example 29 in the book on page 2.83
disp("CHAPTER 2");
disp("EXAMPLE 30");
//VARIABLE INITIALIZATION
f=50; //Hz
rms=20; //in Amp
t1=0.0025; //in sec time to find amplitude
t2=0.0125; //in sec, to find amp after passing through +ve maximum
i3=14.14; //in Amps, to find time when will it occur after passing through +ve maxima
//SOLUTION
//i=Isin(wt)
//solution (a)
w=2*%pi*f;
Im=rms*sqrt(2);
disp(sprintf("The equation would be i=%f. sin(%f.t)", Im,w));
t0=(asin(1)/w); //time to reach maxima in +ve direction
i=Im*sin(w*t1);
disp("SOLUTION (a)");
disp(sprintf("The amplitude at time %f sec is %f Amp", t1,i));
//solution (b)
tx=t0+t2;
i2=Im*sin(w*tx);
disp("SOLUTION (b)");
disp(sprintf("The amplitude at time %f sec is %f Amp", t2,i2));
//solution (c)
ty=(asin(i3/Im))/w;
t3=t0-ty; //since ty is the time starting from 0, the origin needs to be shifted to maxima
disp("SOLUTION (c)");
disp(sprintf("The amplitude of %f Amp would be reached in %f Sec", i3,t3));
disp(" ");
//
//END
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