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+//CHAPTER 2- STEADY-STATE ANALYSIS OF SINGLE-PHASE A.C. CIRCUIT
+//Example 20 // read it as example 19 in the book on page 2.72
+
+disp("CHAPTER 2");
+disp("EXAMPLE 20");
+
+//VARIABLE INITIALIZATION
+L=0.5                           //in Henry
+C=5                             //in mf, multiply by 10^-6 to convert to f
+R=25                            //in ohms 
+//SOLUTION
+
+//solution (i)
+//Resonance frequency f = (1/2π)sqrt((1/LC)-R^2/L^2)
+fr=(1/(2*%pi))*sqrt((1/(L*C*10^-6))-(R^2)/(L^2));
+disp("SOLUTION (i)");
+disp(sprintf("For parallel circuit,Resonant frquency is %3f Hz",  fr)); 
+disp(" ");
+
+//solution (ii)
+//Total circuit impedance at resonance is Z=L/RC
+z=L/(R*C*10^-6);
+disp("SOLUTION (ii)");
+disp(sprintf("Total impedence at resonance is %3f kΩ",  z/1000));
+// 
+//solution (iii)
+//Bandwidth (f2-f1)=R/(2.π.L)
+bw=R/(2*%pi*L);
+disp("SOLUTION (iii)");
+disp(sprintf("Bandwidth is %3f Hz", bw));
+//
+//solution (iv)
+//Quality factor Q=1/R.sqrt(L/C)
+Q=(1/R)*sqrt(L/(C*10^-6));
+disp("SOLUTION (iv)");
+disp(sprintf("Quality Factor is %3f", Q)); 
+//solution in the book is wrong as there is a total mistake in imaginery part 7.2+0.798=11.598
+//
+//END