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// Determine the instantaneous energy stored in the capacitor and inductor
clc;
clear;
Vc=20*sqrt(2);
C=2; // Capacitor
L=1; // Inductor
w=poly(0,'w');
// Impedances in order from let to right (as a function of w)
R1=%i*w;
R2=1/(2*%i*w); // Top
R3=1;// Bottom
Rp=R2*R3/(R2+R3); // Effective resistance of the parallel path
Reff=R1+Rp; // Effective resistance
Reff(2)=Reff(2)*conj(Reff(3));
Reff(3)=Reff(3)*conj(Reff(3));
R=imag(Reff(2))/Reff(3); // Imaginary part of the above equation
//From the above equation we get five roots, three are zero and we take the positive value
w=roots(R(2));
w=abs(w(2)); // Numerical Value
// Impedances in order from let to right (Numerical Value)
R1=%i*w;
R2=1/(2*%i*w); // Top
R3=1;// Bottom
Vcrms=Vc/sqrt(2);
// Taking Vc as reference
Ic=Vcrms/R2; // Current through Capacitor
Ir=Vcrms/R3; // Current through Resistor
Il=Ic+Ir; // Rms value of Current through Inductor
tl=atand(imag(Il)/real(Il)); // Phase angle of Inductor Current
Ilmax=abs(Il)*sqrt(2); // Maximum Current
Eins=C*(Vc^2)/2; // Magnitude of Instaneous energy stored
Ein=L*(abs(Ilmax)^2)/2; // Energy through the inductor
Er=(Ir^2)*R3; // Loss in the resistor
Q0= w*Ein*(1+(1/sqrt(2)))/Er; // Q of the circuit
printf('a) Instaneous Energy Stored in Capacitor = %g (sin(%gt)^2) \n',Eins,w)
printf(' Instaneous Energy Stored in Inductor = %g (sin( %gt + %g)^2) \n',Ein,w,tl)
printf('b) Q of the circuit = %g \n',Q0)
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