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// RLC circuit problems on resonace
clc;
clear;
R=6.28;
L=20*(10^-3);
f=5*(10^3);
w=2*%pi*f;
C=1/(L*(w^2));
Xc=1/(w*C);
Xl=L*w;
Vc=5;
Z=Xc+R+Xl;
I=Vc/Xc // Total current
V=I*R;
// frequency is inversely proportional to square root of capacitance
// So if C is halved; f will increase square root of 2 times more.
fn=sqrt(2)*f;
Xln=2*%pi*fn*L;
Q=Xln/R;
//Note under resonance conditions Vl and Vc are much greater than the supply voltage.
mprintf('i) The value of capacitor = %f micro F \n',(10^6)*C)
mprintf('ii) The supply voltage = %f V \n',V)
mprintf('iii) The frequency of resonance when C is halved = %f Hz \n',fn)
mprintf(' The Q of the new circuit = %f \n',Q)
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