For a series RLC circuit having R=10Ω, L=0.01 H and C=100?F, find the resonant frequency, quality factor and bandwidth.

Answer:

$R= 10 \ \Omega$

$L=0.01 \mathrm H$

$C=100 \mu F$

At resonance, $X_L \ =\ X_c $

$f_r = \dfrac{1}{2\pi \sqrt{L\times C}}$

$f_r = \dfrac{1}{2\pi \sqrt{0.01*100*10^{-6}}}$

$ f_r= 159.154\ \mathrm Hz$

  The resonant frequency is 159.154Hz  

$Q = \dfrac{X_L}{R}$

$Q = \dfrac{2\pi*159.154*0.01}{10}$$\Big[\because X_L = 2\pi \times f_r \times L\Big]$

$ Q= 0.9999$

The Quality factor is 0.9999~1  

$\Delta f = \dfrac{R}{2\pi\times L}$

$\Delta f = \dfrac{10}{2\pi \times 0.01}$

$\Delta f = 159.154Hz$

The bandwidth is 159.154Hz