Given Data:-

• Series R-L-C Circuit
• Resistance,$R=10\ \Omega$
• Inductance,$L=0.014\ H$
• Capacitance,$C=100\mu F=100\times 10^{-6}\ F$
• Lets us consider as series R-L-C circuit as shown in figure below-

(i) Resonance frequency:-

• As we know that for a series R-L-C circuit, resonance frequency is given by,$f_r=\dfrac{1}{2\pi\sqrt{LC}}$

$\therefore\ f_r=\dfrac{1}{2\pi\sqrt{LC}}=\dfrac{1}{2\pi\sqrt{0.014\times 100\times 10^{-6}}}=134.95\ Hz$

(ii)*The Q factor of the circuit:-*

• As we know that Q-factor of the series R-L-C circuit is given by,$Q=\dfrac{\omega L}{R}=\dfrac{2\pi f L}{R}$

$\therefore Q=\dfrac{2\pi f L}{R}=\dfrac{2\pi \times 134.95\times 0.014}{10}=1.19$

(iii) The Bandwidth:-

• Lets us consider the current response with varying frequency for a series R-L-C circuit as shown in figure below-

• As the bandwidth of the circuit in rad/sec is given by,$\Delta \omega=\dfrac{R}{L}=\dfrac{10}{0.014}=714.29\ rad/sec$

Also, $\Delta \omega=2\pi \times \Delta f$

$Where,\ \Delta f \ is \ the \ BW\ in\ Hz$

$\therefore \Delta f=\dfrac{\Delta \omega}{2\pi} =\dfrac{714.29}{2\pi}=113.74\ H_z$

(iii) The half power frequency:-

• As we know that half power frequency are given by,

   $ Lower \ cut-off\ frequency,\ f_1=f_r-\dfrac{\Delta f}{2}$

& 

$Upper \ cut-off\ frequency,f_2 =f_r+\dfrac{\Delta f}{2}$

 $Where,\ \Delta f\ is \ bandwidth\ $

$\therefore \ f_1=f_r-\dfrac{\Delta f}{2}=134.95-\dfrac{113.74}{2}=178.08\ Hz$ $\therefore \ f_2 =f_r+\dfrac{\Delta f}{2}=134.95+\dfrac{113.74}{2}=191.82\ Hz$