Medium
Obtain the equation of bandwidth for an $L-C-R$ series $AC$ circuit and deduce the equation of $Q$ factor.
(N/A) The current amplitude in an $L-C-R$ series circuit is $I_m = \frac{V_m}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}$.
At resonance,$\omega_0 = 1/\sqrt{LC}$,so $I_{max} = V_m/R$.
The half-power frequencies $\omega_1$ and $\omega_2$ occur when $I_m = I_{max}/\sqrt{2}$,which implies $Z = \sqrt{2}R$.
Thus,$R^2 + (\omega L - 1/\omega C)^2 = 2R^2$,leading to $\omega L - 1/\omega C = \pm R$.
For $\omega_2 = \omega_0 + \Delta\omega$,we have $\omega_2 L - 1/(\omega_2 C) = R$.
Substituting $\omega_2 = \omega_0 + \Delta\omega$ and using $\omega_0 L = 1/(\omega_0 C)$,we get $L(2\Delta\omega) = R$,so the bandwidth is $2\Delta\omega = R/L$.
The $Q$ factor is defined as $Q = \omega_0 / (2\Delta\omega) = \omega_0 L / R = \frac{1}{R} \sqrt{\frac{L}{C}}$.
At resonance,$\omega_0 = 1/\sqrt{LC}$,so $I_{max} = V_m/R$.
The half-power frequencies $\omega_1$ and $\omega_2$ occur when $I_m = I_{max}/\sqrt{2}$,which implies $Z = \sqrt{2}R$.
Thus,$R^2 + (\omega L - 1/\omega C)^2 = 2R^2$,leading to $\omega L - 1/\omega C = \pm R$.
For $\omega_2 = \omega_0 + \Delta\omega$,we have $\omega_2 L - 1/(\omega_2 C) = R$.
Substituting $\omega_2 = \omega_0 + \Delta\omega$ and using $\omega_0 L = 1/(\omega_0 C)$,we get $L(2\Delta\omega) = R$,so the bandwidth is $2\Delta\omega = R/L$.
The $Q$ factor is defined as $Q = \omega_0 / (2\Delta\omega) = \omega_0 L / R = \frac{1}{R} \sqrt{\frac{L}{C}}$.
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