Series RLC resonance and quality factor

Problem

A series RLC circuit has $R = 50 \ \mathrm{\Omega}$, $L = 25 \ \mathrm{mH}$, $C = 1{,}0 \ \mathrm{\mu F}$. Calculate: (a) the resonant frequency $f_0$; (b) the quality factor $Q$; (c) the maximum current at resonance if the applied voltage has amplitude $V_0 = 10 \ \mathrm{V}$.

Solution

(a) Resonant frequency. $f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{25\cdot10^{-3}\cdot 1{,}0\cdot10^{-6}}} = \frac{1}{2\pi\cdot 1{,}58\cdot10^{-4}} \approx \ev{1{,}01 \ \mathrm{kHz}}$

(b) Quality factor. $Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{50}\sqrt{\frac{25\cdot10^{-3}}{1{,}0\cdot10^{-6}}} = \frac{1}{50}\sqrt{2{,}5\cdot10^{4}} = \frac{158}{50} \approx \ev{3{,}16}$

(c) Maximum current at resonance. At resonance the reactances cancel ($X_L = X_C$) and the impedance reduces to the resistance alone, $Z = R$: $i_0 = \frac{V_0}{R} = \frac{10}{50} = \ev{0{,}20 \ \mathrm{A}}$

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Topics: Induzione elettromagnetica Concepts: Oscillazioni LC e RLC · Risonanza · Corrente alternata