Quality Factor or Q-Factor

Worked out. A series RLC circuit has a 500 $\mu$H inductor, an $\dfrac{80}{\pi^2}$ pF capacitor and a 628 $\Omega$ resistor, and the resonant frequency and Q-factor are required. The resonant frequency is $f_r=\dfrac{1}{2\pi\sqrt{LC}}$; substituting the given L and C and simplifying the resulting expression gives $f_r\approx2500$ kHz. The Q-factor is $Q=\omega_r L/R = 2\pi f_r L/R = (2\pi)(2500\times10^3)(500\times10^{-6})/628 \approx12.5$. The example is a direct, two-formula numerical drill applying both the resonant-frequency and Q-factor formulas to the sa …

Worked out. A series RLC circuit resonating at 400 kHz has an 80 $\mu$H inductor, a 2000 pF capacitor and a 50 $\Omega$ resistor, and the questions ask for (i) the initial Q-factor, (ii) the new capacitance needed to keep the SAME 400 kHz resonant frequency if the inductance is doubled to 160 $\mu$H, and (iii) the new Q-factor with these doubled-L, adjusted-C values. (i) $Q=\dfrac{1}{R}\sqrt{L/C} = \dfrac{1}{50}\sqrt{(80\times10^{-6})/(2000\times10^{-12})} = 4$. (ii) Keeping $f_r$ fixed while L doubles requires $C_2=1/(4\pi^2f_r^2L_2)$, which works out to $C_2\approx1000$ pF -- exactly HALF the original capacitance, as expected since $f_r\propto1/\sqrt{LC}$ requires the LC product to stay constant. (iii) The new Q-factor is $Q_2=\dfrac{1}{R}\sqrt{L_2/C_2}=\dfrac{1}{50}\sqrt{(160\times10^{-6})/(1000\times10^{-12})}=8$ -- exactly DOUBLE the original Q-factor, showing that doubling L while halving C (to keep …