= Solution
The precise <bandwidth> depends on the output: this is the distinction between <displacement and current bandwidths of a series RLC circuit>. For current output, or voltage across the <resistor>, the <electrical impedance> gives
$$
|i|^2=\frac{|V_0|^2}{R^2+(\omega L-1/(\omega C))^2}.
$$
The half-power points satisfy $|\omega L-1/(\omega C)|=R$. Their positive roots are
$$
\omega_\pm=\frac{\sqrt{R^2+4L/C}\pm R}{2L},\qquad
\boxed{\Delta\omega=\omega_+-\omega_-=R/L},\qquad
\boxed{\Delta f=R/(2\pi L)}.
$$
Changing <capacitance> shifts the centre without changing this absolute width. For the charge response in part (d), the exact half-height points instead obey
$$
\omega_\pm^2=\omega_0^2-\gamma^2/2\pm\gamma\sqrt{\omega_0^2-\gamma^2/4};
$$
when both exist their separation only approaches $\gamma=R/L$ at large <quality factor>. Therefore the printed capacitance-independence claim is exact for current or resistor-voltage output and an approximation for the preceding charge resonance.
For a narrow <band-pass filter>, use current or resistor-voltage output and make $R/L$ small, counting source, load and inductor losses in the total <electrical resistance>. Increase $Q=\sqrt{L/C}/R$; if a specified centre <frequency> is to remain fixed, adjust $C$ with $L$ to keep $LC$ fixed. Capacitor-voltage output alone is not a band-pass response because it has a nonzero DC limit.
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