The bandwidth defined by $f_p$ is
$$\bbox[10px,border:1px solid grey]{BW= f_2 -f_1 = {f_p \over Q_p}}$$
By substituting $Q_p$ from the
previous topic, we can show that
$$\begin{split}
BW&= f_2 -f_1 \approx {f_p \over Q_l} = {f_p \over X_L/R_l}\\
&= {f_p R_l \over 2 \pi f_p L} = {R_l \over 2 \pi L} \end{split}$$
Hence,
$$\bbox[10px,border:1px solid grey]{BW= f_2 -f_1 \approx {R_l \over 2 \pi L}}$$
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