JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Alternating Current

The resonant frequency of a series circuit, ${\omega }_0=\frac{1}{\sqrt{LC}}$, contains no resistance, so changing $R$ leaves it where it is. The sharpness of the resonance does depend on $R$: the bandwidth is $\Delta \omega =\frac{R}{L}$ and the quality factor is $Q=\frac{{\omega }_{0}L}{R}$, so a larger resistance dissipates energy faster, flattens the current peak and lowers $Q$. Raising the resistance increases the band width of the resonance circuit.