With an increase in the frequency of the ac supply, the impedance of…

The correct answer is Option (4) → decreases at first, becomes minimum and then increases

The impedance of a series LCR circuit is \( Z=\sqrt{R^{2}+(X_C-X_L)^{2}} \), with \( X_L=\omega L \) rising and \( X_C=1/\omega C \) falling as the frequency increases.

Option (1) remains constant — Incorrect. Both reactances depend on the frequency, so \( Z \) changes as the frequency is changed.

Option (2) increases — Incorrect. This holds only above the resonant frequency. At low frequencies \( X_C \) is large and falls as the frequency rises, so \( Z \) first goes down.

Option (3) decreases — Incorrect. This holds only below the resonant frequency. Above it, \( X_L \) keeps growing and \( Z \) rises again.

Option (4) decreases at first, becomes minimum and then increases — Correct. The impedance is least at resonance. It is larger on both sides of it.

• Below resonance \( X_C\gt X_L \), and the gap \( X_C-X_L \) shrinks as the frequency rises, so \( Z \) decreases.

• At the resonant frequency \( X_C=X_L \), so \( Z=R \), the minimum. Above it, \( X_L-X_C \) grows and \( Z \) increases.

NCERT (2026-27):
Resonance – “… For an RLC circuit driven with voltage of amplitude v_m and frequency ω, we found that the current amplitude is given by

    \( \displaystyle i_m=\frac{v_m}{Z}=\frac{v_m}{\sqrt{R^{2}+(X_C-X_L)^{2}}} \)

with X_c = 1/ωC and X_L = ωL. So if ω is varied, then at a particular frequency ω₀, X_c = X_L, and the impedance is minimum (\( Z=\sqrt{R^{2}+0^{2}}=R \)).”