To reduce the resonant frequency in an LCR series circuit with a gene…

The correct answer is Option (2) → another capacitor should be added in parallel to the first

The resonant frequency of a series LCR circuit depends only on the inductance and the capacitance of the circuit:

    \( \displaystyle \nu_0 = \frac{1}{2\pi\sqrt{LC}} \)

To reduce \( \nu_0 \), the product \( LC \) must be increased.

Option (1) the generator frequency should be reduced — Incorrect. The generator frequency does not appear in \( \nu_0 \); changing it only moves the circuit away from resonance.

Option (2) another capacitor should be added in parallel to the first — Correct. Capacitors in parallel add, \( C' = C_1 + C_2 \), so \( C \) increases and \( \nu_0 \) falls.

Option (3) the iron core of the inductor should be removed — Incorrect. Removing the iron core lowers \( L \), which raises \( \nu_0 \).

Option (4) dielectric in the capacitor should be removed — Incorrect. Removing the dielectric lowers \( C \) (by the factor \( K \)), which raises \( \nu_0 \).

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 = ω L. 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). This frequency is called the resonant frequency:

    \( \displaystyle X_c = X_L \ \text{or}\ \frac{1}{\omega_0 C} = \omega_0 L \)

    \( \displaystyle \text{or}\ \ \omega_0 = \frac{1}{\sqrt{LC}} \qquad (7.28) \)

At resonant frequency, the current amplitude is maximum; i_m = v_m/R.”