To reduce the resonant frequency in an L-C-R series circuit with a generator, (a) the iron core of the inductor should be removed and the value of the capacitor is kept constant. (b) the dielectric in the capacitor should be removed and the value of the inductor is kept constant. (c) the generator frequency should be reduced. (d) another capacitor should be added in parallel to the first capacitor and the value of the inductor is kept constant.

Answer: (d). f₀ = (1)/(2π√(LC)): to lower f₀, increase L or C. (a) Removing the iron core lowers L, so f₀ rises. (b) Removing the dielectric lowers C, so f₀ rises. (c) Generator frequency does not change f₀, which depends only on L and C. (d) A capacitor in parallel increases C, so f₀ falls. So the answer is (d).

To reduce the resonant frequency in an L-C-R series circuit with a… | MHT CET 2026 (19 May, Shift 2)

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Resonance, Bandwidth & Q Factor

To reduce the resonant frequency in an L-C-R series circuit with a generator,

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Answer: (d)

$f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ : to lower $f_0$ , increase $L$ or $C$ .
(a) Removing the iron core lowers $L$ , so $f_0$ rises.
(b) Removing the dielectric lowers $C$ , so $f_0$ rises.
(c) Generator frequency does not change $f_0$ , which depends only on $L$ and $C$ .
(d) A capacitor in parallel increases $C$ , so $f_0$ falls. So the answer is (d).

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Resonance, Bandwidth & Q Factor question from MHT CET 2026 (19 May, Shift 2)

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