MCQ
To reduce the reasonant frequency in an $\text{LCR}$ series circuit with a generator:
  • A
    The generator frequency should be reduced.
  • Another capacitor should be added in parallel to the first.
  • C
    The iron core of the inductor should be removed.
  • D
    Dielectric in the capacitor should be removed.

Answer

Correct option: B.
Another capacitor should be added in parallel to the first.

Key Concept: Resonant frequency $($Natural frequency$)$
At resonance $\text{X}_\text{L}=\text{X}_\text{C}\Rightarrow\ \omega_0\text{L}=\frac{1}{\omega_0\text{C}}$
$\Rightarrow\ \omega_0=\frac{1}{\sqrt{\text{LC}}}\frac{\text{red}}{\sec}$
$\Rightarrow\ \text{v}_0=\frac{1}{2\pi\sqrt{\text{LC}}}\text{Hz}$
Resonant frequency in an $L-C-R$ circuit is given by
$\text{v}_0=\frac{1}{2\pi\sqrt{\text{LC}}}$
If $L$ or $C$ increases, the resonant frequency will reduce.
To increase capacitance, we must connect another capacitor parallel to the first.

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