Question

To reduce the resonant frequency in an LCR series circuit with a generator.

• A. the generator frequency should be reduced.
• B. dielectric in the capacitor should be removed.
• C. the iron core of the inductor should be removed.
• D. another capacitor should be added in parallel to the first.

Answer 1

The correct option is D another capacitor should be added in parallel to the first.

Formula used:
$v_0=\frac{1}{2\pi \sqrt{LC}}$

As we know, at resonance,

$X_L=X_C\Rightarrow {\omega }_{0}L=\frac{1}{{\omega }_{0}C}$

$\Rightarrow {\omega }_0=\frac{1}{\sqrt{LC}}\frac{rad}{sec}$

$\Rightarrow v_0=\frac{1}{2\pi \sqrt{LC}}Hz$

Resonant frequency in an $LCR$ series circuit is given by,

$v_0=\frac{1}{2\pi \sqrt{LC}}$

So, to reduce resonant frequency, we have to increase either $L$ or to increase $C$. To increase capacitance, another capacitor must be connected in parallel with the first.

Final Answer: Option (b)