Question

Resonance frequency of $LCR$ series $a.c.$ circuit is $f_0$. Now the capacitance is made $4$ times, then the new resonance frequency will become

• A. $f_0/4$
• B. $2f_0$
• C. $f_0$
• D. $f_0/2$

Answer 1

Answer: (D)

In $LCR$ series circuit, resonance frequency $f_0$ is given by:
$L\omega =\frac{1}{C\omega }$
$\Rightarrow {\omega }^2=\frac{1}{LC}$
$\therefore \omega =\sqrt{\frac{1}{LC}}=2\pi f_0$
$\therefore f_0=\frac{1}{2\pi \sqrt{LC}}$
$d\Rightarrow f_0\propto \frac{1}{\sqrt{C}}$
When the capacitance of the circuit is made $4$ times, its resonant frequency become $f_0^{\prime }$
$\therefore \frac{f_0^{\prime }}{f_0}=\frac{\sqrt{C}}{\sqrt{4C}}$
$\Rightarrow f_0^{\prime }=\frac{f_0}{2}$