Question

Comprehension :

An alternating voltage of $260\text{V}$ and $\omega =500{\text{rad s}}^{-1}$ is applied in series $\text{LCR}$ circuit, where $L=0.01\text{H},C=4\times {10}^{-4}\text{F}$ and $R=10\Omega$

$\left(i\right)$ Find the resonance frequency of the circuit (in $\text{Hz}$)-

• A.
  $\frac{25}{\pi }$
• B.
  $\frac{250}{\pi }$
• C.
  $\frac{40}{\pi }$
• D.
  $\frac{200}{\pi }$

Answer 1

The correct option is B

$\frac{250}{\pi }$
Given, $L=0.01\text{H};C=4\times {10}^{-4}\text{F}R=10\Omega ;V=120\text{V};\omega =500{\text{rad s}}^{-1}$

At the resonance condition,

$X_L=X_C$

$\omega L=\frac{1}{\omega C}$

${\omega }^2=\frac{1}{LC}$

$\omega =\frac{1}{\sqrt{LC}}$

Substituting the given data gives,

$\omega =\frac{1}{\sqrt{0.01\times 4\times {10}^{-4}}}$

$\omega =\frac{1}{\sqrt{4\times {10}^{-6}}}$

$\omega =\frac{1}{2\times {10}^{-3}}=500$

$2\pi f=500$

$f=\frac{500}{2\pi }=\frac{250}{\pi }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.