Question

A series LCR circuit consists of an inductor L a capacitor C and a resistor R connected across a source of emf $\epsilon ={\epsilon }_0$ sin $\omega t$. When $\omega L=\frac{1}{\omega C}$ the current in the circuit is $l_0$ and if angular frequency of the source is changed to $\omega$', the current in the circuit becomes $\frac{l_0}{2}$, then the value of $|{\omega }^{\prime }L-\frac{1}{{\omega }^{\prime }C}|$ is

• A. R
• B. $\sqrt{3}$ R
• C. $\sqrt{1}5$ R
• D. Zero

Answer 1

The correct option is B $\sqrt{3}$ R

Let, $Z=|\omega L-\frac{1}{\omega C}|=0$

Then, $I_0=\frac{{\epsilon }_0}{\sqrt{R^2+Z^2}}............\left(1\right)$

If $\omega$ change to $\omega \prime$

$Z^{\prime }=|\omega \prime L-\frac{1}{\omega \prime C}|$

Then, $\frac{I_0}{2}=\frac{{\epsilon }_0}{\sqrt{R^2+Z^{\prime }}}.............\left(2\right)$

from equation $\left(1\right)$ and $\left(2\right)$

$=>2\sqrt{R^2+Z^{\prime }}=\sqrt{R^2+Z^{\prime }}$

$=>Z^{\prime }=\sqrt{3}R$ since $Z=0$ at resonance.