Question

In a series $LCR$ circuit $L=3.0\text{H}$, $C=27\mu \text{F}$ and $R=7.4\Omega$. It is desired to improve the sharpness of the resonance of the circuit by reducing its 'full width at half maximum' by a factor of $2$. Suggest a suitable way.

• A. $R$ should be decreased to $3.7\Omega$.
• B. $L$ should be increased to $6H$.
• C. The resonance frequency of the circuit should be doubled.
• D. $C$ should be increased to $54\mu F$.

Answer 1

The correct option is A $R$ should be decreased to $3.7\Omega$.

We know that:

$\frac{{\omega }_0}{2\Delta \omega }=Q$

If we reduce full width at half maximum $\left(2\Delta \omega \right)$ by a factor of $2$, then the new quality factor,

$Q^{\prime }=2Q$

So, the sharpness of the resonance $\left(Q\right)$ will become $2Q$.

Now, the quality factor is given by:

$Q=\frac{{\omega }_{0}L}{R}$

To increase the quality factor of the circuit by a factor of $2$ without changing the resonance frequency, we have to decrease the value of $R$ to $\frac{R}{2}$.

Therefore, resistance should be decreased to $3.7\Omega$.

Hence, option $\left(A\right)$ is the correct answer.