In the circuit shown in the figure, the ratio of the quality fact… — Solution | JEE Physics

NVAMediumJEE 2023LCR Circuits & Resonance

In the circuit shown in the figure, the ratio of the quality factor and the bandwidth is _____ s.

Answer

Correct answer:10

Standard Method

Given: An LCR circuit is shown.

Find: The ratio $\dfrac{Q}{\Delta \omega}$ in seconds.

For an LCR circuit,

\Delta \omega = \frac{R}{L}, \qquad Q = \frac{\omega_0}{\Delta \omega} = \omega_0 \cdot \frac{L}{R}

The resonant angular frequency is

\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{3 \times 27 \times 10^{-6}}} = \frac{1}{9 \times 10^{-3}}

So,

\omega_0 = 10^3 \, \text{rad/s}

Now,

\frac{Q}{\Delta \omega} = \frac{\omega_0 \cdot L / R}{R / L} = \omega_0 \cdot \frac{L^2}{R^2}

Substituting the values,

\frac{Q}{\Delta \omega} = \sqrt{\frac{1}{LC}} \cdot \frac{L^2}{R^2}

Simplifying further,

\frac{Q}{\Delta \omega} = \frac{1}{9 \times 10^{-3}} \cdot \frac{9}{100} = 10 \, \text{s}

Therefore, the ratio of the quality factor to the bandwidth is $10$ .

Using $Q = \frac{R}{\omega_0 L}$ instead of $Q = \frac{\omega_0 L}{R}$ for a series LCR circuit is incorrect. This reverses the dependence on resistance and gives the wrong final ratio. Use the standard series LCR expression carefully.
Confusing bandwidth in hertz with angular bandwidth is a common mistake. Here the relation used is $\Delta \omega = \frac{R}{L}$ , so the bandwidth is in angular frequency units. Do not replace it directly with $\Delta f$ unless you also convert by $2\pi$ .
Errors in evaluating $\omega_0 = \frac{1}{\sqrt{LC}}$ can spoil the answer. The product $LC = 3 \times 27 \times 10^{-6}$ gives a square root of $9 \times 10^{-3}$ , so $\omega_0 = \frac{1}{9 \times 10^{-3}}$ . Keep powers of ten consistent.

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