Q127 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
In the circuit shown, the ratio of the quality factor and the band width is
(Given: band width = R/L)
(Given: band width = R/L)
- Correct answerA$\sqrt{\dfrac{1}{LC}}\dfrac{L^2}{R^2}$
- B$\sqrt{\dfrac{1}{LC}}$
- C$\sqrt{\dfrac{1}{LC}}\dfrac{L}{R}$
- D$\sqrt{\dfrac{1}{LC}}\dfrac{R}{L}$
Solution
The quality factor $Q$ of a series LCR circuit is given by:
$Q = \dfrac{\omega_0 L}{R}$
where $\omega_0$ is the resonant angular frequency, $\omega_0 = \sqrt{\dfrac{1}{LC}}$.
The bandwidth is given as $\Delta \omega = \dfrac{R}{L}$.
The ratio of the quality factor to the bandwidth is:
$\text{Ratio} = \dfrac{Q}{\Delta \omega} = \dfrac{\dfrac{\omega_0 L}{R}}{\dfrac{R}{L}} = \omega_0 \dfrac{L^2}{R^2}$
Substituting $\omega_0 = \sqrt{\dfrac{1}{LC}}$, we get:
$\text{Ratio} = \sqrt{\dfrac{1}{LC}} \dfrac{L^2}{R^2}$
Answer: $\sqrt{\dfrac{1}{LC}}\dfrac{L^2}{R^2}$
$Q = \dfrac{\omega_0 L}{R}$
where $\omega_0$ is the resonant angular frequency, $\omega_0 = \sqrt{\dfrac{1}{LC}}$.
The bandwidth is given as $\Delta \omega = \dfrac{R}{L}$.
The ratio of the quality factor to the bandwidth is:
$\text{Ratio} = \dfrac{Q}{\Delta \omega} = \dfrac{\dfrac{\omega_0 L}{R}}{\dfrac{R}{L}} = \omega_0 \dfrac{L^2}{R^2}$
Substituting $\omega_0 = \sqrt{\dfrac{1}{LC}}$, we get:
$\text{Ratio} = \sqrt{\dfrac{1}{LC}} \dfrac{L^2}{R^2}$
Answer: $\sqrt{\dfrac{1}{LC}}\dfrac{L^2}{R^2}$
Hence, the correct answer is option: 'A'.