JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2023
- Medium
- Numerical
In the circuit shown in the figure, the ratio of the quality factor and the band width is \(.............\) s.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\(\Delta \omega=\frac{ R }{ L }\)
\(Q =\frac{\omega_0}{\Delta \omega}=\omega_0 \frac{ L }{ R }\)
\(\omega_0=\frac{1}{\sqrt{3 \times 27 \times 10^{-6}}}=\frac{1}{9 \times 10^{-3}}\)
\(\frac{ Q }{\Delta \omega}=\frac{\omega_0 \frac{ L }{ R }}{\frac{ R }{ L }}=\omega_0 \frac{ L ^2}{ R ^2}=\sqrt{\frac{1}{ LC } \frac{ L ^2}{ R ^2}}\)
\(=\frac{1}{9 \times 10^{-3}} \times \frac{9}{100}=10\,s\)
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