JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2021
- Medium
- Numerical
A resonance circuit having inductance and resistance \(2 \times 10^{-4} H\) and \(6.28 \Omega\) respectively oscillates at \(10\, MHz\) frequency. The value of quality factor of this resonator is ......... \([\pi=3.14]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2000\)
Step-by-step Solution
Detailed explanation
Given : \(\quad L =2 \times 10^{-4} H\)
\(R =6.28 \Omega\)
\(f =10 MHz =10^{7} Hz\)
Since quality factor,
\(Q =\omega_{0} \frac{ L }{ R }=2 \pi f \frac{ L }{ R }\)
\(\therefore Q =2 \pi \times 10^{7} \times \frac{2 \times 10^{-4}}{6.28}\)
\(Q =2 \times 10^{3}=2000\)
\(\therefore\) Ans. is 2000
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