JEE Mains · Physics · STD 12 -7. Alternating current
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 ......... \([\pi=3.14]\)
- A \(2000\)
- B \(2500\)
- C \(1600\)
- D \(1800\)
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, \(Q =\omega_{0} \frac{ L }{ R }=2 \pi f \frac{ L }{ R }\) \(\therefore Q =2 \pi \times 10^{7} \times \frac{2 \times 10^{-4}}{6.28}\) \(Q =2 \times 10^{3}=2000\)…
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