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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2021 · 24 February · Morning Shift · Physics · Alternating Current

A resonance circuit having inductance and resistance 2 × 10 ^ - 4 H...

A resonance circuit having inductance and resistance 2×10−4H2 \times {10}^{- 4}H and 6 . 28Ω respectively oscillates at 10MHz10MHz frequency. The value of quality factor of this resonator is_______. [π=3.14]\left[\pi = 3 . 14\right]

Correct answer

0

Step-by-step solution

Given: L=2×10−4HL = 2 \times {10}^{- 4}H

R=6.28ΩR = 6 . 28Ω

f=10MHz=107Hzf = 10MHz = {10}^{7}Hz

Since quality factor,

Q=ω0LR=2\pifLRQ = {\omega}_{0}\dfrac{L}{R} = 2\pif\dfrac{L}{R}

∴Q=2π×107×2×10−46.28∴ Q = 2\pi \times {10}^{7} \times \dfrac{2 \times {10}^{- 4}}{6 . 28}

Q=2×103=2000Q = 2 \times {10}^{3} = 2000

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