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← JEE Main 2021 · 24 Feb · Shift 1
PhysicsInteger+4 / -1Question 26 of 90

JEE Main 2021 · 24 Feb · Shift 1, Physics question 26

A resonance circuit having inductance and resistance 2×10−42\times10^{-4} H and 6.28 Ω respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is _______. [π = 3.14]

Answer and worked solution

Official answer: 2000

Quality factor Q=ωLR=2×3.14×107×2×10−46.28=2000Q=\dfrac{\omega L}R=\dfrac{2\times3.14\times10^7\times2\times10^{-4}}{6.28}=2000.

Answer: 2000