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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Alternating Current

JEE Main 2018 · 8 April · Physics · Alternating Current

For an RLC circuit driven with voltage of amplitude v _ m and...

For an RLC circuit driven with voltage of amplitude vm{v}_{m} and frequency \omega0=1LC{\text{\omega}}_{0} = \dfrac{1}{\sqrt{LC}} the current exhibits resonance. The quality factor, QQ is given by

Options

  1. ACRω0\dfrac{CR}{{\omega}_{0}}
  2. Bω0LR\dfrac{{\omega}_{0}L}{R}
  3. Cω0RL\dfrac{{\omega}_{0}R}{L}
  4. DR(ω0C)\dfrac{R}{{( \omega}_{0}C )}

Correct answer

B. ω0LR\dfrac{{\omega}_{0}L}{R}

Step-by-step solution

The quality factor of an L.C.R. circuit is given by,

(Quality factor) Q=1RLC...(i)Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} . . . ( i )

Where L,R,CL, R, C are the inductance, resistance, and conductance of the L.C.R. circuit. The given resonance frequency is, \omega0=1LC{\text{\omega}}_{0} = \dfrac{1}{\sqrt{LC}}

⇒C=1\omega0L...(ii)⇒ \sqrt{C} = \dfrac{1}{{\text{\omega}}_{0}\sqrt{L}} . . . ( ii )

From (i)( i ) and (ii)( ii )

⇒Q=1RL×\omega0L⇒Q=\omega0LR⇒ Q = \dfrac{1}{R}\sqrt{L} \times {\text{\omega}}_{0}\sqrt{L} ⇒ Q = \dfrac{{\text{\omega}}_{0}L}{R}

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