Question

Quality factor of LCR circuit having resistance R and inductance L at resonance frequency $ (\omega ) $ is given by:

• A. $ {{\left( \frac{\omega L}{R} \right)}^{2}} $
• B. $ {{\left( \frac{\omega L}{R} \right)}^{1/2}} $
• C. $ \frac{R}{\omega L} $
• D. $ \frac{\omega L}{R} $

Answer 1

The Correct Option is D

Here: Resistance $ =R, $ inductance $ =L $ and resonance frequency $ =\omega $ potential difference developed across the inductance $ =I\omega L $ and applied emf $ =IR $ Hence quality factor $ Q=\frac{\text{potential}\,\text{difference}}{\text{applied}\,\text{emf}}=\frac{I\omega L}{IR}=\frac{\omega L}{R} $