A) \[{{\left( \frac{\omega L}{R} \right)}^{2}}\]
B) \[{{\left( \frac{\omega L}{R} \right)}^{1/2}}\]
C) \[\frac{R}{\omega L}\]
D) \[\frac{\omega L}{R}\]
Correct Answer: D
Solution :
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 \[Q=\frac{\text{potential}\,\text{difference}}{\text{applied}\,\text{emf}}=\frac{I\omega L}{IR}=\frac{\omega L}{R}\]
You need to login to perform this action.
You will be redirected in
3 sec
