A) \[{{\left( \frac{\omega L}{R} \right)}^{2}}\]
B) \[{{\left( \frac{R\omega }{L} \right)}^{2}}\]
C) \[\frac{R\omega }{L}\]
D) \[\frac{\omega L}{R}\]
Correct Answer: D
Solution :
The potential difference developed across the inductance is \[=i\,\omega \,L\] The emf applied is \[=iR\] Hence, quality factor \[Q=\frac{Potential\,difference}{Applied\,emf}=\frac{i\,\omega \,L}{iR}=\frac{\omega L}{R}\]
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