A) \[\frac{R}{\omega L}\]
B) \[\frac{R}{{{({{R}^{2}}+{{\omega }^{2}}{{L}^{2}})}^{1/2}}}\]
C) \[\frac{\omega L}{R}\]
D) \[\frac{R}{{{({{R}^{2}}-{{\omega }^{2}}{{L}^{2}})}^{1/2}}}\]
Correct Answer: B
Solution :
From the relation, \[\tan \phi =\frac{\omega L}{R}\] Power factor, \[\cos \phi =\frac{1}{\sqrt{\,1+{{\tan }^{2}}\phi }}\] \[=\frac{1}{\sqrt{\,1+{{\left( \frac{\omega L}{R} \right)}^{2}}}}\] \[=\frac{R}{\sqrt{\,{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\]
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