A) \[R/\omega L\]
B) \[R{{({{R}^{2}}+{{\omega }^{2}}{{L}^{2}})}^{1/2}}\]
C) \[\omega L/R\]
D) \[R/{{({{R}^{2}}-{{\omega }^{2}}{{L}^{2}})}^{1/2}}\]
Correct Answer: B
Solution :
From the relation, \[\tan \phi =\frac{\omega L}{R}\] Power factorcos \[\cos \phi =\frac{1}{\sqrt{1+{{\tan }^{2}}\phi }}\] \[=\frac{1}{\sqrt{1+{{\left( \frac{wL}{2} \right)}^{2}}}}\] \[=\frac{R}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\]You need to login to perform this action.
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