A) \[\frac{1}{\sqrt{3}}\]
B) \[\frac{1}{\sqrt{5}}\]
C) \[\frac{1}{\sqrt{7}}\]
D) \[\frac{1}{\sqrt{11}}\]
Correct Answer: B
Solution :
Power factor, \[\operatorname{Cos}\phi =\frac{1}{\sqrt{2}}=\frac{R}{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}\] Squaring both sides \[\frac{1}{2}=\frac{R}{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}\] \[2{{R}^{2}}={{R}^{2}}+{{\omega }^{2}}{{L}^{2}}\] \[R=\omega L\] When \[\omega \] is double Power factor \[\operatorname{Cos}\phi =\frac{R}{\sqrt{{{R}^{2}}+4{{\omega }^{2}}{{L}^{2}}}}=\frac{R}{\sqrt{{{R}^{2}}+4{{\operatorname{R}}^{2}}}}=\frac{1}{\sqrt{5}}\]
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