A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{4}\]
D) 0
Correct Answer: C
Solution :
\[\cos \varphi =\frac{R}{Z}=\frac{R}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}=\frac{5}{\sqrt{25+{{(50)}^{2}}\times {{(0.1)}^{2}}}}\] \[=\frac{5}{\sqrt{25+25}}=\frac{1}{\sqrt{2}}\,\Rightarrow \,\varphi =\pi /4\]
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