A) 0.32
B) 0.30
C) 0.28
D) 0.24
Correct Answer: B
Solution :
\[\cos \varphi =\frac{R}{Z}=\frac{R}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}\] \[=\frac{12}{\sqrt{{{(12)}^{2}}+4\times {{\pi }^{2}}\times {{(60)}^{2}}\times {{(0.1)}^{2}}}}\]\[\Rightarrow \,\,\cos \varphi =0.30\]
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