A) \[\frac{1}{20\pi }\]
B) \[\frac{1}{\sqrt{2}\times 200\pi }\]
C) \[\frac{1}{\sqrt{2}\times 20\pi }\]
D) \[\frac{1}{200\pi }\]
Correct Answer: B
Solution :
\[\text{Tan}\,{{45}^{o}}=\frac{\omega L}{R}\Rightarrow L=R\] Also, \[Z=\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}=10\] \[\therefore \]\[L=\frac{5\sqrt{2}}{2\pi \times 1000}=\frac{1}{\sqrt{2}\times 200\pi }H\] Hence, the correction option is(b).You need to login to perform this action.
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