A) \[\frac{1}{10\pi }\]
B) \[\frac{1}{20\pi }\]
C) \[\frac{1}{40\pi }\]
D) \[\frac{1}{60\pi }\]
Correct Answer: B
Solution :
\[\tan \phi =\frac{{{X}_{L}}}{R}\] \[\therefore \] \[\tan 4{{5}^{o}}=\frac{{{L}_{\omega }}}{{{L}_{\omega }}}\] \[L=\frac{R}{\omega }=\frac{100}{2\pi \times 100}\] \[(\because \,\tan {{45}^{o}}=1)\] \[L=\frac{1}{20\pi }\]
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