A) \[{{\tan }^{-1}}(\pi )\]
B) \[{{\tan }^{-1}}\left( \frac{\pi }{2} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{\pi }{4} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{\pi }{3} \right)\]
Correct Answer: A
Solution :
\[\tan \phi =\frac{{{X}_{L}}}{R}\] And \[{{X}_{L}}=\omega L=2\pi fL\] \[=2\pi \times 50\times 0.01=\pi \,\Omega \] Also \[R=1\,\Omega \] \[\phi ={{\tan }^{-1}}(\pi )\]
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