A) \[\frac{1}{3\pi }\]H
B) \[\frac{1}{5\pi }H\]
C) \[\frac{1}{7\pi }H\]
D) \[\frac{1}{9\pi }H\]
Correct Answer: B
Solution :
\[R=\frac{P}{i_{rms}^{2}}=\frac{240}{16}=15\Omega \] \[Z=\frac{V}{i}=\frac{100}{4}=25\Omega \] Now \[{{X}_{L}}=\sqrt{{{Z}^{2}}-{{R}^{2}}}=\sqrt{{{(25)}^{2}}-{{(15)}^{2}}}=20\Omega \] \[\therefore 2\pi \nu L=20\Rightarrow L=\frac{20}{2\pi \times 50}=\frac{1}{5\pi }Hz\]
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