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JEE Main & Advanced Sample Paper JEE Main Sample Paper-30

  • question_answer
    An induction coil has an impedance of \[10\,\Omega \]. When an AC signal of frequency \[1000\,Hz\] is applied to the coil, the voltage leads the current by \[{{45}^{o}}\]. The inductance of the coil is :

    A)  \[\frac{1}{2\pi }\]                              

    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 :

    \[\tan 45{}^\circ =\frac{\omega L}{R}\Rightarrow \omega L=R\] Also, \[Z=\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}=10\] \[\therefore L=\frac{5\sqrt{2}}{2\pi \times 1000}=\frac{1}{\sqrt{2}\times 200\pi }Hz\]


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