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NEET AIPMT SOLVED PAPER MAINS 2011

  • question_answer
    A coil has resistance \[30\,\Omega \] and inductive reactance \[20\,\Omega \]at 50 Hz frequency. If an AC source of 200 V, 100 Hz, is connected across the coil, the current in the coil will be                

    A)  4.0 A    

    B)  8.0 A    

    C) \[\frac{20}{\sqrt{12}}A\]              

    D)  2.0 A

    Correct Answer: A

    Solution :

    If\[\omega =50\times 2\pi \]then\[\omega L=20\,\Omega \] Similarly \[\omega '=100\times 2\pi \]then\[\omega '=L=40\,\Omega \]                 \[i=\frac{200}{Z}=\frac{200}{\sqrt{{{R}^{2}}+{{(\omega 'L)}^{2}}}}\]                 \[=\frac{200}{\sqrt{{{(30)}^{2}}+{{(40)}^{2}}}}=4A\]


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