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JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank ac Circuits

  • question_answer
    The reactance of a \[25\,\mu F\] capacitor at the ac frequency of 4000 Hz is

    A)            \[\frac{5}{\pi }\]ohm        

    B)            \[\sqrt{\frac{5}{\pi }}\]ohm

    C)            10 ohm                                   

    D)            \[\sqrt{10}\]ohm

    Correct Answer: A

    Solution :

               \[{{X}_{C}}=\frac{1}{2\pi \nu C}=\frac{1}{2\pi \times 4000\times 25\times {{10}^{-6}}}=\frac{5}{\pi }\Omega \]


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