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

  • question_answer
    Reactance of a capacitor of capacitance \[C\mu F\] for ac frequency \[\frac{400}{\pi }\] Hz is \[25\Omega \]. The value C is [MH CET 2002]

    A)            \[50\mu F\]                          

    B)            \[25\mu F\]

    C)            \[100\mu F\]                        

    D)            \[75\mu F\]

    Correct Answer: A

    Solution :

               \[{{X}_{C}}=\frac{1}{2\pi \nu C}\,\Rightarrow C=\frac{1}{2\pi \nu {{X}_{C}}}=\frac{1}{2\times \pi \times \frac{400}{\pi }\times 25}=50\mu F\]


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