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JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Self Evaluation Test - Alternating Current

  • question_answer
    Resonance frequency of a circuit is f. If the capacitance is made 4 times the initial value, the resonance frequency will become :  

    A) \[f/2\]

    B) 2f

    C) f

    D) f/4

    Correct Answer: A

    Solution :

    [a] \[f=\frac{1}{2\pi \sqrt{LC}}\,\] i.e. \[f\propto \frac{1}{\sqrt{C}}\to \frac{1}{\sqrt{4}}=\frac{1}{2}\] time


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