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

  • question_answer
    A fully charged capacitor C with initial charge \[{{q}_{0}}\] is connected to a coil of self-inductance L at\[t=0.\]The time at which the energy is stored equally between the electric and the magnetic fields is:

    A) \[\frac{\pi }{4}\sqrt{LC}\]

    B) \[2\pi \sqrt{LC}\]

    C) \[\sqrt{LC}\]

    D) \[\pi \sqrt{LC}\]

    Correct Answer: A

    Solution :

    [a] Energy stored in magnetic field = \[\frac{1}{2}\,L{{i}^{2}}\] Energy stored in electric field = \[\frac{1}{2}\,\frac{{{q}^{2}}}{C}\] \[\therefore \,\,\,\,\frac{1}{2}\,L{{i}^{2}}=\frac{1}{2}\,\frac{{{q}^{2}}}{C}\] Also \[q={{q}_{0}}\] cos \[\omega \,t\] and \[\omega =\frac{1}{\sqrt{LC}}\] On solving \[t=\frac{\pi }{4}\,\sqrt{LC}\]


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