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JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    When an AC source of emf \[e={{\text{E}}_{0}}\text{ }sin\text{ (}100t\text{)}\]is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be \[\pi /4\] as shown in the diagram. If the circuit consists possibly only of \[R-C\] or \[R-L\] or \[L-C\] in series, find the relationship between the two elements.

    A)  \[R=1K\Omega ,C=10\mu F\]    

    B)  \[R=1K\Omega ,C=1\mu F\]

    C)  \[8.5\times {{10}^{4}}\]                     

    D)  \[R=1K\Omega ,L=1H\]

    Correct Answer: A

    Solution :

    The circuit is R-C circuit (i leads e by \[\pi /4\]) \[\tan \phi \,=\frac{{{X}_{c}}}{R}\] \[\tan \frac{\pi }{4}\,=\frac{(1/\omega C)}{R}\] \[\omega CR=1\] Given \[\omega =100\,rad/\sec \] So,       \[CR=\frac{1}{100\,}{{(\sec )}^{-1}}\]


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