JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    A circuit connected to an ac source of emf \[e={{e}_{0}}\sin (100t)\]with t in seconds, gives a phase difference of\[\frac{\pi }{4}\] between the emf e and current i. Which of the following circuits will exhibit this?             [JEE Main 8-4-2019 Afternoon]

    A) RC circuit with \[R=1k\Omega \]and \[C=1\mu F\]

    B) RL circuit with \[R=1k\Omega \] and L = 1mH

    C) RL circuit with \[R=1k\Omega \] and L = 10 Mh

    D) RC circuit with \[R=1k\Omega \] and \[C=10\mu F\]

    Correct Answer: D

    Solution :

    Given phase difference \[=\frac{\pi }{4}\]and \[\omega =100\] rad/s \[\Rightarrow \]Reactance (X) = Resistance (R) now by checking options Option \[R=1000\Omega \]and\[{{X}_{C}}=\frac{1}{{{10}^{-6}}\times 100}={{10}^{4}}\Omega \] Option \[R={{10}^{3}}\Omega \,\]and\[\,{{X}_{L}}={{10}^{-3}}\times 100={{10}^{-1}}\Omega \] Option \[R={{10}^{3}}\Omega \] and \[\,{{X}_{L}}=10\times {{10}^{-3}}\times 100=1\Omega \] Option \[R={{10}^{3}}\Omega \]and\[\,{{X}_{C}}=\frac{1}{10\times {{10}^{-6}}\times 100}{{10}^{3}}\Omega \] Clear option matches the given condition


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