JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • question_answer
    Direction: A voltage source \[V={{V}_{0}}\,\sin \,(100t)\] is connected to a black box in which there can he either one element out of L, C, R or any two of them connected in series.
    At steady state the variation of current in the circuit and the source voltage are plotted together with time, using an oscilloscope, as shown.
    Values of the parameters of the elements, present in the black box are

    A)  \[R=50\,\Omega ,\,\,C=200\,\mu F\]

    B)  \[R=50\,\Omega ,\,\,L=2\,m\mu \]

    C) \[R=400\,\Omega ,\,C=50\,\mu F\]

    D)  \[R=50\,\Omega ,\,\,L=02H\]

    Correct Answer: A

    Solution :

     \[\Delta \theta =\omega \Delta t=\frac{\pi }{4}\] \[\tan \,\theta =\frac{X}{R}\] \[\Rightarrow \]            \[X=R\] Since, current leads the voltage the circuit consists of R and C. and      \[{{i}_{0}}=\frac{{{V}_{0}}}{z}\] \[\therefore \]    \[z=\frac{{{V}_{0}}}{{{i}_{0}}}=\frac{100}{\sqrt{2}}=50\sqrt{2}\] Now,    \[R\sqrt{2}=50\sqrt{2}\] \[\Rightarrow \]            \[R={{X}_{C}}=50\] \[{{X}_{C}}=\frac{1}{C\omega }=50\] \[\Rightarrow \]            \[C=\frac{1}{50\,\omega }=200\,\mu F\] \[\tau =RC=50\times 200\times {{10}^{-6}}=1\times {{10}^{-2}}s\]

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