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JEE Main & Advanced Sample Paper JEE Main Sample Paper-33

  • question_answer
    An LCR series circuit with a resistance of \[100\sqrt{5}\Omega \] is connected to an AC-source of 200 V when the capacitor is removed from the circuit, current lags behind emf by \[{{45}^{o}}\]. When the inductor is removed from the circuit, keeping capacitor and resistor in the circuit current leads by an angle of \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]. Then the current in the LCR circuit will be:

    A)  2A    

    B)                                     0.8 A

    C)  5A       

    D)                                     0.1A

    Correct Answer: B

    Solution :

    When c is removed             \[\tan \phi =\tan 45=\frac{{{X}_{L}}}{R}\]             \[{{X}_{L}}=R\] When L is removed             \[\tan \phi =\frac{1}{2}\,=\frac{{{X}_{C}}}{R}\]             \[{{X}_{C}}=\frac{R}{2}\] \[Z=\,\sqrt{{{(50\sqrt{5})}^{2}}+{{(100\,\sqrt{5})}^{2}}}\] \[{{i}_{rms}}\,=\frac{{{v}_{rms}}}{Z}\,=\frac{200}{60\times 5}\,=0.8\,A\]


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