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JCECE Medical JCECE Medical Solved Paper-2013

  • question_answer
    A pure resistive circuit element X when connected to an AC supply of peak voltage 200 V gives a peak current of 5 A. A second current element Y when connected to same AC supply gives the same value of peak current but the current lags behind by \[{{90}^{o}}.\] If series. combination of X and Y is connected to the same supply, what is the impedance of the circuit?

    A) \[40\,\Omega \]

    B) \[80\,\Omega \]    

    C)  \[40\sqrt{2}\,\Omega \]

    D)  \[2\sqrt{40}\,\Omega \]

    Correct Answer: C

    Solution :

     Pure resistive, \[R=\frac{{{E}_{0}}}{{{l}_{0}}}=\frac{200}{5}=40\,\Omega \] As current lags behind the applied voltage by \[{{90}^{o}},\]therefore element Y must be pure inductor. \[{{X}_{L}}=\frac{{{E}_{0}}}{{{l}_{0}}}\] \[=\frac{200}{5}=40\,\Omega \] \[\therefore \] Total impedance, \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] \[=\sqrt{{{40}^{2}}+{{40}^{2}}}=40\sqrt{2}\,\Omega \]


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