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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    A pure resistive circuit element \[X\] when connected to a \[AC\] 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,\[R=\frac{{{E}_{0}}}{{{I}_{0}}}=\frac{200}{5}=40\Omega \] As current lags behind the applied voltage by\[{{90}^{o}}\], therefore element \[Y\] must be pure inductor.                 \[{{X}_{L}}=\frac{{{E}_{0}}}{{{I}_{0}}}=\frac{200}{5}=40\Omega \] \[\therefore \]  Total impedance,                 \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}=\sqrt{{{40}^{2}}+{{40}^{2}}}\]                     \[=40\sqrt{2}\Omega \]


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