BCECE Medical BCECE Medical Solved Papers-2003

  • question_answer
    In a series combination \[R=300\,\,\Omega \], L = 0.9 H, \[C=2.0\,\,\mu F,\,\,\omega =1000\] rad/s, the impedance of the circuit is :

    A)  \[1300\,\,\Omega \]

    B)  \[900\,\,\Omega \]

    C)  \[500\,\,\Omega \]

    D)  \[400\,\,\Omega \]

    Correct Answer: C

    Solution :

    In series RLC circuit, the impedance of the circuit is given by \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\] Also, \[{{X}_{L}}=\omega L,\,\,{{X}_{C}}=\frac{1}{\omega C}\] \[\therefore \] \[Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] Given, \[R=300\,\Omega ,\,\omega =1000\,\,rad/s,\,L=0.9\,H\], \[C=2.0\,\mu F=2\times {{10}^{-6}}F\] Hence, \[Z=\sqrt{{{(300)}^{2}}+{{\left( 1000\times 0.9-\frac{1}{10000\times 2\times {{10}^{-6}}} \right)}^{2}}}\] \[=\sqrt{90000+{{(900-500)}^{2}}}\] \[=\sqrt{90000+16000}\] \[=\sqrt{250000}\] \[=500\,\,\Omega \]


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