JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that \[R=5\Omega ,\,L=25\,mH\] and \[C=1000\,\mu F\]. The total impedance, and phase difference between the voltage across the source and the current will respectively be - [JEE Online 09-04-2017]

    A) \[10\,\Omega \] and \[{{\tan }^{-1}}\,\left( \frac{5}{3} \right)\]

    B)  \[7\,\Omega \] and \[{{45}^{\text{o}}}\]

    C)  \[7\,\Omega \] and \[{{\tan }^{-1}}\left( \frac{5}{3} \right)\]

    D)  \[10\,\Omega \,\] and \[{{\tan }^{-1}}\,\left( \frac{8}{3} \right)\]

    Correct Answer: B

    Solution :

                    \[{{e}_{0}}=283\,\,volt\,\]            \[\omega =320\] \[{{X}_{L}}\,=320\,\times 25\,\times {{10}^{-3}}\,=8\,\Omega \] \[{{X}_{C}}\,=\frac{1}{\omega C}=\,\frac{1}{320\,\times 1000\,\times {{10}^{-6}}}\] \[=\frac{1000}{320}\,=3.1\,\Omega \] \[R=5\,\Omega \] \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\,=\sqrt{50}\,=7\,\Omega \] \[\tan \phi \,=\frac{{{X}_{L}}-{{X}_{C}}}{R}\] \[=1\,\,\,\,\,\,\,\,\,\,\phi ={{45}^{\text{o}}}\]


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