Haryana PMT Haryana PMT Solved Paper-2006

  • question_answer
    If a circuit made up of a resistance \[\Omega \] and inductance 0.01 H, and alternating emf 200 V at SO Hz is connected, then the phase difference           between the current and the emf in the circuit is :

    A)                  \[{{\tan }^{-1}}(\pi )\]                                  

    B)  \[{{\tan }^{-1}}\left( \frac{\pi }{2} \right)\]

    C)                  \[{{\tan }^{-1}}\left( \frac{\pi }{4} \right)\]                         

    D)  \[{{\tan }^{-1}}\left( \frac{\pi }{3} \right)\]

    Correct Answer: A

    Solution :

                    \[\tan \phi =\frac{{{X}_{L}}}{R}\] and \[{{X}_{L}}=\omega L=2\pi \,f\,\,L=2\pi \times 50\times 0.01=\pi \,\Omega \] Also       \[R=1\Omega \phi ={{\tan }^{-1}}(\pi )\]


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