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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    The peak value of an alternating emf E given by \[E={{E}_{0}}\cos \omega t\] is 10 V and its frequency is 50 Hz. At a time \[t=\frac{1}{600}\]s, the instantaneous value of the emf is

    A)  10 V                                     

    B) \[5\sqrt{3}v\]

    C)  5V                         

    D)         1V

    Correct Answer: B

    Solution :

    \[E={{E}_{0}}\cos \omega t=10\cos (2\pi \times ft)\]                 \[=10\cos \left( 2\pi \times 50\times \frac{1}{600} \right)\]                 \[=10\cos \left( \frac{\pi }{6} \right)=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}\,\,V\]


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