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VMMC VMMC Medical Solved Paper-2003

  • question_answer
    The peak value of            an alternating e.m.f. E is given by \[E={{E}_{0}}\,\cos \omega t\] is 10volt and its frequency is 50 Hz. At time\[t=\frac{1}{600}\]sec, the instantaneous e.m.f. is:

    A)  1 V        

    B)                         5 V                        

    C)        10 V                      

    D)        \[5\sqrt{3}V\]

    Correct Answer: D

    Solution :

    Instantaneous value of E is given by \[E={{E}_{0}}\cos \omega t\] Here: \[{{E}_{0}}=10\,\text{volt,}\]\[t=\frac{1}{600}\sec ,\] \[f=50\,Hz\]                 \[\therefore \]  \[E={{E}_{0}}\cos \frac{2\pi t}{T}\]                 or            \[E=10\cos \frac{2\pi \times 50\times 1}{600}\]                                 \[=10\,\cos \frac{5\pi }{3}\times \frac{1}{10}\]                                 \[=10\,\cos \frac{\pi }{6}\]                                 \[=10\cos {{30}^{o}}\]                                 \[=10\times \frac{\sqrt{3}}{2}\]                                 \[=5\sqrt{3}\,\text{volt}\text{.}\]


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