J & K CET Engineering J and K - CET Engineering Solved Paper-2013

  • question_answer
    The alternating current in a circuit is given by \[I=50\,\,\sin \,\,314t\]. The peak value and frequency of the current are

    A)  \[{{I}_{0}}=25\,\,A\]and \[f=100\,\,Hz\]

    B)  \[{{I}_{0}}=50\,\,Hz\]  and \[f=50\,Hz\]

    C)  \[{{I}_{0}}=50A\] and \[f=100\,\,Hz\]

    D)  \[{{\text{I}}_{0}}=25A\]and  \[f=50\,\,Hz\]

    Correct Answer: C

    Solution :

    From standard equation, we have \[I={{I}_{0}}\,\,\sin \omega t\] ?..(i) Given,    \[I=50\,\,\sin \,\,314\,t\] .....(ii) Comparing Eqs. (i) and (ii), we get \[{{I}_{0}}=50A,\,\,\omega =2\pi f=314\] \[\Rightarrow \] \[f=\frac{314}{2\times 3.14}=50Hz\]


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