J & K CET Medical J & K - CET Medical 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\[r=100\,Hz\]

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

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

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

    Correct Answer: C

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner