CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    An AC is given by\[i={{i}_{1}}\cos \omega t+{{i}_{2}}\sin \omega t\]The rms current is given by:

    A)  \[\frac{{{i}_{1}}+{{i}_{2}}}{\sqrt{2}}\]    

    B)         \[\frac{{{i}_{1}}-{{i}_{2}}}{\sqrt{2}}\]     

    C)         \[\sqrt{\left( \frac{i_{1}^{2}+i_{2}^{2}}{2} \right)}\]       

    D)         \[\sqrt{\left( \frac{i_{1}^{2}-i_{2}^{2}}{2} \right)}\]

    E)  \[\frac{{{i}_{1}}{{i}_{2}}}{\sqrt{2}}\]

    Correct Answer: C

    Solution :

    \[{{i}_{rms}}=\sqrt{\frac{i_{1}^{2}+i_{2}^{2}}{2}}\]


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