Manipal Medical Manipal Medical Solved Paper-2004

  • question_answer
    An alternating current is given by\[I={{I}_{1}}\cos \omega t+{{I}_{2}}\sin \omega t\]. The root mean square current is given by:

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

    B)  \[\frac{{{({{I}_{1}}+{{I}_{2}})}^{2}}}{2}\]

    C)  \[\sqrt{\frac{I_{1}^{2}+I_{2}^{2}}{2}}\]

    D)  \[\frac{\sqrt{I_{1}^{2}+I_{2}^{2}}}{2}\]

    Correct Answer: C

    Solution :

     The equation of A.C. is \[I={{I}_{1}}\cos \omega t+{{I}_{2}}\sin \omega t\] The resultant current is given by \[{{I}_{0}}=\sqrt{I_{1}^{2}+I_{2}^{2}}\]          ...(1) Hence, the rms current from relation is \[{{I}_{rms}}=\frac{{{I}_{0}}}{\sqrt{2}}\] \[=\frac{\sqrt{I_{1}^{2}+I_{2}^{2}}}{\sqrt{2}}\]    [from eq.(l)] \[=\sqrt{\frac{I_{1}^{2}+I_{2}^{2}}{2}}\]


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