NEET Sample Paper NEET Sample Test Paper-13

  • question_answer
    The instantaneous values of alternating current and voltages in a circuit are given as\[i=\frac{1}{\sqrt{2}}\sin (100\pi t)\]ampere\[e=\frac{1}{\sqrt{2}}\sin (100\pi t+\pi /3)\,\text{volt}\]. The average power in watts consumed in the circuit is

    A)  \[\frac{1}{4}\]                         

    B)  \[\frac{\sqrt{3}}{4}\]

    C)  \[\frac{1}{2}\]                         

    D)  \[\frac{1}{8}\]

    Correct Answer: D

    Solution :

     Given equations \[i=\frac{1}{\sqrt{2}}\sin (100\pi t)\] and \[e=\frac{1}{\sqrt{2}}\sin (100\pi t+\pi /3)\] \[\therefore \]\[{{i}_{o}}=\frac{1}{\sqrt{2}}\]and \[{{V}_{o}}=\frac{1}{\sqrt{2}}\] We know that average power \[{{P}_{av}}={{V}_{rms}}\times {{i}_{rms}}\cos \phi =\frac{1}{2}\times \frac{1}{2}\times \cos {{60}^{o}}\] \[\left[ \because \,{{i}_{rms}}=\frac{{{i}_{o}}}{\sqrt{2}}\,and\,{{V}_{rms}}=\frac{{{V}_{o}}}{\sqrt{2}} \right]\] \[=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}W\]


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