NEET Physics Alternating Current / प्रत्यावर्ती धारा NEET PYQ-Alternating Current

  • 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 \[e=\frac{1}{\sqrt{2}}\sin (100\pi t+\pi /3)\] volt                                       [AIPMT (M) 2012] 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 \[e=\frac{1}{\sqrt{2}}\sin (100\pi t+\pi /3)\]
    \[\therefore \]      \[{{i}_{0}}=\frac{1}{\sqrt{2}}\]
    and       \[{{V}_{0}}=\frac{1}{\sqrt{2}}\]
    We know that average power        \[{{P}_{av}}={{V}_{rms}}\times {{i}_{rms}}\cos \phi \]
    \[=\frac{1}{2}\times \frac{1}{2}\times \cos {{60}^{o}}\]
    \[\left[ \because {{i}_{rms}}=\frac{{{i}_{0}}}{\sqrt{2}}\text{and}\,{{V}_{rms}}=\frac{{{V}_{0}}}{\sqrt{2}} \right]\]
    \[=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\]\[=\frac{1}{8}W\]


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