JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Self Evaluation Test - Alternating Current

  • question_answer
    The equation of AC voltage is\[E=220\,\sin \,(\omega t+\pi /6)\] and the A.C. current\[I=10\,\sin \,(\omega t+\pi /6)\]. The average power dissipated is

    A) 150 W

    B) 550 W

    C) 250 W

    D) 50 W

    Correct Answer: B

    Solution :

    [b] We know that,  \[Z=\frac{{{E}_{0}}}{{{I}_{0}}}\] Given,  \[{{E}_{0}}=220\] and \[{{I}_{0}}=10\] So  \[Z=\frac{220}{10}=22\,ohm\]   \[\phi =\left[ \frac{\pi }{6}-\left( -\frac{\pi }{6} \right) \right]=\frac{\pi }{3}\] \[{{P}_{a}}=\frac{{{E}_{0}}}{\sqrt{2}}\times \frac{{{I}_{0}}}{\sqrt{2}}\times \cos \]\[\phi =\frac{220}{\sqrt{2}}\times \frac{10}{\sqrt{2}}\times \cos \,\frac{\pi }{3}\]


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