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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    In an AC circuit, V and \[I\] are given by V=150 sin (150 t) volt and \[I\]= 150 \[sin\]\[\left( 150t+\frac{\pi }{3} \right)\]amp. The power dissipated in the circuit is

    A)  106 W        

    B)  150 W

    C)  5625 W         

    D)  zero

    Correct Answer: C

    Solution :

    Given \[V=150\sin (150t)volt\] \[I=150\sin (150t+\pi /3)amp\] and       \[{{I}_{0}}=150amp\] and \[{{V}_{0}}=150\,volt\] \[p=\frac{1}{2}{{V}_{0}}{{I}_{0}}\cos \phi \] \[p=0.5\times 150\times 150\times \cos {{60}^{o}}\] \[p=5625W\]


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