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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    In an AC circuit, V and \[I\] are given by \[V=150\sin \left( 150t \right)\] 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 \[V=150\sin \,(150\,t)\] volt                 \[I=150\sin \,(150\,t+\pi /3)\] amp \[\therefore \]  \[{{I}_{0}}=150\]volt and        \[{{V}_{0}}=150\] volt                 \[P=\frac{1}{2}\,{{V}_{0}}{{I}_{0}}\cos \phi \]                 \[P=0.5\times 150\times 150\times \cos {{60}^{o}}\]                 \[P=5625\,W\]


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