J & K CET Medical J & K - CET Medical Solved Paper-2009

  • question_answer
    The instantaneous values of current and voltage in an AC circuit are given by \[I=6\sin \left( 100\pi t+\frac{\pi }{4} \right),\]\[V=5\sin \left( 100\pi t+\frac{\pi }{4} \right),then\]

    A)  current leads the voltage by \[45{}^\circ \]

    B)  voltage leads the current by \[90{}^\circ \]

    C)  current leads the voltage by \[90{}^\circ \]

    D)  voltage leads the current by \[45{}^\circ \]

    Correct Answer: C

    Solution :

                     The phase difference between instantaneous values of I and Vis\[+\frac{\pi }{4}-\left( -\frac{\pi }{4} \right)=\frac{\pi }{2}\] Hence, current leads the voltage by\[90{}^\circ \].


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