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JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Critical Thinking

  • question_answer
    When 100 volts dc is supplied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts ac is applied across the same coil, the current drops to 0.5 ampere. If the frequency of ac source is 50 Hz, then the impedance and inductance of the solenoid are                                [CPMT 1990]

    A)            200 W and 0.55 henry       

    B)            100 W  and 0.86 henry

    C)            200 W and 1.0 henry          

    D)            100 W and 0.93 henry

    Correct Answer: A

    Solution :

                       For dc, \[R=\frac{V}{i}=\frac{100}{1}=100\,\,\Omega \]                    For ac, \[Z=\frac{V}{i}=\frac{100}{0.5}=200\,\,\Omega \]                    \[\because \,\,\,Z=\sqrt{{{R}^{2}}+{{(\omega L)}^{2}}}\]Þ \[200=\sqrt{{{(100)}^{2}}+4{{\pi }^{2}}{{(50)}^{2}}{{L}^{2}}}\]            \ \[L=0.55\,\,H\]


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