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JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank ac Circuits

  • question_answer
    A coil of 200W resistance and 1.0 H inductance is connected to an ac source of frequency \[200/2\pi \,Hz.\] Phase angle between potential and current will be            [MP PMT 2003]

    A)            30o

    B)            90o

    C)            45o

    D)            0o

    Correct Answer: C

    Solution :

               \[\tan \varphi =\frac{{{X}_{L}}}{R}=\frac{2\pi \nu L}{R}=\frac{2\pi \times \frac{200}{2\pi }\times 1}{200}=1\Rightarrow \varphi ={{45}^{o}}\]


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