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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Mock Test - Electromagnetic Induction

  • question_answer
    A 300\[\Omega \]. resistor is connected in series with a parallel-plate capacitor across the terminals of a 50.0 Hz ac generator. When the gap between the plates is empty, its capacitance is\[70/22\mu F\]. The ratio of the rms current in the circuit when the capacitor is empty to that when ruby mica of dielectric constant k = 5.0 is inserted between the plates, is equal to

    A) 0.1                               

    B) 0.3

    C) 0.6                   

    D) 2.9

    Correct Answer: B

    Solution :

    [b] \[{{i}_{rms}}=\frac{{{E}_{0}}}{\sqrt{2Z}}\Rightarrow \frac{{{i}_{rms,1}}}{{{i}_{rms,2}}}=\frac{{{Z}_{2}}}{{{Z}_{1}}}=\frac{\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega kC} \right)}^{2}}}}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega \,C} \right)}^{2}}}}\]Solving: \[\Rightarrow \frac{{{i}_{rms,1}}}{{{i}_{rms,2}}}=\frac{\pi }{2}\]


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