JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Self Evaluation Test - Alternating Current

  • question_answer
    A series LR circuit is connected to an ac source of frequency \[\omega \] and the inductive reactance is equal to 2R. A capacitance of capacitive reactance equal to R is added in series with L and R. The ratio of the new power factor to the

    A) \[\sqrt{\frac{2}{3}}\]

    B) \[\sqrt{\frac{2}{5}}\]

    C) \[\sqrt{\frac{3}{2}}\]

    D) \[\sqrt{\frac{5}{2}}\]

    Correct Answer: D

    Solution :

    [d] \[Power\text{ }factor{{\text{ }}_{\left( old \right)}}\] \[=\frac{R}{\sqrt{{{R}^{2}}+X_{L}^{2}}}=\frac{R}{\sqrt{{{R}^{2}}+{{(2R)}^{2}}}}=\frac{R}{\sqrt{5R}}\]             \[Power\text{ }facto{{r}_{(new)}}\] \[=\frac{R}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}=\frac{R}{\sqrt{{{R}^{2}}+{{(2R-R)}^{2}}}}=\frac{R}{\sqrt{2}R}\] \[\therefore \,\,\,\frac{New\,power\,factor}{Old\,power\,factor}=\frac{\frac{R}{\sqrt{2}R}}{\frac{R}{\sqrt{5}R}}=\sqrt{\frac{5}{2}}\]


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