JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा JEE PYQ-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 old one is:                     [JEE ONLINE 25-04-2013]

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


You need to login to perform this action.
You will be redirected in 3 sec spinner