JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    A series LR circuit is connected to an ac source of frequency 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 Main Online Paper ( Held On 25  April 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 :

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


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