JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Critical Thinking

  • question_answer
    An ac source of angular frequency w is fed across a resistor r and a capacitor C in series. The current registered is I. If now the frequency of source is changed to w/3 (but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency w   [Roorkee 1996]

    A)            \[\sqrt{\frac{3}{5}}\]        

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

    C)            \[\sqrt{\frac{1}{5}}\]        

    D)            \[\sqrt{\frac{4}{5}}\]

    Correct Answer: A

    Solution :

                       At angular frequency w, the current in RC circuit is given by                    \[{{i}_{rms}}=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}}}}\]                                        ......(i)                    Also \[\frac{{{i}_{rms}}}{2}=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{\frac{\omega }{3}C} \right)}^{2}}}}=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+\frac{9}{{{\omega }^{2}}{{C}^{2}}}}}\]     ......(ii)                    From equation (i) and (ii) we get            \[3{{R}^{2}}=\frac{5}{{{\omega }^{2}}{{C}^{2}}}\Rightarrow \frac{\frac{1}{\omega C}}{R}=\sqrt{\frac{3}{5}}\]Þ \[\frac{{{X}_{C}}}{R}=\sqrt{\frac{3}{5}}\]


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