RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    An AC source of angular frequency \[\omega \] is fed across a resistor rand a capacitor C in series. The current registered is \[i.\] If now the frequency of source is changed to\[\omega /3\] (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency\[\omega \].

    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 to, the current in \[R-C\] 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 Eqs. (i) and (ii), we get                 \[3{{R}^{2}}=\frac{5}{{{\omega }^{2}}{{C}^{2}}}\Rightarrow \frac{\frac{1}{\omega C}}{R}=\sqrt{\frac{3}{5}}\]


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