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AIIMS AIIMS Solved Paper-2008

  • question_answer
    An AC source of angular frequency \[I\] 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 \[\omega /3\] (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 \[\omega \]

    A)  \[I\]     

    B)                        \[\omega \]                      

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

    D)        \[\sqrt{\frac{2}{5}}\]

    Correct Answer: A

    Solution :

    At angular frequency \[({{R}_{2}}>{{R}_{1}})\], the current in RC circuit is given by \[{{R}_{2}}\]                    ?(i) Also, \[R=\frac{{{R}_{2}}\times ({{R}_{1}}+{{R}_{2}})}{({{R}_{2}}-{{R}_{1}})}\]                                                                 ?.?(ii) From Eqs. (i) and (ii), we get \[R={{R}_{2}}-{{R}_{1}}\] \[R=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{1}}+{{R}_{2}})}\]   \[R=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{2}}-{{R}_{1}})}\] \[{{\sin }^{2}}(\omega t)\]  \[2\pi /\omega \]


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