Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    A transistor-oscillator using a resonant circuit      with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency\[f\]. If L is doubled and  C is changed to 4C, the frequency will be

    A) \[f\]/4                                  

    B)  8\[f\]

    C) \[f\]/\[2\sqrt{2}\]          

    D) \[f\]/2

    Correct Answer: C

    Solution :

                    In a series LC circuit, frequency of LC oscillations is given by \[f=\frac{1}{2\pi \sqrt{LC}}\]                 or            \[f\alpha \frac{1}{\sqrt{LC}}\]                 \[\Rightarrow \]               \[\frac{{{f}_{1}}}{{{f}_{2}}}=\sqrt{\frac{{{L}_{2}}{{C}_{2}}}{{{L}_{1}}{{C}_{1}}}}\] Given \[{{L}_{1}}=L,\,{{C}_{1}}=C,\,{{L}_{2}}=2L,\,{{C}_{2}}=4C,\,{{f}_{1}}=f\] \[\therefore \]  \[\frac{f}{{{f}_{2}}}=\sqrt{\frac{2L\times 4C}{LC}}=\sqrt{8}\] \[\Rightarrow \]               \[{{f}_{2}}=\frac{f}{2\sqrt{2}}\]


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