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Punjab Medical Punjab - MET Solved Paper-2008

  • question_answer
    The natural frequency of an\[L-C\] circuit is \[1,25,000\] cycle/s. Then the capacitor \[C\] is replaced by another capacitor with a dielectric medium of dielectric constant\[K\]. In this case, the frequency decreases by\[25kHz\]. The value of \[K\] is

    A)  3.0                                        

    B)  2.1

    C)  1.56                                      

    D)  1.7

    Correct Answer: C

    Solution :

    \[f=\frac{1}{2\pi \sqrt{LC}}\] or            \[f\propto \frac{1}{\sqrt{C}}\] When capacitor\[C\]is replaced by another capacitor \[C\] of dielectric constant\[K\], then                 \[C=KC\] \[\therefore \]  \[\frac{f}{f}=\sqrt{\frac{C}{C}}\] or            \[\frac{125000-25000}{125000}=\sqrt{\frac{C}{KC}}\] or            \[\frac{100}{125}=\frac{1}{\sqrt{K}}\] or            \[K={{\left( \frac{125}{100} \right)}^{2}}=1.56\]


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