2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    The resonance frequency of the tank circuit of an oscillator when\[L=\frac{10}{{{\pi }^{2}}}mH\] and\[C=0.04\text{ }\mu F\]are connected in parallel is

    A)  250 kHz        

    B)         25 kHz 

    C)         2.5 kHz                

    D)         25 MHz

    E)  2.5MHz

    Correct Answer: B

    Solution :

    In parallel resonant circuit resonance frequency \[{{f}_{o}}=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{\frac{10\times {{10}^{-3}}}{{{\pi }^{2}}}\times 0.04\times {{10}^{-6}}}}\] \[=\frac{{{10}^{4}}}{2\times 0.2}=25\,kHz\]


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