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VIT Engineering VIT Engineering Solved Paper-2007

  • question_answer
    A voltage of peak value 283 V and varying frequency is applied to a series L-C-R combination in which \[R=3\,\Omega \], \[\text{L = 25 mH}\]and \[\text{C = 400  }\!\!\mu\!\!\text{ F}\]. The frequency (in Hz) of the source at which maximum power is dissipated in the above, is:

    A)  51.5           

    B)  50.7

    C)  51.1            

    D)  50.3

    Correct Answer: D

    Solution :

    A series resonance circuit admits maximum current, as                                            \[P={{i}^{2}}R\]                                        So, power dissipated is maximum at resonance. So, frequency of the source at which maximum power is dissipated in the circuit is \[v=\frac{1}{2\pi \sqrt{LC}}\] \[=\frac{1}{2\times 3.14\sqrt{25\times {{10}^{-3}}\times 400\times {{10}^{-6}}}}\] \[=\frac{1}{2\times 3.14\sqrt{{{10}^{-5}}}}\] \[=50.3Hz\]


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