Manipal Medical Manipal Medical Solved Paper-2012

  • question_answer
    A capacitor having capacity of 2.0 p F is charged to 200 V and then the plates of the capacitor are connected to a resistance wire. The heat produced in joule will be

    A)  \[2\times {{10}^{-2}}\]

    B)  \[4\times {{10}^{-2}}\]

    C)  \[4\times {{10}^{4}}\]

    D)  \[4\times {{10}^{10}}\]

    Correct Answer: B

    Solution :

     Heat produced in a wire is equal to energy stored in capacitor. \[H=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}\times (2\times {{10}^{-6}})\times {{(200)}^{2}}\] \[={{10}^{-6}}\times 200\times 200\] \[=4\times {{10}^{-2}}J\]


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