BVP Medical BVP Medical Solved Paper-2008

  • question_answer
    A capacitor having capacity of 2.0\[\mu F\] 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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