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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    A capacitor of capacitance 2.0 \[\mu \]F is charged upto 200 V and then the plates of capacitor are connected to a resistance wire. The heat released in joule will be

    A)  \[2\times {{10}^{-2}}\]          

    B)  \[4\times {{10}^{-2}}\]

    C)  \[4\times {{10}^{4}}\]            

    D)  \[2\times {{10}^{10}}\]

    Correct Answer: B

    Solution :

     Energy released \[H=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times (2\times {{10}^{-6}})\times {{(200)}^{2}}\] \[={{10}^{-6}}\times 200\times 200\] \[=4\times {{10}^{-2}}J\]


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