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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    A 1 µF capacitance of TV is subjected to 4000 V potential difference. The energy stored in the capacitor is

    A)  8 J                                         

    B)  16 J

    C)  \[4\times {{10}^{-3}}\text{J}\]                                 

    D)  \[2\times {{10}^{-3}}\text{J}\]

    Correct Answer: A

    Solution :

    Given \[C=1\mu F,\,V=4000volt.\] Energy stored in the capacitor,                 \[U=\frac{1}{2}C{{V}^{2}}\]                 \[=\frac{1}{2}\times 1\times {{10}^{-6}}\times {{(4000)}^{2}}\]                 \[=8J.\]


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