RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    700 pF capacitor is charged by 50 V battery. Electrostatic energy is stored by it will be:

    A)  \[17.0\times {{10}^{-8}}J\]         

    B)  \[13.0\times {{10}^{-9}}J\]

    C)  \[8.7\times {{10}^{-7}}J\]   

    D)         \[6.7\times {{10}^{-7}}J\]

    Correct Answer: C

    Solution :

    Here: Capacitance \[C=700\,pF=700\times {{10}^{-12}}F\] Source voltage\[V=50\,V\] Electrostatic energy is given by                                 \[E=\frac{1}{2}C{{V}^{2}}\] \[=0.5\times 700\times {{10}^{-12}}\times {{(50)}^{2}}\] \[=8.7\times {{10}^{-7}}J\]


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