Manipal Medical Manipal Medical Solved Paper-2002

  • 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 \[V=50\text{ }V\] 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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