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VMMC VMMC Medical Solved Paper-2008

  • question_answer
    A capacitor of capacitance C has charge Q and stored energy is W. If the charge is increased to 2Q the stored energy will be

    A) \[\frac{W}{4}\] 

    B)                                        \[\frac{W}{2}\]                 

    C)         2 W                                      

    D)         4 W

    Correct Answer: D

    Solution :

    From the formula, \[W=\frac{{{Q}^{2}}}{2C}\]                                                          ?(i) \[W=\frac{{{(2Q)}^{2}}}{2C}\]                    \[(\because \,Q=2Q)\]                 \[\Rightarrow \]               \[W=4\frac{{{Q}^{2}}}{2C}\] \[\Rightarrow \]               \[W=4W\]                           [From (i)]


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