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BVP Medical BVP 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) \[W=\frac{(2Q)}{2C}\]   (\[\because \]  \[Q=2Q\])                 \[\Rightarrow \]               \[W=4\frac{{{Q}^{2}}}{2C}\]                 \[\Rightarrow \]               \[W=4W\]           [ From (i)]


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