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BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    A capacitor of\[1\mu F\]initially charged to 10 V is connected across an ideal inductor of 0.1 mH. The maximum current in the circuit is

    A)  0.5 A                                    

    B)  1 A

    C)  1.5 A                                    

    D)  2 A

    Correct Answer: B

    Solution :

                     \[\frac{1}{2}C{{V}^{2}}=\frac{1}{2}L{{i}^{2}}\] \[\therefore \]  \[{{10}^{-6}}\times {{(10)}^{2}}=0.1\times {{10}^{-3}}\times {{I}^{2}}\]                 \[{{10}^{-6}}\times 100=0.1\times {{10}^{-3}}\times {{I}^{2}}\]                 \[{{I}^{2}}=\frac{100\times {{10}^{-6}}}{0.1\times {{10}^{-3}}}=1\]


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