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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 38

  • question_answer
    In the circuit shown in the figure, the emf of battery is \[50V,\] the resistance is \[250\Omega \]. and the capacitance is\[0.5\mu F\]. The switch S is closed for a long time and no voltage is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum value of\[150\text{ }V\]. Calculate the inductance, in \[mH,\]to the nearest three digit integer.

    A) \[281.25\,mH\]  

    B) \[312.75\,mH\]

    C) \[450.00\,mH\]              

    D) \[225.50\,mH\]

    Correct Answer: A

    Solution :

    [a] \[\frac{1}{2}C{{V}^{2}}=\frac{1}{2}Li_{0}^{2}\] \[L=\frac{(0.5\times {{10}^{-6}}){{(150)}^{2}}{{(250)}^{2}}}{{{(50)}^{2}}}\] \[L=281.25mH\]


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