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Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    A coil of resistance \[10\Omega \] and an inductance \[5\,\,H\] is connected to a \[100\,\,volt\] battery. Then energy stored in the coil is

    A) \[125\,\,erg\]                   

    B) \[125\,\,J\]

    C) \[250\,\,erg\]                   

    D) \[250\,\,J\]

    Correct Answer: D

    Solution :

    Final current\[=i=\frac{E}{R}\]                                    ... (1) (here:\[E=100\,\,V,\]   \[R=10\Omega \]) from eq. (1)                 \[i=\frac{100}{10}=10\,\,A\] Energy stored in magnetic field is given by\[U=\frac{1}{2}L{{i}^{2}}\]                       \[(Here:L=5H)\] So,                          \[U=\frac{1}{2}\times 5\times {{(10)}^{2}}=250\,\,J\]


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