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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    In a coil current changes from \[4A\]to \[2A\] in\[0.05\text{ }sec,\] if induced e.m.f. is 8 volt then self-inductance of the coil is:

    A)  \[0.5H\]              

    B)                         \[0.35H\]           

    C)         \[0.2H\]              

    D)                         \[2mH\]

    Correct Answer: C

    Solution :

    For coefficient of self induction (L) \[e=L\frac{\Delta i}{\Delta t}\]                 \[\Rightarrow \]               \[8=L\times \frac{2}{0.05}\]                                 \[L=0.2H\]


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