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Punjab Medical Punjab - MET Solved Paper-2007

  • question_answer
    In the inductive circuit given in the figure, the currents rises after the switch is closed. At instant when the current is \[15mA\], then potential difference across the inductor will be

    A)  zero

    B)  240 V

    C)  180 V

    D)  60 V

    Correct Answer: C

    Solution :

    At any instant, the emf of the battery is equal to the sum of potential drop across the resistor and emf developed-in the induction coil. Therefore, we have                 \[E=IR+L\frac{dI}{dt}\] or            \[L\frac{dI}{dt}=E-IR\] Substituting\[E=240V,\,\,I=15mA\] \[=15\times {{10}^{-3}}A,\,\,R=4000\Omega \], we get                 \[L\frac{dI}{dt}=180V\] Hence, from Faradays law, we have the potential developed in the induction coil is                 \[E\,\,=L\frac{dI}{dt}=180V\]


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