BCECE Medical BCECE Medical Solved Papers-2008

  • question_answer
    A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R, the value of R is

    A)  \[40\,\,k\Omega \]

    B)  \[4\,\,k\Omega \]

    C)  \[200\,\,k\Omega \]

    D)  \[400\,\,k\Omega \]

    Correct Answer: D

    Solution :

    The term LED is abbreviated as Light Emitting Diode. It is forward-biased p-n junction which emits spontaneous radiation. Current in the circuit \[=10\,mA=10\times {{10}^{-3}}A\] and voltage in the circuit \[=6-2=4\,V\] From Ohms law,                 \[V=IR\] \[\therefore \] \[R=\frac{V}{I}=\frac{4}{10\times {{10}^{-3}}}=400\,\,\Omega \]


You need to login to perform this action.
You will be redirected in 3 sec spinner