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VMMC VMMC Medical Solved Paper-2011

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

    A)  40\[k\Omega \]              

    B)                         4\[k\Omega \]                

    C)  200\[\Omega \]                              

    D)         400\[\Omega \]

    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 \[=10\,mA=10\times {{10}^{-3}}A\] \[\text{Voltage}=6-2=4V\] From Ohms law \[V=IR\] \[R=\frac{V}{I}=\frac{4}{10\times {{10}^{-3}}}=400\,\Omega \]


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