RAJASTHAN PMT Rajasthan - PMT Solved Paper-2007

  • question_answer
    A light emitting diode (I.ED) has a voltage drop of 2 V across it and passes a current of 10 n-I A. 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 \[\,\Omega \]          

    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 \[=10\,\,mA=10\times {{10}^{-3}}A\] and voltage in the circuit\[=6-2=4\,\,V\]. From Ohms law,                 \[V=IR\] \[\therefore \]  \[R=\frac{V}{I}=\frac{4}{10\times {{10}^{-3}}}=400\Omega \]


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