JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    A Light Emitting Diode (LED) has a voltage drop of 2 V across it and passes a current of10 mA. When it operates with a 6 V batery 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 abbreviation of 'Light Emitting Diode'. It is forward-biased p-n junction with emits spontaneous radiation. Current in the circuit \[=10\,mA=10\times {{10}^{-3}}A\] and voltage in the circuit = 6 - 2 = 4V. From Ohm's law, \[V=IR\] \[\therefore \]    \[R=\frac{V}{I}=\frac{4}{10\times {{10}^{-3}}}=400\,\,\Omega \]

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