BVP Medical BVP Medical Solved Paper-2014

  • question_answer
    A Light Emitting Diode (LED) has a voltage drop of 2V across it and passes a current of 10mA, when it operates with a 6V battery through a limiting resistor R, the value of     R is

    A)  \[\frac{\omega (M+2M)}{M}\]                

    B)  \[{{a}_{c}}={{k}^{2}}{{r}^{2}}{{t}^{2}},\]

    C)  \[2mk{{r}^{2}}t\]                            

    D)  \[mk{{r}^{2}}{{t}^{2}}\]                             

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


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