JEE Main & Advanced Physics Semiconducting Devices Question Bank Junction Transistor

  • question_answer
    For a transistor the parameter b = 99. The value of the parameter a is                                    [Pb CET 1998]

    A)            0.9 

    B)            0.99

    C)            1    

    D)            9

    Correct Answer: B

    Solution :

               \[\alpha =\frac{\beta }{1+\beta }=\frac{99}{1+99}=0.99\].


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