JEE Main & Advanced Physics Communication System / संचार तंत्र Question Bank Mock Test - Electronic Devices Communication Systems

  • question_answer
    The transfer ratio of a transistor is 50. The input resistant of the transistor when used in the common-emittej configuration is 1 k\[\Omega \]. The peak value for an AC input voltage of 0.01 V peak is

    A) \[100\mu A\]     

    B) \[0.01mA\]

    C) \[0.25mA\]        

    D) \[500\mu A\]  

    Correct Answer: D

    Solution :

    [d] \[\beta =50,\]\[{{R}_{i}}=1000\Omega ,\]\[{{V}_{i}}=0.01V\] \[\beta =\frac{{{i}_{c}}}{{{i}_{b}}}\] and \[ib=\frac{{{V}_{i}}}{{{R}_{i}}}=\frac{0.01}{{{10}^{3}}}={{10}^{-5}}A\] Hence \[{{i}_{c}}=50\times {{10}^{-5}}A=500\mu A.\]


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