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JEE Main & Advanced Physics Semiconducting Devices Question Bank Junction Transistor

  • question_answer
    The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 KW. The peak value for an A.C input voltage of 0.01 V peak is                                   [CBSE PMT 1998]

    A)            100 mA                                    

    B)            0.01 mA

    C)            0.25 mA                                  

    D)            500 mA

    Correct Answer: D

    Solution :

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


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