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  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    For a transistor amplifier in common emitter configuration for load impedance of \[1\,k\Omega ({{h}_{fe}}=50\]and \[{{h}_{oe}}\,=25\,\,\,\mu A/V\]), the current gain is

    A) \[-5.2\]                

    B)        \[-15.7\]              

    C)        \[-24.8\]              

    D)        \[-48.78\]

    Correct Answer: D

    Solution :

    For a transistor amplifier in common emitter configuration, current gain \[{{A}_{i}}=-\frac{{{h}_{fe}}}{1+{{h}_{oe}}{{R}_{L}}}\] where,\[{{h}_{fe}},{{h}_{ce}}\]are hybrid parameters of a transistor. \[\therefore \]\[{{A}_{i}}=-\frac{50}{1+25\times {{10}^{-6}}\times 1\times {{10}^{3}}}=-48.78\]


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