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BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer
    The transfer ratio p of a transistor is 50. The input resistance of the transistor when used in the common emitter mode is\[1\,k\Omega \]. The peak value of the collector alternating current for an input peak voltage of 0.01 V is

    A) \[100\text{ }\mu A\]                     

    B)  \[\text{500 }\mu A\]

    C)  \[0.01\text{ }\mu A\]                   

    D)  \[0.25\text{ }\mu A\]

    Correct Answer: B

    Solution :

                      Input current\[=\frac{{{V}_{i}}}{{{R}_{i}}}=\frac{0.01}{{{10}^{3}}}={{10}^{-5}}A\]                 \[\therefore \]Output collector current                 \[{{I}_{C}}=\beta {{I}_{B}}=50\times {{10}^{-5}}A\]                 \[{{I}_{C}}=500\,\mu A\]


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